67486 + Interview Questions in Secondary School in Physics Page 11 InterviewSolution

501.	The focal length ſ =(10cm). At what distance object should be placed to get an image twice size of object
Answer» ANSWER: Answer is GIVEN in the PIC. HOPE this HELPS you

501.

The focal length ſ =(10cm). At what distance object should be placed to get an image twice size of object

Answer»

ANSWER:

Answer is GIVEN in the PIC.

HOPE this HELPS you

Discussion

502.	Igskhsyelhdgkdisyjgsgskgsigkskgskgs
Answer» ANSWER: nnbbwnan ananananannanansnnwnw

502.

Igskhsyelhdgkdisyjgsgskgsigkskgskgs

Answer»

ANSWER:

nnbbwnan ananananannanansnnwnw

Discussion

503.	HEYLO!up for c..h.a.t ?wzymrrvxpp
Answer» ANSWER: he he he he he he he he he he he he

503.

HEYLO!up for c..h.a.t ?wzymrrvxpp

Answer»

ANSWER:

he he he he he he he he he he he he

Discussion

504.	Find the electric field intensity due to cylindrical wire.
Answer» ANSWER: Electric Field:CYLINDER of Charge Considering a Gaussian surface in the form of a cylinder at radius r > R, the electric field has the same MAGNITUDE at every POINT of the cylinder and is directed OUTWARD. The electric flux is then just the electric field times the area of the cylinder

504.

Find the electric field intensity due to cylindrical wire.

Answer»

ANSWER:

Electric Field:CYLINDER of Charge

Considering a Gaussian surface in the form of a cylinder at radius r > R, the electric field has the same MAGNITUDE at every POINT of the cylinder and is directed OUTWARD. The electric flux is then just the electric field times the area of the cylinder

Discussion

505.	At given t taken a body at rest which then moves with an acceleration, after 3 sec its momentum
Answer» Answer: We know force F=mass×acceleration-------(1) Given time = 2 seconds We can rewrite the EQUATION (1) as Force=mass× Time(t) Final(v)-Initial VELOCITY(u) --------(2) Also change in momentum dp=p final −p initial = mass×change in velocity Equation (2) becomes F= dt dp Fdt=dp Integrating the above equation we get: F∫ 0 2 dt=∫ p i p f dp 30×2 = P f −P i Therefore change in momentum = 60 kg m/s

505.

At given t taken a body at rest which then moves with an acceleration, after 3 sec its momentum

Answer»

Answer:

We know force F=mass×acceleration-------(1)

Given time = 2 seconds

We can rewrite the EQUATION (1) as

Force=mass×

Time(t)

Final(v)-Initial VELOCITY(u)

--------(2)

Also change in momentum dp=p

final

−p

initial

= mass×change in velocity

Equation (2) becomes F=

dt

dp

Fdt=dp

Integrating the above equation we get:

F∫

0

2

dt=∫

p

i

p

f

dp

30×2 = P

f

−P

i

Therefore change in momentum = 60 kg m/s

Discussion

506.	Replace the "C" in Covid-19 by the first letter of your name.Mine will be Rovid-19.
Answer» ANSWER: OHH haaa Explanation: very nice please MARK me as BRAINLIST

506.

Replace the "C" in Covid-19 by the first letter of your name.Mine will be Rovid-19.

Answer»

ANSWER:

OHH haaa

Explanation:

very nice

please MARK me as BRAINLIST

Discussion

507.	Can there be an electric circuit be complete without a key or switch?
Answer» If a switch is absent because it has been physically removed, leaving dangling WIRES (power wiring) or EMPTY solder pads (electronics) , then current will not FLOW- it's functionally equivalent to a switch always in its "off" state. Wires are cheaper than switches.

Discussion

508.	Answer the below question
Answer» EXPLANATION: xkgpxoydHIDIEZIFYZUSGDLHXDICHCOGCLGXOSGZFOFOFGFFFOOOFTPTTTzpgzOGXOFZ

508.

Answer the below question

Answer»

EXPLANATION:

xkgpxoydHIDIEZIFYZUSGDLHXDICHCOGCLGXOSGZFOFOFGFFFOOOFTPTTTzpgzOGXOFZ

Discussion

509.	Answer the below given question
Answer» OPEN the ATTACHMENT PLEASE MARK BRAINLIEST

509.

Answer the below given question

Answer»

OPEN the ATTACHMENT

PLEASE MARK BRAINLIEST

Discussion

510.	Explain angle of friction and it's relation with co-efficient of friction
Answer» okkkkkkkk I'll okkkkkkkkkkkk okkkkkkkkkkkk

Discussion

511.	Define conduction ? give one example where heat is transferred by the process of conduction .
Answer» CONDUCTION is the transfer of energy in the form of heat or electricity from one ATOM to another within an OBJECT by direct contact. ... Some examples of conduction of heat are accidentally touching a hot pot, or when a heating PAD is applied to you DIRECTLY and warms your muscles

Discussion

512.	एक अच्छा खिलाड़ी सफलता पाने के लिए क्या करता है ? *(a) ਸਖ਼ਤ ਮਿਹਨਤ ਕਰਦਾ ਹੈ/ कठिन परिश्रम करता है(b) ਨਵੀਂ ਤਕਨੀਕ ਸਿੱਖਣ ਦੀ ਕੋਸ਼ਿਸ਼ ਕਰਦਾ ਹੈ / वह नई तकनीक सीखने की केशिश करता है ।He tries to learn new techniques.(c)ਆਪਣੀ ਖੇਡ ਦਾ ਚੰਗਾ ਪ੍ਰਦਰਸ਼ਨ ਅਤੇ ਜਿੱਤਣ ਲਈ ਪੂਰਾ ਜ਼ੋਰ ਲਗਾਉਂਦਾ ਹੈ/वह खेल का बढ़िया प्रदर्शन करता है और जीतने की पूरी कोशिश करता है /He does his best to win the game.(d)ਉਪਰੋਕਤ ਸਾਰੇ/ ऊपर के सभी/ All of the above
Answer» ANSWER: OPTION (D) is RIGHT All of the above

512.

