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pogonyaev
2 years ago
13

Derive an equation that relates the initial release height hxhx of block xx and the speed vsvs of the two-block system after the

collision in terms of mxmx, mymy, and fundamental constants, as appropriate.
Physics
1 answer:
erik [133]2 years ago
3 0

Answer:

1 point is earned for stating that the conservation of energy should be applied to this situation.

1 point is earned for stating that the conservation of momentum should be applied to this situation.

Example Response:

Students will need to use both conservation of momentum (for the collision) and conservation of energy (for the slide down the ramp) to be able to determine the relationship between the release height of block X and the speed at which the two-block system travels after they collide and stick together.

Explanation:

Please say thank you.

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The result of a wave generator traveling faster than the speed of a wave is:
Vesnalui [34]
The result of a wave generator traveling faster than the speed of a wave is called as a boom. If the wave is a sound wave, it is called a sonic boom. However, if the wave is light, it is called as a luminal boom.  Luminal bloom happens in some industries and is commonly called as the Cherenkov radiation.
4 0
3 years ago
Read 2 more answers
A 4.4 kg mess kit sliding on a frictionless surface explodes into two 2.2 kg parts, one moving at 2.9 m/s, due north, and the ot
sp2606 [1]

Answer:

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

Explanation:

Let north represent positive y axis and east represent positive x axis.

Here momentum is conserved.

Let the initial velocity be v.

Initial momentum = 4.4 x v = 4.4v

Velocity of 2.2 kg moving at 2.9 m/s, due north = 2.9 j m/s

Velocity of 2.2 kg moving at 6.8 m/s, 35° north of east = 6.9 ( cos 35i + sin35 j ) = 5.62 i + 3.96 j m/s

Final momentum = 2.2 x 2.9 j + 2.2 x (5.62 i + 3.96 j) = 12.364 i + 15.092 j kgm/s

We have

         Initial momentum = Final momentum

         4.4v = 12.364 i + 15.092 j

         v =2.81 i + 3.43 j

Magnitude

        v=\sqrt{2.81^2+3.43^2}=4.43m/s

Direction

       \theta =tan^{-1}\left ( \frac{3.43}{2.81}\right )=50.67^0

       50.67° north of east.

Original speed of the mess kit = 4.43 m/s at 50.67° north of east.

6 0
3 years ago
If mechanical energy is conserved, then a pendulum that has a potential energy of 20 J at its highest point and 0.5 J at its low
liraira [26]
At the highest point: kinetic energy is 0 due to the speed  is 0

So the total mechanical energy is 20

Assume no frictions present, then the mechanical energy is conserved

So at the lowest point, kinetic energy = mechanical energy - potential energy

Answer will be 20 - 0.5 = 19.5 J
4 0
3 years ago
The graph represents the reaction 2H2 + 02 32H20 as it reaches
Alex

Answer:

C and D

Explanation:

5 0
3 years ago
Read 2 more answers
Please help I literally don’t understand
Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0
3 years ago
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