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velikii [3]
3 years ago
12

A Meteorite Strikes On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the tru

nk.If the meteorite struck the car with a speed of 130 m/s, what was the magnitude of its deceleration, assuming it to be constant?
Physics
1 answer:
vagabundo [1.1K]3 years ago
7 0

Answer:

Acceleration of the meteorite, a=-38409.09\ m/s^2

Explanation:

It is given that,

A Meteorite after striking struck a car, v = 0

Initial speed of the Meteorite, u = 130 m/s

Distance covered by Meteorite, s = 22 cm = 0.22 m

We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{-(130)^2}{2\times 0.22}

a=-38409.09\ m/s^2

So, the deceleration of the Meteorite is -38409.09\ m/s^2. Hence, this is the required solution.

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d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
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Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
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ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
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Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
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