एक अच्छा खिलाड़ी सफलता पाने के लिए क्या करता है ? *(a) ਸਖ਼ਤ ਮਿਹਨਤ ਕਰਦਾ ਹੈ/ कठिन परिश्रम करता है(b) ਨਵੀਂ ਤਕਨੀਕ ਸਿੱਖਣ ਦੀ ਕੋਸ਼ਿਸ਼ ਕਰਦਾ ਹੈ / वह नई तकनीक सीखने की केशिश करता है ।He tries to learn new techniques.(c)ਆਪਣੀ ਖੇਡ ਦਾ ਚੰਗਾ ਪ੍ਰਦਰਸ਼ਨ ਅਤੇ ਜਿੱਤਣ ਲਈ ਪੂਰਾ ਜ਼ੋਰ ਲਗਾਉਂਦਾ ਹੈ/वह खेल का बढ़िया प्रदर्शन करता है और जीतने की पूरी कोशिश करता है /He does his best to win the game.(d)ਉਪਰੋਕਤ ਸਾਰੇ/ ऊपर के सभी/ All of the above

Answer»

ANSWER:

OPTION (D) is RIGHT All of the above

Discussion

513.	Find total resistance for parallel combination
Answer» Answer: The sum of the currents through each path is EQUAL to the total current that FLOWS from the source. You can FIND total RESISTANCE in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +... If one of the parallel PATHS is broken, current will continue to flow in all the other paths.

513.

Find total resistance for parallel combination

Answer»

Answer:

The sum of the currents through each path is EQUAL to the total current that FLOWS from the source. You can FIND total RESISTANCE in a Parallel circuit with the following formula: 1/Rt = 1/R1 + 1/R2 + 1/R3 +... If one of the parallel PATHS is broken, current will continue to flow in all the other paths.

Discussion

514.	Redraw the circuit of Questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter ?
Answer» Answer: An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, Ohm’s LAW can be used to obtain the readings of ammeter and voltmeter. ACCORDING to Ohm’s law, V = IR, Where, Potential difference, V = 6 V Current flowing through the circuit/resistors = I Resistance of the circuit, R = 5 + 8 + 12 = 25ohm I = V/R = 6/25 = 0.24 A Potential difference across 12ohm resistor = V1 Current flowing through the 12ohm resistor, I = 0.24 A Therefore, using Ohm’s law, we obtain V1 = IR = 0.24 x 12 = 2.88 V Therefore, the reading of the ammeter will be 0.24 A. The reading of the voltmeter will be 2.88 V

514.

Redraw the circuit of Questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter ?

Answer»

Answer:

An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel,

Ohm’s LAW can be used to obtain the readings of ammeter and voltmeter. ACCORDING to Ohm’s law,

V = IR,

Where,

Potential difference, V = 6 V

Current flowing through the circuit/resistors = I

Resistance of the circuit, R = 5 + 8 + 12 = 25ohm

I = V/R = 6/25 = 0.24 A

Potential difference across 12ohm resistor = V1

Current flowing through the 12ohm resistor, I = 0.24 A

Therefore, using Ohm’s law, we obtain

V1 = IR = 0.24 x 12 = 2.88 V

Therefore, the reading of the ammeter will be 0.24 A.

The reading of the voltmeter will be 2.88 V

Discussion

515.	Find the angles respectively of a ∆PQR. For each of the following cases, state whether EF \|\| QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cmKoi mujhse friendship karega :)
Answer» Explanation: E and F are two points on side PQ and PR in △PQR. (i) PE=3.9 cm, EQ=3 cm and PF=3.6 cm, FR=2.4 cm Using BASIC proportionality theorem, ∴ EQ PE = 3 3.9 = 30 39 = 10 13 =1.3 FR PF = 2.4 3.6 = 24 36 = 2 3 =1.5 EQ PE  = FR PF So, EF is not parallel to QR. (II) PE=4 cm, QE=4.5 cm, PF=8 cm, RF=9 cm Using Basic proportionality theorem, ∴ QE PE = 4.5 4 = 45 40 = 9 8 RF PF = 9 8 QE PE = RF PF So, EF is parallel to QR. (iii) PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm, PF=0.36 cm Using Basic proportionality theorem, EQ=PQ−PE=1.28−0.18=1.10 cm FR=PR−PF=2.56−0.36=2.20 cm EQ PE = 1.10 0.18 = 110 18 = 55 9 ... (i) FR PE = 2.20 0.36 = 220 36 = 55 9 ... (ii) ∴ EQ PE = FR. PF So, EF is parallel to QR.

515.

Find the angles respectively of a ∆PQR. For each of the following cases, state whether EF || QR: (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cmKoi mujhse friendship karega :)

Answer»

Explanation:

E and F are two points on side PQ and PR in △PQR.

(i) PE=3.9 cm, EQ=3 cm and PF=3.6 cm, FR=2.4 cm

Using BASIC proportionality theorem,

∴

EQ

PE

=

3

3.9

=

30

39

=

10

13

=1.3

FR

PF

=

2.4

3.6

=

24

36

=

2

3

=1.5

EQ

PE



=

FR

PF

So, EF is not parallel to QR.

