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juin [17]
3 years ago
10

Newton’s Third Law of Motion states that for every action there is an equal and opposite reaction. Which of these illustrates Ne

wton’s Third Law?
A: Two students of similar mass run into each other. They bounce off each other when they collide.

B: Two students of similar mass run side by side into a wall and punch holes of equal size before emerging on the other side.

C: Two students of similar mass run in opposite directions, but one accelerates at twice the rate of the other.

D: Two students of similar mass run at the same speed in opposite directions on a track. They will collide with 400 Newtons of force in 400 meters.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
4 0

Answer:

A

Explanation:

An equal and opposite reaction happens when they run into eachother and bounce off. What makes it equal is when they have the same amount of force applied when they collided.

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in case you dont want to read the answer is B

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A 60 kg student is standing atop a spring in an elevator that is accelerating upward at 3.0 m/s2. The spring constant is 2.5 x 1
dimaraw [331]

the spring will be compressed by 0.3072 m

Explanation:

acceleration of elevator=3 m/s²

mass of student= 60 Kg

spring constant=2.5 x 10³ N/m

the force on the student is given by F = m ( g +a)

F=60 (9.8+3)

F=768 N

now the formula for spring force is given by

F= k x

768= 2.5 x 10³ (x)

x=0.3072 m

6 0
3 years ago
The charge an electron has spinning around a nucleus.<br><br>(What is it?) <br><br>​
Evgen [1.6K]

Answer:

Negative charge

Explanation:

Electron is an negative charge that spins around the nucleus.

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3 years ago
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It's nighttime, and you've dropped your goggles into a 3.1-mm-deep swimming pool. If you hold a laser pointer 1.2 mm above the e
arlik [135]

Answer:

the googles are 4.33mm from the edge

Explanation:

In this question, we are asked to calculate the distance of a set of googles from the edge of a pool.

We proceed as follows;

depth of pool , d = 3.1mm

Now, let i be the angle of incidence

i = arctan(1.8/1.2)

i = 56.31 degree

Using snell's law ,

n1 * sin(i) = n2 * sin(r)

1 * sin(56.3) = 1.33 * sin(r)

r = 38.72 degrees

Now,

distance of googles = 1.8+ d*tan(r)

distance of googles = 1.8+ 3.1 * tan(38.72)

distance of googles = 4.3 mm

5 0
3 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
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