Rocks within Earth both expand and contract as P waves pass
Explanation:
Rocks within the earth both expands and contracts as P-waves passes through them. P-waves are elastic waves.
- Elastic waves behaves in such a way that they do not cause permanent deformation of rocks.
- They can be said to cause elastic deformation when they travel through rocks.
- They simply temporarily expand and contract the rock within a short period by causing the vibration of particles of the medium.
- After a short while, the rock returns back to its original position as if nothing has happened to it.
- These elastic waves are better called seismic waves.
- P-waves are primary waves that can travel through any medium.
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A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction. The proton will be subject to a magnetic pull that is directed into the page. Option B is correct.
<h3>What is the right-hand thumb rule?</h3>
Hold a current-carrying conductor in your right hand with your thumb pointing in the direction of the current then wrap your fingers around the conductor and orient them in the direction of the magnetic field lines.
A proton travels through a constant magnetic field in the negative y-direction while moving in the negative x-direction.
The proton will be subject to a magnetic pull that is directed into the page.
Hence, option B is correct.
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If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.
If you apply 1 newton to the machine's input, the output force is
(2.5 x 1 newton) = 2.5 newtons.
If you apply 120 newtons to the machine's input, the output force is
(2.5 x 120 newtons) = 300 newtons.
Answer:
mechanical advantage!
Explanation:
The Mechanical advantage of a machine is the factor by which the machine changes the input force.
When a a machine multiplies an input force, that's called a mechanical advantage.
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Defenition of Mechanical Advantage
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Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.