(II) PE=4 cm, QE=4.5 cm, PF=8 cm, RF=9 cm

Using Basic proportionality theorem,

∴

QE

PE

=

4.5

4

=

45

40

=

9

8

RF

PF

=

9

8

QE

PE

=

RF

PF

So, EF is parallel to QR.

(iii) PQ=1.28 cm, PR=2.56 cm, PE=0.18 cm, PF=0.36 cm

Using Basic proportionality theorem,

EQ=PQ−PE=1.28−0.18=1.10 cm

FR=PR−PF=2.56−0.36=2.20 cm

EQ

PE

=

1.10

0.18

=

110

18

=

55

9

... (i)

FR

PE

=

2.20

0.36

=

220

36

=

55

9

... (ii)

∴

EQ

PE

=

FR.

PF

So, EF is parallel to QR.

Discussion

516.	an athlete completes one round of a circular track of diametre 50 m in 50 s. What will be the distance covered and displacement in 4 min?
Answer» Answer: GIVEN : Diameter of the circular track D=2R=200 m Time t=2 minutes 20 s =140 s As the athlete COMPLETES one round in 40 seconds, thus he will be at point B after 140 seconds by covering three and HALF rounds. ∴ Displacement of the athlete s′=2R=200 m Distance covered d=3.5×(2πR) d=3.5×722×200=2200 m

516.

an athlete completes one round of a circular track of diametre 50 m in 50 s. What will be the distance covered and displacement in 4 min?

Answer»

Answer:

GIVEN : Diameter of the circular track D=2R=200 m

Time t=2 minutes 20 s =140 s

As the athlete COMPLETES one round in 40 seconds, thus he will be at point B after 140 seconds by covering three and HALF rounds.

∴ Displacement of the athlete s′=2R=200 m

Distance covered d=3.5×(2πR)

d=3.5×722×200=2200 m

Discussion

517.	As water is made of atoms having protons (charge +e) and electrons (charge 2:e), does the water flowing through pipe carry an electric current? Explain.
Answer» Answer: The net charge on proton is +e. This can be OBTAINED by p=(32+32−31)e=+e Hence, proton has two 'up' quark and a 'down quark'. The net charge on neutron is 0. This can be obtained by N=(32−31−31)=0 Hence, neutron has ONE 'up' quark and two 'down' quarks.

517.

As water is made of atoms having protons (charge +e) and electrons (charge 2:e), does the water flowing through pipe carry an electric current? Explain.

Answer»

Answer:

The net charge on proton is +e. This can be OBTAINED by p=(32+32−31)e=+e

Hence, proton has two 'up' quark and a 'down quark'.

The net charge on neutron is 0. This can be obtained by N=(32−31−31)=0

Hence, neutron has ONE 'up' quark and two 'down' quarks.

Discussion

518.	Four-point charges q1 = 2 micro C, q2 = -5 micro C, q3 = 2 micro C, and q4 = -5 micro Care located add the corners of a square ABCD of side 10 cm. what is the force on a charge of1 micro C placed at the centre of square?
Answer» $\underline{\blacksquare\:\:\:\footnotesize{\purple{\textsf{ \textbf{{SOLUTION:-}}}}}}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ • Side of square (a) = 10 cm = 0.1 m • Five point charges : q₁ = 2μC = 2 × 10⁻⁶ C q₂ = -5μC = 5 × 10⁻⁶ C q₃ = 2μC = 2 × 10⁻⁶ C q₄ = -5μC = 5 × 10⁻⁶ C q₅ = 1 μC = 1 × 10⁻⁶ C (Point charge present at the CENTRE) • Half of the diagonal of the square = a/√2 = 0.1/√2 m $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ See the attached FIGURE! • First of all we will find the ELECTRIC field along diagonal AC DUE to point charge q₁ = 2μC and q₃ = 2μC. • Due to point Charge q₁ = 2μC, the electric field will be away from the centre. • Due to point Charge q₃ = 2μC, the electric field will be away from the centre. $\footnotesize{\longrightarrow \rm E_{(AC)}=E_{1} - E_{3} } \\$ $\footnotesize{\longrightarrow \rm E_{(AC)}=\dfrac{Kq_{1} }{r^2} - \dfrac{Kq_{3} }{r^2} }$ $\footnotesize{\longrightarrow \rm E_{(AC)}=\dfrac{9 \times {10}^{9} \times 2 \times {10}^{ - 6} }{ \bigg( \frac{0.1}{ \sqrt{2} } \bigg)^{2} } - \dfrac{9 \times {10}^{9} \times 2\times {10}^{ - 6} }{ \bigg(\frac{0.1}{ \sqrt{2} } \bigg)^{2}} }$ • The magnitude of the charges are equal therefore they will cancel each other and electric field along diagonal AC will be 0. $\footnotesize{\longrightarrow \underline{ \boxed{ \red{\bf E_{(AC)} = 0 \: N/C}}}}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ •Now, let's find the electric field along diagonal BD. • Due to point charge q₂ = -5μC ,the electric field will be directed towards the centre. • Due to point charge q₄ = -5μC, the electric field will be directed towards centre. • From the observation we find that the magnitude of the charges are equal therefore they will cancel each other. • Hence, electric field along diagonal BD will be 0. $\footnotesize{\longrightarrow \underline{ \boxed{ \orange{\bf E_{(BD)} = 0 \: N/C}}}}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$ At the centre there are two electric field which are perpendicular to each other, so net electric field can be calculated by using superposition principle: $\footnotesize{\longrightarrow \rm E_{(Net)} = \sqrt{E_{(AC)} ^{2} + E_{(BD)}^{2} + 2 E_{(AC)} ^{2} E_{(BD)}^{2} \cos( {90}^{ \circ} ) }} \\$ $\footnotesize{\longrightarrow \rm E_{(Net)} = \sqrt{E_{(AC)} ^{2} + E_{(BD)}^{2} }}$ $\footnotesize{\longrightarrow \rm E_{(Net)} =E \sqrt{2 }}$ $\footnotesize{\longrightarrow \underline{ \boxed{ \gray{\bf E_{(Net)} =0 \: N/C}}}}$ $\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

518.

