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professor190 [17]
2 years ago
10

A 4500 kg car is pushed by 3 boys who exert a force totaling 1500 N. When the boys begin to push the car it is traveling 4.5 m/s

. What is the velocity of the car after 3.0 seconds?
Physics
1 answer:
Brilliant_brown [7]2 years ago
7 0

Answer:

The velocity of the car after 3 seconds is 5.5 m/s

Explanation:

Mathematically, we can get the acceleration of the car first;

F = m•a

where m is the mass, F is the applied force and a is the acceleration

from the question;

F = 1500 N , m = 4500 kg and a = ?

By rewriting the formula;

a = F/m

a = 1500/4500 = 1/3 m/s^2

Now to get the velocity of the car after 3 seconds;

Mathematically;

V = U + at

where V is the final

velocity, U is the initial velocity,

From the question, we are trying to get V;

where U = 4.5 m/s

a = 1/3 m/s^2 and t = 3 seconds

Thus;

V = 4.5 + (1/3)(3)

V = 4.5 + 1

V = 5.5 m/s

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The given parameters;

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The moment of inertia of the rod about the end is given as;

I_i = \frac{1}{3} ML^2

The moment of inertia of the rod between the middle and the end is calculated as;

I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} =  \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}

Apply the principle of conservation of angular momentum as shown below;

I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

3 0
2 years ago
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ΔT = 0.12 K

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