13753 + Interview Questions in Physics QA in Physics Page 185 InterviewSolution

9201.	Date ..Pagey Find the kinetic energy of an electron whosede broglie wavelength is same as that of alooke x-ray.
Answer»

Discussion

9202.	Define excitation and ionization potential.
Answer» Excitation and Ionization Energy The minimum energy required to excite an electron from the ground state of an atom to any excited state is called excitation energy. Thus, the energy required to excite an electron in the ground state to the first excited state is called the first excitation energy and so on. The excitation energy of an energy state is the difference between the energy of the excited state and that of the ground state. For example, for the hydrogen atom, the ground state energy(energy of the first orbit) is E1= -13.6 eV and energy of the second orbit is E2= -3.4 eV. This means to excite hydrogen atom, the energy required to be given to it is E2– E1= -3.4 + 13.6 = 10.2 eV. Hence, 10.2 eV is the first excitation energy of the hydrogen atom. Similarly, the II, the III excitation energy of hydrogen atom are E3– E1, E4– E1etc. Ionization Energy Exciting an electron in an atom from its ground state (i.e. n = 1 state) to the infinite state (i.e. nααstate) is called ionization. Thus, the energy required to excite an atom from its ground state to the infinite state is called ionization energy. The ionization energy of an atom, in fact, is the difference of energy when it is in infinite state and in the ground state. If E1and E∞∞represent the energy of the atom when it is in the ground state and in the infinite state respectively, than the ionization energy of the atom is given by E∞∞– E1= 0 + 13.6 = 13.6 eV. Excitation and Ionization Potential Excitation Potential By definition, if an electron is accelerated by 1 volt of potential difference, it acquires 1 eV of energy. So if the electron is accelerated through a p.d. of 10.2 volts, it acquires 10.2 eV energy. If such an extra electron collides with a ground state hydrogen atom, the hydrogen atom may be excited to the first excited energy state. Hence, the 10,2 volt of potential difference is the first excitation potential for a hydrogen atom. Second excitation potential for the hydrogen atom is given by (E3in eV – E1in eV) volt and so on. Thus, the excitation potential of an energy of an atom is the potential, which is required for an electron to jump from the ground state to any one of its excited states. Ionization Potential The potential difference through which the extra electron is to be in acceleration in order for it to cause the ionization of an atom is called the ionization potential of the atom. For example, for the hydrogen atom, the ionization energy is 13.6 eV. By definition of 1 eV, an electron acquires 13.6 eV energy when it is accelerated through a potential difference of 13.6 volts. Thus, ionization potential is the minimum potential to be applied in order to remove the electron completely from its ground state to infinity.

9202.

Define excitation and ionization potential.

Answer»

Excitation and Ionization Energy

The minimum energy required to excite an electron from the ground state of an atom to any excited state is called excitation energy. Thus, the energy required to excite an electron in the ground state to the first excited state is called the first excitation energy and so on. The excitation energy of an energy state is the difference between the energy of the excited state and that of the ground state. For example, for the hydrogen atom, the ground state energy(energy of the first orbit) is E1= -13.6 eV and energy of the second orbit is E2= -3.4 eV. This means to excite hydrogen atom, the energy required to be given to it is E2– E1= -3.4 + 13.6 = 10.2 eV. Hence, 10.2 eV is the first excitation energy of the hydrogen atom. Similarly, the II, the III excitation energy of hydrogen atom are E3– E1, E4– E1etc.

Ionization Energy

Exciting an electron in an atom from its ground state (i.e. n = 1 state) to the infinite state (i.e. nααstate) is called ionization. Thus, the energy required to excite an atom from its ground state to the infinite state is called ionization energy. The ionization energy of an atom, in fact, is the difference of energy when it is in infinite state and in the ground state. If E1and E∞∞represent the energy of the atom when it is in the ground state and in the infinite state respectively, than the ionization energy of the atom is given by E∞∞– E1= 0 + 13.6 = 13.6 eV.

Excitation and Ionization Potential

Excitation Potential

By definition, if an electron is accelerated by 1 volt of potential difference, it acquires 1 eV of energy. So if the electron is accelerated through a p.d. of 10.2 volts, it acquires 10.2 eV energy. If such an extra electron collides with a ground state hydrogen atom, the hydrogen atom may be excited to the first excited energy state. Hence, the 10,2 volt of potential difference is the first excitation potential for a hydrogen atom. Second excitation potential for the hydrogen atom is given by (E3in eV – E1in eV) volt and so on.

Thus, the excitation potential of an energy of an atom is the potential, which is required for an electron to jump from the ground state to any one of its excited states.

Ionization Potential

The potential difference through which the extra electron is to be in acceleration in order for it to cause the ionization of an atom is called the ionization potential of the atom. For example, for the hydrogen atom, the ionization energy is 13.6 eV. By definition of 1 eV, an electron acquires 13.6 eV energy when it is accelerated through a potential difference of 13.6 volts.