Four-point charges q1 = 2 micro C, q2 = -5 micro C, q3 = 2 micro C, and q4 = -5 micro Care located add the corners of a square ABCD of side 10 cm. what is the force on a charge of1 micro C placed at the centre of square?

Answer»

$\underline{\blacksquare\:\:\:\footnotesize{\purple{\textsf{ \textbf{{SOLUTION:-}}}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

• Side of square (a) = 10 cm = 0.1 m

• Five point charges :

q₁ = 2μC = 2 × 10⁻⁶ C
q₂ = -5μC = 5 × 10⁻⁶ C
q₃ = 2μC = 2 × 10⁻⁶ C
q₄ = -5μC = 5 × 10⁻⁶ C
q₅ = 1 μC = 1 × 10⁻⁶ C (Point charge present at the CENTRE)

• Half of the diagonal of the square = a/√2 = 0.1/√2 m

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

See the attached FIGURE!

• First of all we will find the ELECTRIC field along diagonal AC DUE to point charge q₁ = 2μC and q₃ = 2μC.

• Due to point Charge q₁ = 2μC, the electric field will be away from the centre.

• Due to point Charge q₃ = 2μC, the electric field will be away from the centre.

$\footnotesize{\longrightarrow \rm E_{(AC)}=E_{1} - E_{3} } \\$

$\footnotesize{\longrightarrow \rm E_{(AC)}=\dfrac{Kq_{1} }{r^2} - \dfrac{Kq_{3} }{r^2} }$

$\footnotesize{\longrightarrow \rm E_{(AC)}=\dfrac{9 \times {10}^{9} \times 2 \times {10}^{ - 6} }{ \bigg( \frac{0.1}{ \sqrt{2} } \bigg)^{2} } - \dfrac{9 \times {10}^{9} \times 2\times {10}^{ - 6} }{ \bigg(\frac{0.1}{ \sqrt{2} } \bigg)^{2}} }$

• The magnitude of the charges are equal therefore they will cancel each other and electric field along diagonal AC will be 0.

$\footnotesize{\longrightarrow \underline{ \boxed{ \red{\bf E_{(AC)} = 0 \: N/C}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

•Now, let's find the electric field along diagonal BD.

• Due to point charge q₂ = -5μC ,the electric field will be directed towards the centre.

• Due to point charge q₄ = -5μC, the electric field will be directed towards centre.

• From the observation we find that the magnitude of the charges are equal therefore they will cancel each other.

• Hence, electric field along diagonal BD will be 0.

$\footnotesize{\longrightarrow \underline{ \boxed{ \orange{\bf E_{(BD)} = 0 \: N/C}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

At the centre there are two electric field which are perpendicular to each other, so net electric field can be calculated by using superposition principle:

$\footnotesize{\longrightarrow \rm E_{(Net)} = \sqrt{E_{(AC)} ^{2} + E_{(BD)}^{2} + 2 E_{(AC)} ^{2} E_{(BD)}^{2} \cos( {90}^{ \circ} ) }} \\$

$\footnotesize{\longrightarrow \rm E_{(Net)} = \sqrt{E_{(AC)} ^{2} + E_{(BD)}^{2} }}$

$\footnotesize{\longrightarrow \rm E_{(Net)} =E \sqrt{2 }}$

$\footnotesize{\longrightarrow \underline{ \boxed{ \gray{\bf E_{(Net)} =0 \: N/C}}}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{0.5mm}\put(1,1){\line(1,0){6.8}}\end{picture}$

Discussion

519.	A ball of mass 150 g is acted upon by a force of 250 dynes for 10 seconds. If the ball is initially at rest, then calculate the final velocity.
Answer» Answer:Given that, Force F=100dyn Mass m=5g Time t=10s Initial VELOCITY u=0 Now, the acceleration is F=ma a= m F a= 5×10 −3 100×10 −5 a=0.20m/s 2 Now, from EQUATION of motion v=u+at v=0+0.20×10 v=2m/s v=200cm/s Hence, the velocity produced is 200 cm/s Explanation:

519.

A ball of mass 150 g is acted upon by a force of 250 dynes for 10 seconds. If the ball is initially at rest, then calculate the final velocity.

Answer»

Answer:Given that,

Force F=100dyn

Mass m=5g

Time t=10s

Initial VELOCITY u=0

Now, the acceleration is

F=ma

a=

m

F

a=

5×10

−3

100×10

−5

a=0.20m/s

2

Now, from EQUATION of motion

v=u+at

v=0+0.20×10

v=2m/s

v=200cm/s

Hence, the velocity produced is 200 cm/s

Explanation:

Discussion

520.	Aaj kal koi baat bhi nahi karte, thanks bhi nahi dete, F0LL0W bhi nahi karte.Aisa kya kiya maine ?
Answer» Answer: OK Explanation: ok bro L will give you so MUCH thanks SEE notifications PLEASE mark me as brainlist

520.

