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stellarik [79]
3 years ago
5

1. What is the purpose of the lab, the importance of the topic, and the question you are trying to answer?

Physics
1 answer:
notsponge [240]3 years ago
6 0

Answer:

1.The main purpose of lab is to perform different experiments and importance of topic is that it supports essay's thesis statement.

Explanation:

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A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is ex
Delvig [45]

Answer:

Explanation:

Impulse results in a change of momentum

FΔt = mΔV

F = mΔV/Δt

The impulse acting on the hammer will equal the impulse acting on the nail

If we assume upward is the positive direction

F = m(vf - vi)/t

F = 1.2(1.0 - (-1.5)) / 0.001

F = 3000 N

7 0
2 years ago
3 m/s north is an example of a(n) ____.
alexandr1967 [171]
Rate of speed (3 m/s north is three miles per second north, so it's a rate of speed)
4 0
3 years ago
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We can catch a rolling ball but not a flying bullet?? give reason
Triss [41]

Answer:

yeah this statement is tru

Explanation:

it is because the speed of the bullet is more than the speed of rolling ball .so from this reason we cannot catch a bullet.

3 0
3 years ago
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How does frequency change as wavelength increases?
FrozenT [24]
"Frequency decreases" is the one way among the following choices given in the question that <span>frequency change as wavelength increases. The correct option among all the options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
5 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}&#10; 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
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