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Savatey [412]
4 years ago
14

A 15-g bullet moving at 300 m/s passes through a 2.0 cm thick sheet of foam plastic and emerges with a speed of 90 m/s. Let's as

sume that the speed of the bullet changes takes place uniformly (uniform deceleration). Find the change in momentum of the bullet. Find the time the bullet is in contact with the plastic, Delta t. Find the average force impeded by the bullet's motion through the plastic.
Physics
1 answer:
Shkiper50 [21]4 years ago
8 0

Answer:

Explanation:

a) Change in momentum, Δp = mΔv = m(v - u) = (15 * 10-3) * (90 - 300) = -3.15 kg-m2

b) Acceleration of the bullet, a = (v2 - u2) / 2s = (902 - 3002) / (2 * 0.02) = -2047500 m/s2

So, the bullet is in contact with the plastic for the time,  \bigtriangleupt = \frac{(v - u)}{a} =\frac{(90 - 300)}{(-2047500)} = 1.03 \times 10^{-4} s

c) Average force, F_{avg} =\frac{\bigtriangleup p}{\bigtriangleup t} =\frac{(-3.15)}{(1.03 \times 10^{-4})} = 30.6 kN

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4 0
3 years ago
Match the letter and term.
rosijanka [135]

Answer:

1. Nitrogen

2. Oxygen

3. Carbon dioxide

4. Water vapor

5. Ozone

Explanation:

The atmosphere composes of 78% nitrogen which occupies the largest percentage followed by oxygen which takes up 21%, Argon takes up 1% then other components such as water vapor occupy between 0-7% and ozone takes 0.0-0.01. Moreover, 0.01-0.1 is occupied by carbon dioxide. Therefore, the answers for 1-5 are as follows.

1. Nitrogen

2. Oxygen

3. Carbon dioxide

4. Water vapor

5. Ozone

4 0
3 years ago
Two sound waves of equal amplitude interfere so that the compression of one wave falls on the rarefaction of the other.Which sta
MAVERICK [17]
<span>Two sound waves of equal amplitude interfere so that the compression of one wave falls on the rarefaction of the other. The statement which is true is that A. no sound is heard.
These waves interfere, which means that their equal amplitudes match and thus cancel each other out, which is why there is no sound.
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8 0
3 years ago
Read 2 more answers
An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th
kolbaska11 [484]

Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

mg=8*10^{2},

m=(8*10^{2})/g,

since we now that g=9.8m/s^{2} we get that the mass is

m=81.6kg.

Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

where m is the mass of the astronaut and M is the mass of the earth. From Newton's second law we know that

F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

mg=200.4 N.

5 0
3 years ago
What is the efficiency of a ramp if the input work is 96 J and the output work is 24 J?
iren2701 [21]

Answer:It is 200

Explanation: GOOGLE

7 0
3 years ago
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