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NNADVOKAT [17]
3 years ago
6

A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 450 V. 1) If an electron leaves the negative plate

, starting from rest, how fast is it going when it hits the positive plate

Physics
1 answer:
Bumek [7]3 years ago
8 0

Answer:

Explanation:

this is the answer to your question

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An object with kinetic energy k explodes into two pieces, each of which moves with twice the speed of the original object.
zlopas [31]
<span>Assuming that the momenta of the two pieces are equal: when they have equal velocities, then the masses of the two pieces are also equal. Since there is no force from outside of the system, the center of mass moves on with the same velocity as before the equation. So the two pieces must fly at the side side of the mass center, i.e., they must always be at 90° to the side of the mass center. Otherwise it would not be the mass center, respectively the pieces would not have equal velocities. This is only possible, when the angle of their velocity with the initial direction is 60°. Because, cos (60°) = 1/2 = v/(2v).</span>
6 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

t=11.2\,\mathrm s

At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0
3 years ago
The image below shows four points on the orbit of a comet around the Sun.
Firlakuza [10]

Answer:

<u>Point </u><u>A</u>

Explanation:

And the strongest force of gravity is point C

3 0
3 years ago
How can we protect space shuttles or astronauts from space radiation in the absence of the atmospheric layer?
Arisa [49]
Earth's protective magnetic bubble, called the magnetosphere, deflects most solar particles, but in the absence of atmospheric layer, Polyethylene is a good shielding material because it has high hydrogen content, and hydrogen atoms are good at absorbing and dispersing radiation.

The Earth’s atmospheres are kept in place by gravity. The air near the ground is pulled on by gravity and compressed by the air higher in the sky. This causes the air near the ground to be denser and creating different layers with different qualities in which are the atmosphere.

Hope that helps ^^
7 0
3 years ago
on an unknown temperature scale, the freezing point of water is -15°U and the boiling point is +60°U. develop a linear conversio
shutvik [7]

Answer:

Since this is a linear equation

y = m x + b     or

U = m F + b     is a linear equation

when ΔF = (212 - 32) = 180

and ΔU = (60 - (-15)) = 75

m = 75 / 180 = 2.4 if converting F to U and a = .417

U = .417 F + b

If F = 32 then U = -15 and

-15 = .417 * 32 + b

b = -15 - 13.3 = -28.3 and our equation becomes

U = .417 F - 28.3

Check: let F = 212

U = .417 * 212 - 28.3 = 60          as it should

5 0
1 year ago
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