Thus, ionization potential is the minimum potential to be applied in order to remove the electron completely from its ground state to infinity.

Discussion

9203.	2. The largest producer of cocoa
Answer» brazil is the largest product of cocoa

Discussion

9204.	One body of mass m is suspended from three springsas shown in figure each spring has spring constantk. If mass m is displaced slightly then time periodoscillation isofIm(a) 223k(b) 2z, 3m2k2m(c) 2Ď(d) 2TSk
Answer»

Discussion

9205.	Problem 1.05. An object is placed at a distance of t40 cm from a concave mirror of focal lengtth 15 cm. If the robject is displaced through a distance of 20 cm towards themirror, by how much distance is the image displaced? 1(P.S.S.C.E. 1998)t
Answer» It just needs to be done as two separate calculations then the difference of the two answers is found. In the initial setup, the object distance is 40 cm, which gives 24 cm image distance given the focal length of 15 cm. (equation: 1/15= 1/40 + 1/i) The second equation changes object distance to 20 cm (40–20), which results in 60 cm as the image distance. (equation: 1/15= 1/20 + 1/i) The difference between 60 and 24 is 36 cm (in the direction away from mirror).

9205.

Problem 1.05. An object is placed at a distance of t40 cm from a concave mirror of focal lengtth 15 cm. If the robject is displaced through a distance of 20 cm towards themirror, by how much distance is the image displaced? 1(P.S.S.C.E. 1998)t

Answer»

It just needs to be done as two separate calculations then the difference of the two answers is found.

In the initial setup, the object distance is 40 cm, which gives 24 cm image distance given the focal length of 15 cm. (equation: 1/15= 1/40 + 1/i)

The second equation changes object distance to 20 cm (40–20), which results in 60 cm as the image distance. (equation: 1/15= 1/20 + 1/i)

The difference between 60 and 24 is 36 cm (in the direction away from mirror).

Discussion

9206.	7(i) A bike and a car have exactly same value of momentum. Whose kinetic energyis more, and why?
Answer» 𝑻𝒉𝒆 𝒃𝒊𝒌𝒆 𝒉𝒂𝒔 𝒎𝒐𝒓𝒆 𝒌𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 𝒂𝒔 𝒊𝒕 𝒉𝒂𝒔 𝒍𝒆𝒔𝒔 𝒎𝒂𝒔𝒔. P² KE = ------ 2m So, we know the mass of the car is more than a bike so, as momentum is the same.And the above expression tells that kinetic energy is inversely proportional to mass.The bike has more kinetic energy as it has less mass.

9206.

7(i) A bike and a car have exactly same value of momentum. Whose kinetic energyis more, and why?

Answer»

𝑻𝒉𝒆 𝒃𝒊𝒌𝒆 𝒉𝒂𝒔 𝒎𝒐𝒓𝒆 𝒌𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 𝒂𝒔 𝒊𝒕 𝒉𝒂𝒔 𝒍𝒆𝒔𝒔 𝒎𝒂𝒔𝒔.

P² KE = ------ 2m

So, we know the mass of the car is more than a bike so, as momentum is the same.And the above expression tells that kinetic energy is inversely proportional to mass.The bike has more kinetic energy as it has less mass.

Discussion

9207.	2Calculate the current in a circuit if 500 C of charge passes through it in 10 minutes.
Answer» Q= 500ct= 10 min => 10*60 = 600 secso, current I= q/t=> 500/600=>0.83 A

Discussion

9208.	(b) Skow that the momentum of bodty having mass m and kinetic energy E is/2mE
Answer»

Discussion

9209.	A ball whose kinetic energy is E, is thrown at anangle of 45° with the horizontal, its kinetic energyat the highest point of its flight will be :-AIPMT 1997, 2000](4) zero2
Answer» Given:let v be initial velocity.K.E= E =1/2 mV²VELOCITY OF PARTICLE AT HIGHEST POINT = vcos 45kinetic energy of particle at highest point is =1/2 m(vcos45)²=1/2(mv²)(1/2)=1/2 E

Discussion

9210.	A ball whose kinetic energy is E, is thrown at anangle of 45 with the horizontal, its kineticenergy at the highest point of its flight will be-[AIPMT 1997, 2000]) (4) zero
Answer» let v be initial velocity.K.E= E =1/2 mV²VELOCITY OF PARTICLE AT HIGHEST POINT = vcos 45kinetic energy of particle at highest point is =1/2 m(vcos45)²=1/2(mv²)(1/2)=1/2 E

