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Mars2501 [29]
3 years ago
8

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

1.2 kg is attached to the free end of the spring. The glider is pulled toward the right along a frictionless air track, and then released. Now the glider is moving in simple harmonic motion with amplitude 0.045 m. The motion is horizontal (one-dimensional).
Suddenly, Slimer holding an apple flies in and approaches the glider. Slimer drops the apple vertically onto the glider from a very small height. The apple sticks to the glider. The mass of the apple is 0.18 kg.
Recall that the total mechanical energy is E = 1/2 mv^2 + 1/2 kx^2 = 1/2 kA^2 = constant
(a) Calculate the new amplitude of the motion of the glider with apple if the apple is dropped at the moment when the glider passes through its equilibrium position, x = 0 m.
Hints: The total energy of the glider just before the collision is E = 1/2mglider v^2 = 1/2KA^2
The apple sticks to the glider in a completely inelastic collision. The glider is now moving with the apple but at a lower speed. The linear momentum is conserved. Write a corresponding equation.
Also, assume that the collision is very short, so just before the collision the glider is at x = 0 m, and just after the collision the glider and apple are still at x = 0 m. Therefore, the total energy of the glider just after the collision is Enew = 1/2mglider + applev^2new = 1/2kA^2new
(b) Calculate the period of the motion of the glider and the period of the motion of the glider with apple.
Hint: it's a very simple question.

Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

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ruslelena [56]
How does the cat walk-
4 0
3 years ago
A 55 kg person falling with a velocity of 0.6
siniylev [52]

Answer:

What are we supposed to find, if it is kinetic energy then this is the solution.

K.E=1/2mv^2

K.E= kinetic energy

M=mass

V=velocity

K.E =0.5*55*0.6^2

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Explanation:

3 0
2 years ago
A proton moving at 8.00 106 m/s through a magnetic field of magnitude 1.72 T experiences a magnetic force of magnitude 7.20 10-1
gladu [14]

Answer:

19.1 deg

Explanation:

v = speed of the proton = 8 x 10⁶ m/s

B = magnitude of the magnetic field = 1.72 T

q = magnitude of charge on the proton = 1.6 x 10⁻¹⁹ C

F = magnitude of magnetic force on the proton = 7.20 x 10⁻¹³ N

θ = Angle between proton's velocity and magnetic field

magnitude of magnetic force on the proton is given as

F = q v B Sinθ

7.20 x 10⁻¹³ = (1.6 x 10⁻¹⁹) (8 x 10⁶) (1.72) Sinθ

Sinθ = 0.327

θ = 19.1 deg

4 0
3 years ago
Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.
ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0
2 years ago
Pleaseeeee HELPPPP THIS IS TIMED ALSO,
TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)

Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.

8 0
2 years ago
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