A 1500-kg car travels at Constant speed of 22 m/s around a circular track that h

Question

A 1500-kg car travels at Constant speed of 22 m/s around a circular track that has a radius of 85 m. Determine the magnitude of the net force that acts on the car. 1.5 Times 10^4 N 8.5 Times 10^3 N zero newtons 1800 N net torque applied to rigid body object always tends to produce linear acceleration rotational equilibrium angular acceleration rotational inertia none of these An inelastic collision is one in which momentum is conserved but kinetic energy is conserved, total mass is not conserved but momentum is conserved, neither kinetic energy and momentum is conserved, momentum is conserved but kinetic energy is not conserved, total impulse is equal to change in kinetic energy A wheel rotating at a constant angular speed of 8 rad/s The time it (in sec) required to rotate through 80 rev is : 10J 5. 10. 20. 40 A certain wheel has rotational inertia of 12 kg m^2. As it turns through 5 rev, its angular velocity Increases from 5 rad/s to 6 rad/s. The net torque is constant it value is 0.016 Nm. 018 Nm. 0.57 Nm. 2.1 Nm. 3 6 Nm The rotational inertia of the wheel does not depend upon diameter, mass, distribution of mass, speed of rotation, material composition What is the length of the simple pendulum when its time period is 2 sec at the surface of the earth 1 92 m. 2 92 m. 0.0092m 0 92 m 0 092 A simple pendulum of length is 1.2 m and has 30 collisions in 48 seconds : its period and gravity are: 1.6 sec, 18.5 m/s^2 1.6 sec, 1.8 m/s^2 1.6 sec, 18.5 m/s 1.6, 9.8 m/s^2 none of these

Explanation / Answer

8.

Fc = mv^2/r = (1500*22^2)/85 = 8.5*10^3 N

9.

= I

Thus torque produces angular acceleration .

10.

In inelastic collision momentum is conserved but kinetic energy is not conserved.

11.

= t

t= / = (80*2*3.14)/8 = 62.8s

12.

f^2 = i^2 + 2

(6)^2 = (5)^2 + 2(5*2*3.14)   => =0.175 rad/s^2

= I = 12*0.175 = 2.1Nm

13.

Rotational inertia of wheel = I= mr^2 , does not depend on speed of rotation

14.

T = 2*sqrt(L/g)

2 = 2*3.14*sqrt(L/9.8)   => L= 0.99m

15.

T =48s/30osci = 1.6s

T = 2*sqrt(L/g)

1.6 = 2*3.14*sqrt(1.2/g)   => g= 18.5 m/s^2

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