Discussion

9211.	en projected at an angle of projection for maximum range. Its K.E. at theA body has kinetic energy E whhighest point of its path will be:219.(A) E2(D) zero
Answer» For maximum range the value of angle is 45° , assume initial velocity as V. so, E = kV².. k is some constant. and at the maximum height, only horizontal part of velocity is left that is Vcos∅ = Vcos45 =V/√2. now, K.E at top is E' = k(V/√2)² = kV²/2. but kV² = E so E' = E/2. option C

9211.

en projected at an angle of projection for maximum range. Its K.E. at theA body has kinetic energy E whhighest point of its path will be:219.(A) E2(D) zero

Answer»

For maximum range the value of angle is 45° , assume initial velocity as V.

so, E = kV².. k is some constant.

and at the maximum height, only horizontal part of velocity is left that is Vcos∅ = Vcos45 =V/√2.

now, K.E at top is E' = k(V/√2)² = kV²/2. but kV² = E

so E' = E/2.

option C

Discussion

9212.	A block of mass m is slightly displaced fromtop of a hemisphere of mass M as shown in figThe displacement of the hemisphere whenblock reaches the angular position 'e', (given fricis absent everywhere)KR. ImRcose(1)m+MmR sinem +M
Answer» Option 1 is the answer option 2 is the right one (2) is right anawer of this question. please like my answer option 2) is the correct answer is option a is the correct answer of the given question mrcos thita/n+m...... option 2 is the correct answer for this question as it's vertical velocity is usintheta It's vertical component is sin. Option 2 is right. Option 2 is the right answer for your question option 2 is the answer option 2 is answer to your question (ii) is the correct answer to this question . It is vertical components sign option 2 is true kadpskkckdkccnjvsejdkkefkekkvekjvsjvjjsjvwkkvkeg option 2 is the right answer 1 this is right answer I don't know this answer because I study in class 6 and I think this sum is class 10 so I can't answer this question option 2 is right as it is vertical compnent options (A) is the correct answer arggjh socthec vertebrates of human beings arrc a is the right answer option 1 is correct answer a is the correct ans. a option is the correct ans . the correct answer is 2 the correct answer is b options 2 is correct answers option a is the correct answer your question I don't know answer correctly options b is the correct answer It's vertical component is sin.option ( 2 ) is the correct answerplease like my answer 2 option is a right answer option 2 is the correct answer option 2 is the right answer Option 2 is correct answaer. the write answer is option number 2 option 2 is the best answer option A is the best answer hello mai apko es sawal ka anser bata skata hu 2 is the correct answer option 2 is the correct answer option 2 is correct answer option ( a ) is the correct answerof the questionplease like my answer option 2 is the right answer option 2 is the correct answer option 2 is the right answer option 2 is the right answer option 2 is the right answer option 2 is the right answer option 2 is the right answer option 2 is the right answer option a is the correct answer option 2 is the correct answer a is the correct option correct option right 2 A is the best answer (2) is right answer option a is the right answer option 2. is the right answer dusra answer hoga dusra answer hoga (2) mRsinA/m+n this right answer 1] be the correct answer was this question it's vertical components is sin. option 2 is tha right answers MR sin theta\m+M is correct answer option 2 is the correct answer Option 2 is the best answer

9212.

A block of mass m is slightly displaced fromtop of a hemisphere of mass M as shown in figThe displacement of the hemisphere whenblock reaches the angular position 'e', (given fricis absent everywhere)KR. ImRcose(1)m+MmR sinem +M

Answer»

Option 1 is the answer

option 2 is the right one

(2) is right anawer of this question. please like my answer

option 2) is the correct answer is

option a is the correct answer of the given question

mrcos thita/n+m......

option 2 is the correct answer for this question as it's vertical velocity is usintheta

It's vertical component is sin. Option 2 is right.

Option 2 is the right answer for your question

option 2 is the answer

option 2 is answer to your question

(ii) is the correct answer to this question .

It is vertical components sign option 2 is true

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option 2 is the right answer

1 this is right answer

I don't know this answer because I study in class 6 and I think this sum is class 10 so I can't answer this question

option 2 is right as it is vertical compnent

options (A) is the correct answer

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options 2 is correct answers

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Option 2 is correct answaer.

the write answer is option number 2

option 2 is the best answer

option A is the best answer

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it's vertical components is sin. option 2 is tha right answers

MR sin theta\m+M is correct answer

option 2 is the correct answer

Option 2 is the best answer

Discussion

9213.	4.(A) What is Avogadro constant?
Answer» The Avogadro constant is the same no. of atoms,molecules or ions (6.022*10 to the power 23) in 1 mole of the substance. Avogadro constant=Avagadro Number=6.022×10 to the power 23

9213.

4.(A) What is Avogadro constant?

Answer»

The Avogadro constant is the same no. of atoms,molecules or ions (6.022*10 to the power 23) in 1 mole of the substance.

