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Vedmedyk [2.9K]
2 years ago
9

7. A child pulls a sled up a snow-covered hill. The child does 360 J of

Physics
2 answers:
Arlecino [84]2 years ago
8 0

Answer:

The child fell off and died.

Aleksandr-060686 [28]2 years ago
5 0

Answer:

.6 m

Explanation:

My friend told me

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Which phrase best describes the outer planets?
pav-90 [236]

Answer:

Astronomers have divided the eight planets of our solar system into the inner planets and the outer planets. The 4 inner planets are the closest to the Sun, and the outer planets are the other four – Jupiter, Saturn, Uranus, and Neptune. The outer planets are also called the Jovian planets or gas giants.

Explanation:

6 0
3 years ago
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He __________ uses a series of gears to adjust the output speed of the engine.
Verizon [17]
Hey /人 ◕ ‿‿ ◕ 人\
The answer is transmission

uses a series of gears to transmit power to facilitate changes in speed .


GLAD TO HELP


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4 0
3 years ago
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AysviL [449]
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8 0
2 years ago
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A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision
marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

x \approx 103,46x10^{9}m.

B) The space time coordinate t of the collision in Earth's reference frame is

t=377,29s

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

                          x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                                y'=y

                                z'=z

                          t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

                       x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                           y=y'

                           z=z'

                        t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

First we calculate the expression in the denominator

                            \frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}

                                \sqrt{1-\frac{v^{2}}{c^{2}}} =0,714

then we calculate t

                      t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                      t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}

                      t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}

                      t=\frac{190s+79,388s}{0,714}

finally we get that

                                     t=377,29s

then we calculate x

                         x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

                         x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}

                         x=\frac{3,4x10^{10}m+39872396914m}{0,714}}

                         x=\frac{73872396914m}{0,714}}

                         x=103462740775,91m

finally we get that

                                     x \approx 103,46x10^{9} m

5 0
3 years ago
If each of the charges is increased by two times and the distance between them is also increased by two times, the electromagnet
PIT_PIT [208]

Answer: The force does not change.

Explanation:

The force between two charges q₁ and q₂ is:

F = k*(q₁*q₂)/r^2

where:

k is a constant.

r is the distance between the charges.

Now, if we increase the charge of each particle two times, then the new charges will be: 2*q₁ and 2*q₂.

If we also increase the distance between the charges two times, the new distance will be 2*r

Then the new force between them is:

F = k*(2*q₁*2*q₂)/(2*r)^2 = k*(4*q₁*q₂)/(4*r^2) = (4/4)*k*(q₁*q₂)/r^2 = k*(q₁*q₂)/r^2

This is exactly the same as we had at the beginning, then we can conclude that if we increase each of the charges two times and the distance between the charges two times, the force between the charges does not change.

8 0
3 years ago
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