Aaj kal koi baat bhi nahi karte, thanks bhi nahi dete, F0LL0W bhi nahi karte.Aisa kya kiya maine ?

Answer»

Answer:

OK

Explanation:

ok bro L will give you so MUCH thanks SEE notifications PLEASE mark me as brainlist

Discussion

521.	In the given circuit, B1, B2 and B3 are three bulbs connected with a battery of6 V.(a) What will happen to the glow of the other two bulbs if the bulb B2, gets fused?(b) If the wattage of each bulb is 3 W, what reading will ammeter A show when all the three bulbs glow simultaneously.(c) Find the total resistance of the circuit.
Answer» ANSWER: PLEASE MARK as BRAINLIST

521.

In the given circuit, B1, B2 and B3 are three bulbs connected with a battery of6 V.(a) What will happen to the glow of the other two bulbs if the bulb B2, gets fused?(b) If the wattage of each bulb is 3 W, what reading will ammeter A show when all the three bulbs glow simultaneously.(c) Find the total resistance of the circuit.

Answer»

ANSWER:

PLEASE MARK as BRAINLIST

Discussion

522.	Define Displacement
Answer» Explanation: Displacement:–The SHORTEST distance between the initial POSITION and the final position of a moving object in the given interval of TIME from initial to the final position of the object is KNOWN as the displacement of the object. Hope this helps you ❤️ pls mark brainlist and follw me

522.

Define Displacement

Answer»

Explanation:

Displacement:–The SHORTEST distance between the initial POSITION and the final position of a moving object in the given interval of TIME from initial to the final position of the object is KNOWN as the displacement of the object.

Hope this helps you ❤️

pls mark brainlist and follw me

Discussion

523.	A Trucks movie with velocity 36 km per hour and gains its speed to 72 km per hour in 5 seconds. Find the acceleration of the truck.a) 20m/s²b) 2m/s²c) 20m/s²d) 0.20m/s²
Answer» Answer: Provided that: Initial velocity = 36 kmph Final velocity = 72 kmph Time = 5 seconds To calculate: The acceleration Solution: The acceleration = 2 m/s² Using concepts: • We can use either acceleration formula or first EQUATION of motion to calculate the acceleration. • Formula to CONVERT kmph-mps. Using FORMULAS: • Acceleration formula: ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$ • First equation of motion: ${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$ • Formula to convert kmph-mps: ${\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}$ Where, a DENOTES acceleration, u denotes initial velocity, v denotes final velocity, t denotes time taken. Required solution: ~ Firstly let us convert kmph-mps! Converting 72 kmph-mps. $:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{72} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$ Converting 36 kmph-mps. $:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$ Therefore, Initial velocity = 10 mps Final velocity = 20 mps ~ Now let's FIND out the acceleration! By using acceleration formula. $:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20 - 10}{5} \\ \\ :\implies \sf a \: = \dfrac{10}{5} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$ By using first equation of motion. $:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 20 = 10 + a(5) \\ \\ :\implies \sf 20 - 10 = 5a \\ \\ :\implies \sf 10 = 5a \\ \\ :\implies \sf a \: = \dfrac{10}{5} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

523.

A Trucks movie with velocity 36 km per hour and gains its speed to 72 km per hour in 5 seconds. Find the acceleration of the truck.a) 20m/s²b) 2m/s²c) 20m/s²d) 0.20m/s²

Answer»

Answer:

Provided that:

Initial velocity = 36 kmph
Final velocity = 72 kmph
Time = 5 seconds

To calculate:

The acceleration

Solution:

The acceleration = 2 m/s²

Using concepts:

• We can use either acceleration formula or first EQUATION of motion to calculate the acceleration.

• Formula to CONVERT kmph-mps.

Using FORMULAS:

• Acceleration formula:

${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

• First equation of motion:

${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$

• Formula to convert kmph-mps:

${\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}$

Where, a DENOTES acceleration, u denotes initial velocity, v denotes final velocity, t denotes time taken.

Required solution:

~ Firstly let us convert kmph-mps!

Converting 72 kmph-mps.

$:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{72} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Converting 36 kmph-mps.

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \: (Cancelling) \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

Initial velocity = 10 mps
Final velocity = 20 mps

~ Now let's FIND out the acceleration!

By using acceleration formula.

$:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{20 - 10}{5} \\ \\ :\implies \sf a \: = \dfrac{10}{5} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

By using first equation of motion.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 20 = 10 + a(5) \\ \\ :\implies \sf 20 - 10 = 5a \\ \\ :\implies \sf 10 = 5a \\ \\ :\implies \sf a \: = \dfrac{10}{5} \\ \\ :\implies \sf a \: = 2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = 2 \: ms^{-2}$

Discussion

524.	The capacity of isolated conducting charge sphere of radius r is propertional to
Answer» ANSWER: C CHORUS okkkkkkkkkkkkkkkkkkk

524.