Avogadro constant=Avagadro Number=6.022×10 to the power 23

Discussion

9214.	A liquid contained in a calorimeter cools from 80°C to 60°C in 10s. If the temperature of thesurrounding is 30°C, calculate the time it will take to cool further to 45'C
Answer» According to Newton's law of cooling, we havemc (T1- T2)/t = k [(T1+ T2)/2 - T0)] ---------------(1)mc/k = [(T1+ T2)/2 - T0)]/[(T1- T2)/t]mc/k =[(80+50)/2 - 20)]/[(80- 50)/5] = 45/6Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in°C) in Eq. (1), we getmc (60- 30)/t = k [(60+ 30)/2 - 20)]30 mc/t = 25 kt = 30 mc/(25k) = (30/25) x (45/6) = 9 minThe body takes 9 min to cool from60°Cto30°C. but the answer is 13.5 seconds

9214.

A liquid contained in a calorimeter cools from 80°C to 60°C in 10s. If the temperature of thesurrounding is 30°C, calculate the time it will take to cool further to 45'C

Answer»

According to Newton's law of cooling, we havemc (T1- T2)/t = k [(T1+ T2)/2 - T0)] ---------------(1)mc/k = [(T1+ T2)/2 - T0)]/[(T1- T2)/t]mc/k =[(80+50)/2 - 20)]/[(80- 50)/5] = 45/6Substituting T1 = 60 and T2 = 30 and T0 = 20 (all in°C) in Eq. (1), we getmc (60- 30)/t = k [(60+ 30)/2 - 20)]30 mc/t = 25 kt = 30 mc/(25k) = (30/25) x (45/6) = 9 minThe body takes 9 min to cool from60°Cto30°C.

but the answer is 13.5 seconds

Discussion

9215.	. When starter motor of a car is switched on for08 s, a charge 24 C passes through the coil of themotor. Calculate the current in the coil.Ans. 30 A
Answer» Answer:Electric current (i) = Charge (q) /Time (t) = 24 C/0.8 s = 30 C/s = 30 amp

Discussion

9216.	The tyres of aircrafts are not made of ordinary(which is an insulator) but of a special rubberis slightly conducting. Why?
Answer» tyresof anaircraftis motmadeup of ordinary rubbers because a great force is applied on thetyresduring landing which an ordinarytyrecan not sustain. during landing due to friction plane gaet charged and also emf induced in plane due to earth magnetic field sotyreof plane aremadeslightelyconductive.

Discussion

9217.	10A solid sphere is given a kinetic energy E. Whatfraction of kinetic energy is associated withrotation?(a) 3/7(b) 5/7(d) 2/7
Answer» If the sphere is rolling at velocity v, the kinetic energy is KE = 0.5mv² + 0.5Iω² I (moment of inertia of the sphere) = (2/5)mr² ω = v/r KE = 0.5mv² + 0.5(2/5)mr²(v/r)² KE = 0.5mv² + (1/5)mv² Total KE = (7/10)m*v² The fraction that is rotational is (1/5) / (7/10) = 2/7 or 0.286 (d) is correct option

9217.

10A solid sphere is given a kinetic energy E. Whatfraction of kinetic energy is associated withrotation?(a) 3/7(b) 5/7(d) 2/7

Answer»

If the sphere is rolling at velocity v, the kinetic energy is

KE = 0.5*m*v² + 0.5*I*ω²

I (moment of inertia of the sphere) = (2/5)*m*r² ω = v/r

KE = 0.5*m*v² + 0.5*(2/5)*mr²*(v/r)²

KE = 0.5*m*v² + (1/5)*m*v²

Total KE = (7/10)*m*v²

The fraction that is rotational is (1/5) / (7/10) = 2/7 or 0.286

(d) is correct option

Discussion

9218.	mercury to tbarometer is 75 cm at sea level and S0 em at the top of a hill. Ratio of densitya) 250 mThe height of a mercury 75 cof mercury to that of air is 10. The height of the hill is750 m
Answer»

Discussion

9219.	(g)How can we say that air exerts pressure?
Answer» The molecules are continually colliding with each other and with the walls of the container. When a molecule collides with the wall, theyexertsmall force on the wall Thepressure exerted by the gas is due to the sum of all these collision forces.The more particles that hit the walls, the higher thepressure

Discussion

9220.	A uniform chain of length 2L is hanging in equilibrium position, if end B is given aslightly downward displacement the imbalance causes an acceleration. Here pulley issmall and smooth & string is inextensibleThe acceleration of end B when it has been displaced by distance x, is2x(B)-gD) g(A)-gA B
Answer»

Discussion

9221.	Consider the barometer shown in figure. If a small hole is made at point 'P' in the barometer tube. Than whichof following option is correct.A) The mercury will come out form this holeB) The mercury will not come out from this holeC) The mercury level inside the tube decreasesD) The mercury level inside the tube increases
Answer»

