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rusak2 [61]
2 years ago
14

5) Find the initial velocity for a 700 kg car that

Physics
1 answer:
Serhud [2]2 years ago
5 0

Answer:

Δv = 12 m/s, but we are not given the direction, so there are really an infinite number of potential solutions.

Maximum initial speed is 40.6 m/s

Minimum initial speed is 16.6 m/s

Explanation:

Assume this is a NET impulse so we can ignore friction.

An impulse results in a change of momentum

The impulse applied was

p = Ft = 1400(6.0) = 8400 N•s

p = mΔv

Δv = 8400 / 700 = 12 m/s

If the impulse was applied in the direction the car was already moving, the initial velocity was

vi = 28.6 - 12 = 16.6 m/s

if the impulse was applied in the direction opposite of the original velocity, the initial velocity was

vi = 28.6 + 12 = 40.6 m/s

Other angles of Net force would result in various initial velocities.

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A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v
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Answer:

<em>600N.</em>

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

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a = v-u/t

F = m(v-u)/t

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Recall that: F = m(v-u)/t

F  = 300(30-20)/5

F = 60(30-20)

F = 60(10)

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An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec
jolli1 [7]

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

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