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Initially,Ei = - GMem/4Re

The change in T.E =gmRe/8

Substituting the values , we get=3.13*10^9

The K.E will be reduced and the P.E will become twice.

as we know ,v=u+at

since initial speed was u in negative direction ,and final speed is zerogives 0=u+atand a=-u/t.

1)shows that acceleration will act in positive direction2)if velocity is positive so to reduce it's magnitude acceleration should be in negative direction.

Solution :-

The difference of K.E in first case is 1/2 m 400 - 1/2 m 100 = 1.2 m 300

The difference in K.E. in second case is 1/2 m 100 - 1/2 m 0 = 1/2 m 100

So answer is 3 times.

i Don't ⛔ understand

For firing angle 30°v= 400/(1+cos30)= 400/(1+√3/2)800/3.372=237.24 volt

Based on Work - Energy Principle :

The Change in the Kinetic Energy of the Cycle is equal to the Net Work done on the Object.

W net = mv^2/2-Mu^2/2

Where u is the initial velocity and v is the final velocity

Given the mass of the Cycle = 30 kg

Intial Velocity (u) = 54km/hr = 54000/3600 = 15m/sec

Final Velocity (v) = 0 (because we are stopping the cycle)

Substituting all the values in our Net Work done Equation we get :

⇒ Work done to stop the Cycle = 0 - (30 × 15²)/2 = -15³ = -3375 Joules

(Negative sign show's that We are doing Work on the Cycle to stop

Velocity = 90 km/hr = 25 m/s.

Power = 25 kW = 25,000 W.

Work = Kinetic energy

∴ W =1/2 mv²

⇒ W = 375000

Since, Power = Work/time.

⇒ Time = Work/Power

⇒ Time = 375000/25000

∴ t = 15 s.

As we know thatcontinuously conduct is theresopower will be 24*2/4= 12 w

the correct answer is 12w

this question is correct answer b

Here power= 48/24*6= 12w

Force = Ma= mass = 10kgv= 9u= 5F= m(v-u)/t= 10(9-5)/2= 10*4/2= 20NOption A

Let the force be F,Mass is M,time for which it is accelerated = 2svelocity is increased from 3m/s to 7m/sso, acceleration is , a = (7-3)/2 = 2m/s²The force applied is, F = ma = (5)(2) = 10 NNow, the force is applied for 5s, the velocity after 5s will be,v=u+at=> v =3+(2)(5)=>v=13m/s , this is final velocity

Initial speed = 54Km/h = 54×5/18 = 15m/sfinial speed = 0time = 90sv = u + at0 = 15 + 90aa = -1/6let distance = sv² = u² + 2as0 = 15² - 2×s/6s = 225×3s = 675 m

given that ,

》mass of block = 2kg

》force applied = 2.5

》coefficient of static friction, µs= 0.4

☆Maximum static frictional force,

fmax = µs× m × g = 0.4 × 2 × 9.8

=> fmax= 7.84 N

Since,

the applied force is less than the maximum static frictional force,

the frictional force on the block is equal to the applied force = 2.5 N.

☆it is due to the fact that static friction is a self adjusting force.

option b

Here we will apply relative conceptlet the speed of the car isu then relative speed of car with respect to car c is (u+30)

Distance is 5 kilometres

Time is 4minutes= 4÷ 60 hours

Speed= distance/ time

U+30= 5/4/60

U+30=5×60/4

On solving above equation we get

U+30 = 75 then u= 75 - 30= 45 km/h

Convex mirrorsare commonlyusedas rear-view (wing)mirrorsinvehiclesbecause they give an erect, virtual, full size diminished image of distant objects with a wider field of view. Thus,convex mirrorsenable the driver to view much larger area than would be possible with a planemirror.

B. 4 The ratio is 4:1

How to start the MS - paint

Banking of road is done to ensure the safety of cars on curved roads. As the speed of the vehicle increases, the centripetal force required for circular motion also increases. The maximum speed of the vehicle depends only on the coefficient of friction between the wheels and road. However, the friction coefficient reduces if the roads are wet or cover with oil. Thus, to increase the safety of the vehicle, the outer edge is raised above the inner one, making an inclination with the horizontal. The banking of road provides the necessary centripetal force for the motion of the vehicle on curved roads.

Optimum speed = v₀ = [RgtanΘ]¹⁻₂

R= 300theta = 15⁰g= 10

so, v₀= √300 x 10 x tan 15= 28.06 ms⁻²

2. max speed= Vm=√Rg u+tan theta/ 1- utan thetaas u=0.2

=√300 x 10 0.2+tan15/1- 0.2tan 15

= 38.13 ms⁻¹

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For the car with 60kmph to overtake the first car it should travel a relative distance of 10 m

(5m for catching up and 5m for getting clear ahead)

therefore Srel = 10 m

Vrel = 60-42 = 18km/hr = 5 m/s

Relative acceleration= 0 .since both are moving with constant velocities.

Now using relative motion, S = Vrel × t

therefore time ,t = S / Vrel = 10/5 =2s

Distance used to overtake is 60× 5/18× 2= 100/3 = 33.33 m

( D) Positive terminal of

the correct answer is option (c) because the positive terminal should be attached to the negative terminal of the next cell

what is a dielectric ?

DISTANCE COVERED BY CAR WILL EQUAL TO THE DISTANCE COVERED BY BUS.

SO,

UT + 1/2 AT² = 50T

0×T+1/2×5×T² = 50T

5/2 T² = 50T

5/2 T = 50

T = 50×2/5

T = 20 SECONDS

DISTANCE = SPEED × TIME

DISTANCE = 50 × 20

DISTANCE = 1000 METRES

USING 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION,

V² = U² + 2AS

V² = 0² + 2×5×1000

V² = 10×1000

V = √10000

V = 100 M/S

thanks

what is the full form of w.r.t

If deflection takes place in opposite direction, the device is not properly connected in the circuit.We must interchange the terminal connection, so that devices can be used properly in the circuit.

When a vehicle moves round a horizontal curve on the road with sufficient speed, the necessary centripetal force is provided by the friction between the tyres and the road. So, to increase the centripetal force, friction should be increased which causes wear and tear of the tyres. If the road is banked on the curve, the necessary centripetal force is provided by the horizontal component of normal resistance.

i.e. Fc = Rsinθ

Here, centripetal force can be increased be increasing the angle of banking. So, roads are banked on curved path.

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When the body placed on some surface is just on the point of sliding then in that condition theanglebetween applied force and normal force is called theangle of friction.