Discussion

9222.	An archer pulls back 0.75 m on a bowwhich has a stiffness of 200 N/m. Thearrow weighs 50 g. What is the velocityof the arrow immediately after release?
Answer» This can be solved using an energy method. We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow. The bow can be treated as a type of spring. The potential energy of a spring is: (1/2)kx2, where k is the stiffness and x is the amount the spring is stretched, or compressed. Therefore, the potential energy PE of the bow is: PE = (1/2)(200)(0.75)2 = 56.25 J The kinetic energy of a particle is: (1/2)mv2, where m is the mass and v is the velocity. The arrow can be treated as a particle since it is not rotating upon release. Therefore, the kinetic energy KE of the arrow is: KE = (1/2)(0.05)v2 If we assume energy is conserved, then PE = KE Solving for the velocity of the arrow v we get v = 47.4 m/s

9222.

An archer pulls back 0.75 m on a bowwhich has a stiffness of 200 N/m. Thearrow weighs 50 g. What is the velocityof the arrow immediately after release?

Answer»

This can be solved using an energy method.

We can solve this by equating the potential energy of the bow to the kinetic energy of the arrow.

The bow can be treated as a type of spring. The potential energy of a spring is:

(1/2)kx2, where k is the stiffness and x is the amount the spring is stretched, or compressed.

Therefore, the potential energy PE of the bow is:

PE = (1/2)(200)(0.75)2 = 56.25 J

The kinetic energy of a particle is:

(1/2)mv2, where m is the mass and v is the velocity.

The arrow can be treated as a particle since it is not rotating upon release.

Therefore, the kinetic energy KE of the arrow is:

KE = (1/2)(0.05)v2

If we assume energy is conserved, then

PE = KE

Solving for the velocity of the arrow v we get

v = 47.4 m/s

Discussion

9223.	by Write one sighmcant eA car travels from stop A to stop B with a speed of 30 km/h and then returns back to Awith a speed of 50 km/h. Find(aDisplacement of the car.(b) Distance travelled by the car.Average speed of the car.
Answer» a) displacement is zerob) distance cannot be calculated without time.3). average speed is (50+30)/2 = 40 kmph average speed of 59+30/2=40kmph displacement is zero distance cannot be calculated without time

9223.

by Write one sighmcant eA car travels from stop A to stop B with a speed of 30 km/h and then returns back to Awith a speed of 50 km/h. Find(aDisplacement of the car.(b) Distance travelled by the car.Average speed of the car.

Answer»

a) displacement is zerob) distance cannot be calculated without time.3). average speed is (50+30)/2 = 40 kmph

average speed of 59+30/2=40kmph displacement is zero distance cannot be calculated without time

Discussion

9224.	1. From lihe sop of a tower of height 40 m, e ball is projected up wards with a speed of 20 rmsat an angle of elevation of30 Then the ratio of the total time taken by the ball to hit the ground to the fime taken to hall come af sarme level autop of towerA)2:1B) 3:1C) 3:2D) 4 I
Answer» Time taken to reach maximum height in a horizontal projection : H = 20²sin²30° / 2 × 10 H = (400 × 0.25) / 20 = 5m Total height is : 5 + 40 = 45 m Time of flight = (2 × 20 × Sin 30) / 10 = 2 seconds Time of free fall is √(2 × 45)/10 = √9 = 3 seconds. The total time taken to reach the ground is thus : 1 + 3 = 4 Seconds The ratio is thus 4:2

9224.

1. From lihe sop of a tower of height 40 m, e ball is projected up wards with a speed of 20 rmsat an angle of elevation of30 Then the ratio of the total time taken by the ball to hit the ground to the fime taken to hall come af sarme level autop of towerA)2:1B) 3:1C) 3:2D) 4 I

Answer»

Time taken to reach maximum height in a horizontal projection :

H = 20²sin²30° / 2 × 10

H = (400 × 0.25) / 20 = 5m

Total height is :

5 + 40 = 45 m

Time of flight = (2 × 20 × Sin 30) / 10 = 2 seconds

Time of free fall is √(2 × 45)/10 = √9 = 3 seconds.

The total time taken to reach the ground is thus :

1 + 3 = 4 Seconds

The ratio is thus 4:2

Discussion

9225.	A boy goes to market at an average speed of29 kmph. Finding the market closed, he immediatelyreturns back by the same path with an average speedof 40 km h-1. Find the (a) average speed (b) averagevelocity for the complete journey.
Answer» Distance = s in both caseT1 = s/v1 =s/29 hT2=s/v2 =s/40 hA) Av speed for whole trip =total distance /total time = s+s/t1+t2=2s km/ s/29+ s/40 h=2s/69s/1160 km/h=2× 1160/69 =33.62 km/hAnd B)Displacement=0T1=0/29 km/hT2=0/40 km/hAverage velocity=displacement/time takenAv=0/0/1160=0km/h because displacement is 0

Discussion

9226.	4. An object travels 16 m in 4 s and then another 16 m2 s, then what would be its average speed?
Answer» total time is = 4s +2s = 6s , total distance = 16+16 = 32m so, avg speed = total distance/time = 32/6 = 16/3 m/s thanc thanx

9226.

