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Aliun [14]
3 years ago
14

Suppose, the same angular momentum is transferred to two rotating bodies of different moment of inertia , how will you compare t

he angular velocities of the two bodies as a result of angular momentum transfer.
Physics
1 answer:
Sonbull [250]3 years ago
4 0

Answer:

As per the law of conservation of angular momentum, the angular velocity will be higher for the body with a lower moment of inertia and vice versa.

Explanation:

Angular momentum L of a body is given by:

L=I\times \omega

Now when the same angular momentum is transferred to two different bodies with different moment of inertia, the body with a higher moment of inertia will have lower angular velocity and vice versa.

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Explain why there is no acceleration horizontally on an object in Projectile Motion. A. There is no acceleration horizontally be
Jobisdone [24]

Answer:

Option  A is correct.

Explanation:

The force of gravity does not affect the horizontal component in projectile motion; a projectile maintains a constant horizontal velocity. So there is no acceleration ( a=0 ) in horizontal direction as velocity remains constant.

4 0
3 years ago
A car drives on a circular road of radius R. The distance driven by the car is given by = + [where a and b are constants, and t
pishuonlain [190]

Answer:

The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"

answers:

a.18a m/s^{2}

b. a_{rad}=\frac{(27a +b)^{2}}{R}

Explanation:

First let state the mathematical expression for the tangential acceleration and the radial acceleration.

a. tangential acceleration is express as

a_{tan}=\frac{d|v|}{dt} \\

since the distance is expressed as

d=at^{3}+bt

the derivative is the velocity, hence

V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\

hence when we take the drivative of the velocity we arrive at

a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2

b. the expression for the radial acceleration is expressed as

a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}

4 0
3 years ago
How much force is needed to accelerate a 1,800kg car at rate of 1.5 m/s2?
topjm [15]

<u>Given data:</u>

acceleration (a) = 1.5 m/s² ,

           mass (m) = 1800 Kg ,

        Determine F = ?

         <em>From Newtons II law</em>

                          F = m.a   N

                              = 1800× 1.5

                             = 2700 N

<em>2700 N force needed to accelerate the car</em>

7 0
3 years ago
Does gravitational potential energy increase when getting closer to planet universal gravitation law.
LuckyWell [14K]

Answer:

Yes

Explanation:

When the planet moves farther away, the speed and kinetic energy decrease, and the gravitational potential energy increases.

4 0
2 years ago
show how three identical 6 resistors must be connected tho have the following effective resistance values 9 and 4 ohms​
SpyIntel [72]

Answer:

connect two 9 ohms resistance in series now it becomes 18 ohm

5 0
3 years ago
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