The capacity of isolated conducting charge sphere of radius r is propertional to

Answer»

ANSWER:

C CHORUS okkkkkkkkkkkkkkkkkkk

Discussion

525.	the resultant resistance of a circuit where 10ohm, 20ohm, 30ohm, and 40ohm, are connected in parallel
Answer» ANSWER: RP= 5.45 OHM Explanation: 1/Rp = 1/R1 + 1/R2 + 1/R3 1/Rp = 1/10 + 1/20 + 1/30 1/Rp = 0.1833 Rp= 1/ 0.18333 Rp = 5.45 Ohm

525.

the resultant resistance of a circuit where 10ohm, 20ohm, 30ohm, and 40ohm, are connected in parallel

Answer»

ANSWER:

RP= 5.45 OHM

Explanation:

1/Rp = 1/R1 + 1/R2 + 1/R3

1/Rp = 1/10 + 1/20 + 1/30

1/Rp = 0.1833

Rp= 1/ 0.18333

Rp = 5.45 Ohm

Discussion

526.	Explain giving two examples to show that inertia of a body changes due to change in its mass.
Answer» Answer: mark me as brainliest Explanation: OBJECTS want to STAY in rest or MOTION unless an outside force causes a change. For example, if you roll a BALL, it will continue rolling unless friction or something else stops it by force. You can also think about the WAY that your body keeps moving forward when you hit the brake on your bike.

526.

Explain giving two examples to show that inertia of a body changes due to change in its mass.

Answer»

Answer:

mark me as brainliest

Explanation:

OBJECTS want to STAY in rest or MOTION unless an outside force causes a change. For example, if you roll a BALL, it will continue rolling unless friction or something else stops it by force. You can also think about the WAY that your body keeps moving forward when you hit the brake on your bike.

Discussion

527.	16 একটি জলের ট্যাংকের দৈর্ঘ্য, প্রস্থ ও উচ্চতা যথাক্রমে 3m, 2m ও 1m। ট্যাংকটি অর্ধেক জলপূর্ণ। ট্যাংকের জলের আয়তন হল- A 6000 L ৪) 30000 L C 3000 L D 6000
Answer» Answer: HELLO FRIEND GOOD morning have a great day

527.

16 একটি জলের ট্যাংকের দৈর্ঘ্য, প্রস্থ ও উচ্চতা যথাক্রমে 3m, 2m ও 1m। ট্যাংকটি অর্ধেক জলপূর্ণ। ট্যাংকের জলের আয়তন হল- A 6000 L ৪) 30000 L C 3000 L D 6000

Answer»

Answer:

HELLO FRIEND

GOOD morning

have a great day

Discussion

528.	Reactance depends on
Answer» EXPLANATION: As the frequency APPROACHES infinity the inductors reactance would also increase to infinity ACTING like an open CIRCUIT. However, as the frequency approaches zero or DC, the inductors reactance would decrease to zero, acting like a SHORT circuit. This means then that inductive reactance is “proportional” to frequency.

528.

Reactance depends on

Answer»

EXPLANATION:

As the frequency APPROACHES infinity the inductors reactance would also increase to infinity ACTING like an open CIRCUIT. However, as the frequency approaches zero or DC, the inductors reactance would decrease to zero, acting like a SHORT circuit. This means then that inductive reactance is “proportional” to frequency.

Discussion

529.	The refractive index of transparent medium is equal to __________
Answer» Answer: 1 ONE Transparent medium become invisible in vacuum if its refractive index becomes EQUAL to 1, since refractive index of vacuum is also unity because of which refraction of light will not TAKE PLACE in the transparent medium. Explanation: PLEASE mark me as brainlist

529.

The refractive index of transparent medium is equal to __________

Answer»

Answer:

1 ONE

Transparent medium become invisible in vacuum if its refractive index becomes EQUAL to 1, since refractive index of vacuum is also unity because of which refraction of light will not TAKE PLACE in the transparent medium.

Explanation:

PLEASE mark me as brainlist

Discussion

530.	A student performs an experiment on measuring friction forces in different trials. For trial A, she pulls a wooden block across a horizontal surface with a constant speeD) For trial B, she inclines the same surface at angle ? with respect to the horizontal. Which of the following is true about the friction force between the block and the surface?
Answer» ANSWER: For trial hope it's HELP you!!!

530.

A student performs an experiment on measuring friction forces in different trials. For trial A, she pulls a wooden block across a horizontal surface with a constant speeD) For trial B, she inclines the same surface at angle ? with respect to the horizontal. Which of the following is true about the friction force between the block and the surface?

Answer»

ANSWER:

For trial

hope it's HELP you!!!

Discussion

531.	If there is an attractive force between all objects, why do we not feel ourselves gravitating towardmassive buildings in our vicinity?
Answer» Explanation: Gravitational force is a weak force unless large masses are involved. The masses of HUMANS are quite small and thus the magnitude of gravitational force is also very small and negligible as compared to that SHOWN by Earth. Thus, gravitational force exists between TWO people sitting CLOSE by but it is negligible and cannot be felt.

531.

If there is an attractive force between all objects, why do we not feel ourselves gravitating towardmassive buildings in our vicinity?

Answer»

Explanation:

Gravitational force is a weak force unless large masses are involved. The masses of HUMANS are quite small and thus the magnitude of gravitational force is also very small and negligible as compared to that SHOWN by Earth. Thus, gravitational force exists between TWO people sitting CLOSE by but it is negligible and cannot be felt.

Discussion

532.	The velocity of the car is 15 m/sec and the acceleration is zero.how can you express such situation ?
Answer» ANSWER: EXPLANATION: EASY

532.

The velocity of the car is 15 m/sec and the acceleration is zero.how can you express such situation ?

Answer»

ANSWER:

EXPLANATION:

EASY

Discussion

533.	Find in supeepotion theroem2.mesh and nodal analysis.
Answer» ANSWER: what is the answer TELL US

533.

Find in supeepotion theroem2.mesh and nodal analysis.

