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erica [24]
2 years ago
14

A student jumps off a sled toward the West after it stops at the bottom of an icy hill. Based on the law of action-reaction, in

what direction will the sled move as the student jumps off?
East

North

South

West
Physics
1 answer:
Aneli [31]2 years ago
5 0
We know the newton 3rd law as action and reaction. 
Mathmatically= F(action) -F(reaction)
this law states that every action is thier own reaction which is equal in magnitude but opposite in direction.
so the answer is equal to East..


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Fnety = (FT)(sin 32°) – Fg Or the answer B, I checked it.
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The speed of a bus increases uniformly from 15 ms per second to 60 ms per second in 20 seconds. calculate 1. the average speed 2
joja [24]
Vf= 60m/s
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t=20s
a = vf-vs/t
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The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c
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64.59kpa

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6 0
2 years ago
Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^
ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

T \approx e^{-2CL}

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)

Next, you divide the equation (3) into (4), and finally, you solve for L':

\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0
2 years ago
The drag force F on a boat varies jointly with the wet surface area A of the boat and the square of the speed s of the boat. A b
Advocard [28]

Answer:

Wet surfaces areaA=+25.3ft^2

Explanation:

Using F= K×A× S^2

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K = F/AS^2=996/(83×20^2)

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7 0
2 years ago
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