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m_a_m_a [10]
1 year ago
9

A 1,200 kg car accelerates at 3 m/s2. What net force is the car experiencing during thisacceleration?0 3600 NO 2000 N0 2400 NOON

Physics
1 answer:
Mama L [17]1 year ago
5 0

m = mass = 1,200 kg

A = acceleration = 3 m/s^2

Apply Newton's second law:

Force = mass x acceleration

F = 1,200 x 3 =3600 N

The net force the car experiences is 3600 N

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Answer:

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Explanation:

Here we will call:

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2. E_2: The energy after the mass is liberated by the spring

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4. E_4: The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. E_1 = E_2

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1. equation 1 is equal to:

\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2

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\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2

Solving for velocity, we get:

V_2 = 65.319 m/s

2. Now, equation 2 is equal to:

\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd

where M is the mass, V_2 the velocity in the situation 2, V_3 is the velocity in the situation 3, U_k is the coefficient of the friction, N the normal force and d the distance, so:

\frac{1}{2}(0.3)(65.319)^2-\frac{1}{2}(0.3)V_3^2 = (0.16)(0.3*9.8)(2)

Volving for V_3, we get:

V_3 = 65.27 m/s

3. Finally, equation 3 is equal to:

\frac{1}{2}MV_3^2 = \frac{1}{2}K_2X_2^2

where K_2 is the constant of the second spring and X_2 is the compress of the second spring, so:

\frac{1}{2}(0.3)(65.27)^2 = \frac{1}{2}(2)X_2^2

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