Answer»

ANSWER:

what is the answer TELL US

Discussion

534.	Accuracy of the value 26.4 mm isa) 0,1mm b) 0,01cm c) Both (a) and (b) d) None of these
Answer» ANSWER: d) none of the above EXPLANATION: as the above MENTIONED value is not equal to 26.4mm and if we CONVERT it into cm then 2.6 cm is your answer so d is CORRECT

534.

Accuracy of the value 26.4 mm isa) 0,1mm b) 0,01cm c) Both (a) and (b) d) None of these

Answer»

ANSWER:

d) none of the above

EXPLANATION:

as the above MENTIONED value is not equal to 26.4mm and if we CONVERT it into cm then 2.6 cm is your answer so d is CORRECT

Discussion

535.	Solve this question and plz explain this question
Answer» ANSWER: don't KNOW sooooorrrrrrrrrrryyyyyyyyyy

535.

Solve this question and plz explain this question

Answer»

ANSWER:

don't KNOW sooooorrrrrrrrrrryyyyyyyyyy

Discussion

536.	1. An object is placed 20 cm from (a) a converging lens, & (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.2. A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image. Solve this questions. Whoever will answer my questions ?Thank "U"In Return, Your answers will be gonna rain of thanks ✨Have a nice day,Stay safe happy & healthy to everyone.
Answer» PLEASE see the attachment. Hope it helps u my DEAR di. God bless you and your family and FRIENDS. $genius \: girl$

536.

1. An object is placed 20 cm from (a) a converging lens, & (b) a diverging lens, of focal length 15 cm. Calculate the image position and magnification in each case.2. A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image. Solve this questions. Whoever will answer my questions ?Thank "U"In Return, Your answers will be gonna rain of thanks ✨Have a nice day,Stay safe happy & healthy to everyone.

Answer»

PLEASE see the attachment.

Hope it helps u my DEAR di.

God bless you and your family and FRIENDS.

$genius \: girl$

Discussion

537.	What are protons ?.... ...Mai to sanskari huclass bunk ni marta...
Answer» ANSWER: Protons are the POSITIVELY CHARGED particles which are present in the nucleus of a HYDROGEN atom. .. Explanation:

537.

What are protons ?.... ...Mai to sanskari huclass bunk ni marta...

Answer»

ANSWER:

Protons are the POSITIVELY CHARGED particles which are present in the nucleus of a HYDROGEN atom. ..

Explanation:

Discussion

538.	Accuracy of the value 26.4 mm is . a) 0,1mm b) 0,01cm c) Both (a) and (b) d) None of these
Answer» Answer: d NONE of the above EXPLANATION: as when we CONVERT 26.4mm into CM then the answer will be 2.6cm which is not GIVEN so d is correct

538.

Accuracy of the value 26.4 mm is . a) 0,1mm b) 0,01cm c) Both (a) and (b) d) None of these

Answer»

Answer:

d NONE of the above

EXPLANATION:

as when we CONVERT 26.4mm into CM then the answer will be 2.6cm which is not GIVEN so d is correct

Discussion

539.	'किनशिप ऑर्गेनाइजेशन इन इण्डिया' नामक पुस्तक के लेखक कौन हैं? ऑब्जेक्ट
Answer» This is not PHYSICS this is HINDI

Discussion

540.	What id the full form of dav
Answer» ANSWER: Dayanand ANGLO VEDIC (DAV) Explanation: MARK me as BRAINLIEST

540.

What id the full form of dav

Answer»

ANSWER:

Dayanand ANGLO VEDIC (DAV)

Explanation:

MARK me as BRAINLIEST

Discussion

541.	Friction opposes the____ between the surface in contact with eachother.
Answer» ANSWER: MOVEMENT... EXPLANATION: HOPE it HELPS

541.

Friction opposes the____ between the surface in contact with eachother.

Answer»

ANSWER:

MOVEMENT...

EXPLANATION:

HOPE it HELPS

Discussion

542.	A ball is thrown vertically upward with a velocity of 98 metre per second how high does the ball rise
Answer» Answer:xf=xi + vi(t) + (at^2)/2 Make the vertical axis the x-axis. COULD have USED y, but I don't want to retype the formula. xf= 0 + 98(t) -4.9t^2 vf= vi + at ….the ball is projected up. Once the ball leaves the cannon, only gravity and air resistance act UPON it. We’re neglecting air resistance, so we’re considering only gravity. As the ball rises it continues to decelerate until it reaches its maximum height, at which point it momentarily stops, and its velocity is zero. So, vf = 0. We couldn't make this assumption if the ball was dropped, since we don't know its velocity when it hits the ground. 0 = 98 - 9.81(t) t= 10 s xf = 980 -490 = 490m This is the maximum height. Explanation:

542.

A ball is thrown vertically upward with a velocity of 98 metre per second how high does the ball rise

Answer»

Answer:xf=xi + vi(t) + (at^2)/2

Make the vertical axis the x-axis. COULD have USED y, but I don't want to retype the formula.

xf= 0 + 98(t) -4.9t^2

vf= vi + at ….the ball is projected up. Once the ball leaves the cannon, only gravity and air resistance act UPON it. We’re neglecting air resistance, so we’re considering only gravity. As the ball rises it continues to decelerate until it reaches its maximum height, at which point it momentarily stops, and its velocity is zero. So, vf = 0. We couldn't make this assumption if the ball was dropped, since we don't know its velocity when it hits the ground.

0 = 98 - 9.81(t)

t= 10 s

xf = 980 -490 = 490m This is the maximum height.

Explanation:

Discussion

543.	Solve this questionthis question
Answer» ANSWER: PIC not CLEAR make SURE that UR pic is clear

543.