4. An object travels 16 m in 4 s and then another 16 m2 s, then what would be its average speed?

Answer»

total time is = 4s +2s = 6s , total distance = 16+16 = 32m

so, avg speed = total distance/time = 32/6 = 16/3 m/s

thanc

thanx

Discussion

9227.	8.1 An object travels 16 m in 4 sleampand then another 16 m in 2the average speed of the object?s. What isSolution:Total distance travelled by the object-16 m + 16 m 32 mTotal time taken4 s+2 s- 6 s
Answer» Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s Average speed = Total distance / Total time = 32/6 = 5.33 m/s

9227.

8.1 An object travels 16 m in 4 sleampand then another 16 m in 2the average speed of the object?s. What isSolution:Total distance travelled by the object-16 m + 16 m 32 mTotal time taken4 s+2 s- 6 s

Answer»

Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s

Average speed = Total distance / Total time = 32/6 = 5.33 m/s

Discussion

9228.	ampleand then8.1 An object travels 16 m in 4 sother 16 m in 2 s. What is3the average speed of the ob
Answer» Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s Average speed = Total distance / Total time = 32/6 = 5.33 m/s

9228.

ampleand then8.1 An object travels 16 m in 4 sother 16 m in 2 s. What is3the average speed of the ob

Answer»

Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s

Average speed = Total distance / Total time = 32/6 = 5.33 m/s

Discussion

9229.	1.An object travels a distance of 18min 4s and then another 18m in 2s. What is the average speed ofthe object?
Answer»

Discussion

9230.	xample 8.1 An object travels 16 m in 4sand then another 16 m in 2 s. What isthe average speed of the object?
Answer» Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s Average speed = Total distance / Total time = 32/6 = 5.33 m/s

9230.

xample 8.1 An object travels 16 m in 4sand then another 16 m in 2 s. What isthe average speed of the object?

Answer»

Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s

Average speed = Total distance / Total time = 32/6 = 5.33 m/s

Discussion

9231.	Example 8.1 An object travels 16 m in 4 sand then another 16 m in 2 s. What isthe average speed of the object?
Answer» what is your question Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s Average speed = Total distance / Total time = 32/6 = 5.33 m/s I hope this was helpful ;) 1 Secondary SchoolPhysics5 points An object travels 16m in 4s and then another 16m in 2s.what is the average speed of the object? Ask for detailsFollowReport22.08.2015 Answers THE BRAINLIEST ANSWER! praneeth1803Ambitious Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s Average speed = Total distance / Total time = 32/6 = 5.33 m/s I hope this was helpful ;) 4.6 227 votes THANKS 829 CommentsReport lovingheartVirtuoso Answer: The average speed of the object is 5.33 m/s. Explanation: Velocity V = Distance traveled by an object in part 1, D1 = 16 m Time taken by an object (in part 1), T1 = 4 seconds Distance traveled by an object in part 2, D2 = 16 m Time taken by in part 2, T2 = 2 seconds Average speed of an object = 16 + 16 / 4 + 2 = 5.33 m/s 5.33 is answer of this question

9231.

Example 8.1 An object travels 16 m in 4 sand then another 16 m in 2 s. What isthe average speed of the object?

Answer»

what is your question

Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s

Average speed = Total distance / Total time = 32/6 = 5.33 m/s

I hope this was helpful ;)

1

Secondary SchoolPhysics5 points

An object travels 16m in 4s and then another 16m in 2s.what is the average speed of the object?

Ask for detailsFollowReport22.08.2015

Answers

THE BRAINLIEST ANSWER!

praneeth1803Ambitious

Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s

Average speed = Total distance / Total time = 32/6 = 5.33 m/s

I hope this was helpful ;)

4.6

227 votes

THANKS

829

CommentsReport

lovingheartVirtuoso

Answer:

The average speed of the object is 5.33 m/s.

Explanation:

Velocity V =

Distance traveled by an object in part 1, D1 = 16 m

Time taken by an object (in part 1), T1 = 4 seconds

Distance traveled by an object in part 2, D2 = 16 m

Time taken by in part 2, T2 = 2 seconds

Average speed of an object = 16 + 16 / 4 + 2 = 5.33 m/s

5.33 is answer of this question

Discussion

9232.	Esmple 8.1 An object travels 16 m in 4 sand then another 16 m in 2 s. What isthe average speed of the object?
Answer» Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s Average speed = Total distance / Total time = 32/6 = 5.33 m/s

9232.

