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lara [203]
2 years ago
8

In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired

are equal in magnitude, why do the bullet and gun have very different accelerations?" Think about Newton's Second Law (F = ma) and answer his question.
Physics
1 answer:
Sauron [17]2 years ago
4 0
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)
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Han and Greedo fire their blasters at each other. The blasts are loud, and the intensity of the sound spreads through the cantin
noname [10]
I will say it is B; the Inverse square law. 
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2 years ago
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A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp
Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}

1. Finding Ф from the initial conditions:

-0.02 = 0.02cos(0 + \phi) => \phi = \pi

2. Finding time t at position y = 1 cm:

0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}

3. Find velocity v at time t from equation 2:

v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))

5 0
2 years ago
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Lucy and her bike together have a mass of 120kg. She slows down from 4.5m/s to 3.5m/s. How much kinetic energy does she lose?
vovangra [49]
The kinetic energy of a moving object is given by
K= \frac{1}{2}mv^2
where m is the object's mass and v its velocity.

In our problem, the initial kinetic energy is:
K_i =  \frac{1}{2} m v_i^2 = \frac{1}{2}(120 kg) (4.5 m/s)^2=1215 J

while the final kinetic energy is:
K_f =  \frac{1}{2}mv_f^2 =  \frac{1}{2}(120 kg)(3.5 m/s)^2= 735 J

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7 0
2 years ago
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I believe is 10 lb if not it's 9 lb.
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Charge q1 is distance r from a positive point charge Q. Charge q2=q1/3 is distance 2r from Q. What is the ratio U1/U2 of their p
worty [1.4K]

We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

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From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

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Therefore

The Equations of U1 and U2 is

For U1

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For U2

U2=\frac{-k*q_1Q}{3*2r}

Since

U is a function of q and  q2=q1/3

Therefore

U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

brainly.com/question/23379286?referrer=searchResults

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