Solve this questionthis question

Answer»

ANSWER:

PIC not CLEAR

make SURE that UR pic is clear

Discussion

544.	A boy travels a 3 km road stops and comes back at center? What will be the distance and displacement?
Answer» ANSWER: ● Distance of the boy = 6 km. ● Displacement of the boy = 0 km. Explanation: A boy travels a distance of 3km then stops and COMES back to the point from where he STARTS travelling. ☆ Total Distance covered by the boy. = 2 × 3 km. = 6 km. ☆ Displacement of the boy = 0 km. = ( Because INITIAL position and the FINAL position of the boy is same ) .

544.

A boy travels a 3 km road stops and comes back at center? What will be the distance and displacement?

Answer»

ANSWER:

● Distance of the boy = 6 km.

● Displacement of the boy = 0 km.

Explanation:

A boy travels a distance of 3km then stops and COMES back to the point from where he STARTS travelling.

☆ Total Distance covered by the boy.

= 2 × 3 km.

= 6 km.

☆ Displacement of the boy = 0 km.

= ( Because INITIAL position and the FINAL position of the boy is same ) .

Discussion

545.	A body thrown up vertically with a velocity of 19.6 ms-1. Find the maximum height reached and time of ascent.
Answer» ANSWER: A BODY PROJECTED VERTICALLY upwards with a velocity of 19.6ms−1 reaches a HEIGHT of 19.6m on earth. If it is projected vertically up with the same velocity on moon, then the maximum height reached by it is (gearth=g ms−2, gmoon=6g ms−2)

545.

A body thrown up vertically with a velocity of 19.6 ms-1. Find the maximum height reached and time of ascent.

Answer»

ANSWER:

A BODY PROJECTED VERTICALLY upwards with a velocity of 19.6ms−1 reaches a HEIGHT of 19.6m on earth. If it is projected vertically up with the same velocity on moon, then the maximum height reached by it is (gearth=g ms−2, gmoon=6g ms−2)

Discussion

546.	3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s^-1. Calculate the initial recoil velocity of the rifle.
Answer» Answer: 63kg hope it's HELP you!!!

546.

3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s^-1. Calculate the initial recoil velocity of the rifle.

Answer»

Answer:

63kg

hope it's HELP you!!!

Discussion

547.	Find the maximum velocity for slipping a car moving on a circular track of radius 100m. The coefficient of friction between the road and tire is 0.2:A. 0.14m/sB. 140 m/s C.1.4km/s D. 14m/s
Answer» ANSWER: NOPE HOPE DOPE BRO i don't

547.

Find the maximum velocity for slipping a car moving on a circular track of radius 100m. The coefficient of friction between the road and tire is 0.2:A. 0.14m/sB. 140 m/s C.1.4km/s D. 14m/s

Answer»

ANSWER:

NOPE HOPE DOPE BRO i don't

Discussion

548.	What is the net force
Answer» ANSWER: Net force is the vector sum of forces acting on a PARTICLE or body. The net force is a single force that replaces the EFFECT of the ORIGINAL forces on the particle's motion. It gives the particle the same acceleration as all those actual forces together as DESCRIBED by Newton's second law of motion

548.

What is the net force

Answer»

ANSWER:

Net force is the vector sum of forces acting on a PARTICLE or body. The net force is a single force that replaces the EFFECT of the ORIGINAL forces on the particle's motion. It gives the particle the same acceleration as all those actual forces together as DESCRIBED by Newton's second law of motion

Discussion

549.	Is there someone who knows physics and can help me with some questions ?
Answer» ANSWER: okaayyyyyy Explanation: okayyyyyyyyyyyyyyyyyyyy

549.

Is there someone who knows physics and can help me with some questions ?

Answer»

ANSWER:

okaayyyyyy

Explanation:

okayyyyyyyyyyyyyyyyyyyy

Discussion

550.	Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following : (i) Which among iron and mercury is a better conductor ? (ii) Which material is the best conductor ?
Answer» EXPLANATION: (1) IRON (2) silver is best conductor

550.

Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following : (i) Which among iron and mercury is a better conductor ? (ii) Which material is the best conductor ?

Answer»

EXPLANATION:

(1) IRON

(2) silver is best conductor

Discussion

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The capacity of isolated conducting charge sphere of radius r is propertional to

The refractive index of transparent medium is equal to __________

Find in supeepotion theroem2.mesh and nodal analysis.

Accuracy of the value 26.4 mm isa) 0,1mm b) 0,01cm c) Both (a) and (b) d) None of these

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What are protons ?.... ...Mai to sanskari huclass bunk ni marta...

Accuracy of the value 26.4 mm is . a) 0,1mm b) 0,01cm c) Both (a) and (b) d) None of these

'किनशिप ऑर्गेनाइजेशन इन इण्डिया' नामक पुस्तक के लेखक कौन हैं? ऑब्जेक्ट

What id the full form of dav

Friction opposes the____ between the surface in contact with eachother.

A ball is thrown vertically upward with a velocity of 98 metre per second how high does the ball rise

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A boy travels a 3 km road stops and comes back at center? What will be the distance and displacement?

A body thrown up vertically with a velocity of 19.6 ms-1. Find the maximum height reached and time of ascent.

3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s^-1. Calculate the initial recoil velocity of the rifle.

Find the maximum velocity for slipping a car moving on a circular track of radius 100m. The coefficient of friction between the road and tire is 0.2:A. 0.14m/sB. 140 m/s C.1.4km/s D. 14m/s

What is the net force

Use the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following : (i) Which among iron and mercury is a better conductor ? (ii) Which material is the best conductor ?