Esmple 8.1 An object travels 16 m in 4 sand then another 16 m in 2 s. What isthe average speed of the object?

Answer»

Total distance = 16m + 16m = 32mTotal time = 4s + 2s = 6s

Average speed = Total distance / Total time = 32/6 = 5.33 m/s

Discussion

9233.	ole 8.1 An object travels 16 m in 4 sand then another 16 m in 2 s. What isthe average speed of the object?
Answer» Total distance= 16+16= 32total time taken = 4+2 = 6 soas we knowspeed= distance/time= 32/6= 5.3m/$

Discussion

9234.	Q. 7. Prove the relation âœ“ = x7, where thesymbols have their usual meanings.
Answer»

Discussion

9235.	For a prism, prove the relation i + e = A + δ wherethe symbols have their usual meanings.
Answer»

Discussion

9236.	Deduce graphically v= u+ at, where the symbols have their usual meanings.
Answer» Consider the velocity – time graph of a body shown in the below Figure. The body has an initial velocityuat pointAand then its velocity changes at a uniform rate fromAtoBin timet. In other words, there is a uniform acceleration 'a' fromAtoB, and after timetits final velocity becomes 'v' which is equal toBCin the graph. The timetis represented byOC. To complete the figure, we draw the perpendicularCBfrom pointC, and drawADparallel toOC. BEis the perpendicular from pointBtoOE. Now, Initial velocity of the body,u=OA....(1)And, Final velocity of the body,v=BC......(2)But from the graphBC=BD +DC.Therefore,v=BD + DC......... (3)AgainDC=OASo,v=BD + OANow, From equation(1),OA=uSo,v=BD+u........... (4)We should find out the value ofBDnow. We know that the slope of a velocity – time graph is equal to acceleration,a. Thus, Acceleration,a=slope of lineAB ora=BD/ADButAD=OC = t,so puttingtin place ofADin the above relation, we get:a=BD/torBD=atNow, putting this value ofBDin equation (4) we get :v=at+uThis equation can be rearranged to give: v=u + atAnd this is the first equation of motion. It has been derived here by the graphical method. thanks

9236.

Deduce graphically v= u+ at, where the symbols have their usual meanings.

Answer»

Consider the velocity – time graph of a body shown in the below Figure.

The body has an initial velocityuat pointAand then its velocity changes at a uniform rate fromAtoBin timet.

In other words, there is a uniform acceleration 'a' fromAtoB, and after timetits final velocity becomes 'v' which is equal toBCin the graph. The timetis represented byOC.

To complete the figure, we draw the perpendicularCBfrom pointC, and drawADparallel toOC.

BEis the perpendicular from pointBtoOE.

Now, Initial velocity of the body,u=OA....(1)And, Final velocity of the body,v=BC......(2)But from the graphBC=BD +DC.Therefore,v=BD + DC......... (3)AgainDC=OASo,v=BD + OANow, From equation(1),OA=uSo,v=BD+u........... (4)We should find out the value ofBDnow. We know that the slope of a velocity – time graph is equal to acceleration,a.

Thus, Acceleration,a=slope of lineAB ora=BD/ADButAD=OC = t,so puttingtin place ofADin the above relation, we get:a=BD/torBD=atNow, putting this value ofBDin equation (4) we get :v=at+uThis equation can be rearranged to give:

v=u + atAnd this is the first equation of motion. It has been derived here by the graphical method.

thanks

Discussion

9237.	δQ1 : For a prism, prove the relation i + e =A + deltawhere the symbol have their usual meanings.
Answer»

Discussion

9238.	7. Explain what is meant by the equationg=GX-R2where the symbols have their usual meanings
Answer»

Discussion

9239.	electronsBxample I.3 How much guustive and negative charge is there in acup of water?olution Let us ame tet the
Answer» Wateris H2O. Each hydrogen atom contains on electron and eachwatermolecule contains 2 hydrogens. Each oxygen contains 8 electrons

Discussion

9240.	pulsion betiieenQg CA’\V\‘\*}\\ n७७ही 23011 (एक &,\| poaitively chornged il chang5 a2 xS ™. whan लेन पर. depanated by QAbdonee af 5% Mow ooy elacknany ant13004. ही each ian
Answer» Electrostatic force F = q1q2/(4πϵor2)Let q1 = q2 = ne where n is the number of electrons missing from each atom and e the charge on an electron = 1.6 x 10^-19C.1/(4πϵo) = 9 x 10^9N.m2/C2.r = 5 x 10^-10m F = 3.7 x 10^-9N = n2. (1.6 x 10-19C)2x 9 x109N.m2/C2/(5 x 10-10m)2. ⇒ n2=3.7 x 10^-9N x(5 x 10-10m)^2/(1.6 x 10-19C)2x 9 x109N.m2/C2. = 4⇒ n = 2 The number of electrons missing from each atom = 2

9240.

pulsion betiieenQg CA’\V\‘\*}\\ n७७ही 23011 (एक &,| poaitively chornged il chang5 a2 xS ™. whan लेन पर. depanated by QAbdonee af 5% Mow ooy elacknany ant13004. ही each ian

Answer»

Electrostatic force F = q1q2/(4πϵor2)Let q1 = q2 = ne where n is the number of electrons missing from each atom and e the charge on an electron = 1.6 x 10^-19C.1/(4πϵo) = 9 x 10^9N.m2/C2.r = 5 x 10^-10m

F = 3.7 x 10^-9N = n2. (1.6 x 10-19C)2x 9 x109N.m2/C2/(5 x 10-10m)2.

⇒ n2=3.7 x 10^-9N x(5 x 10-10m)^2/(1.6 x 10-19C)2x 9 x109N.m2/C2. = 4⇒ n = 2

The number of electrons missing from each atom = 2

Discussion

9241.	Find dy/dxa ) (x+y)^2 = 4 b) x²y+xy²=6
Answer» a ) (x+y)^2 = 4 Differentiating both sides w.rt x d ( (x+y)^2) /dx = d4/dx2(x+y ) ( 1+ dy/dx) = 0 1+dy/dx = 0 dy/dx = -1 b) Differentiating both sides w.rt x 2xy + x^2 (dy/dx) + y^2 + 2xy (dy/dx) = 0 (x^2 + 2xy) (dy/dx) = -2xy- y^2 dy/dx = (-2xy- y^2)/(x^2+2xy)

9241.

Find dy/dxa ) (x+y)^2 = 4 b) x²y+xy²=6

Answer»

a ) (x+y)^2 = 4 Differentiating both sides w.rt x d ( (x+y)^2) /dx = d4/dx2(x+y ) ( 1+ dy/dx) = 0 1+dy/dx = 0 dy/dx = -1

b) Differentiating both sides w.rt x 2xy + x^2 (dy/dx) + y^2 + 2xy (dy/dx) = 0 (x^2 + 2xy) (dy/dx) = -2xy- y^2 dy/dx = (-2xy- y^2)/(x^2+2xy)

Discussion

9242.	Explain first law of thermodynamics.
Answer» Ans :- The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as ΔU=q+w with ΔU is the total change in internal energy of a system,q is the heat exchanged between a system and its surroundings, andw is the work done by or on the system. PLEASE LIKE THE ANSWER

9242.

Explain first law of thermodynamics.

Answer»

Ans :- The First Law of Thermodynamics states that energy can be converted from one form to another with the interaction of heat, work and internal energy, but it cannot be created nor destroyed, under any circumstances. Mathematically, this is represented as

ΔU=q+w

with

ΔU is the total change in internal energy of a system,q is the heat exchanged between a system and its surroundings, andw is the work done by or on the system.

PLEASE LIKE THE ANSWER

Discussion

9243.	What is First law of thermodynamics?
Answer» The heat supplied to the system is equal to the sum of the increase in internal energy of the system and the external work done by the system. First law of thermodynamics is only another statement of the Law of Conservation of Energy.

Discussion

9244.	what is First law of thermodynamics
Answer» The heat supplied to the system is equal to the sum of the increase in internal energy of the system and the external work done by the system. First law of thermodynamics is only another statement of the Law of Conservation of Energy.

Discussion

9245.	Ä°Ä°K that air exerts.3. State two experiences that made y(other than those given in the text
Answer»

Discussion

9246.	State and explain first law of thermodynamics.
Answer»

Discussion

9247.	3.Under what conditions is the first law of thermodynamics valid?
Answer»

Discussion

9248.	1What is a Vernier Calliper? Why is it so called?
Answer» A vernier scale is a visual aid to take an accurate measurement reading between two graduation markings on a linear scale by using mechanical interpolation; thereby increasing resolution and reducing measurement uncertainty by using Vernier acuity to reduce human estimation error. I am student of class 9

9248.

1What is a Vernier Calliper? Why is it so called?

Answer»

A vernier scale is a visual aid to take an accurate measurement reading between two graduation markings on a linear scale by using mechanical interpolation; thereby increasing resolution and reducing measurement uncertainty by using Vernier acuity to reduce human estimation error.

I am student of class 9

Discussion

9249.	First law of newrtons first law?
Answer» Newton's First Law statesthat an object will remain at rest or in uniformmotionin a straight line unless acted upon by an external force. It may be seen as a statement about inertia, that objects will remain in theirstateofmotionunless a force acts to change themotion.

Discussion

9250.	2While the piston is at a distance 2L from the top, the hole at the top is sealed. The piston is thenreleased, to a position where it can stay in equilibrium. In this condition, the distance of the piston fromthe top is2PR(2L)
Answer»

Discussion

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