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Bingel [31]
1 year ago
10

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi

th a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are
Physics
1 answer:
Lelu [443]1 year ago
4 0

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

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1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin
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Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

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3 years ago
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Which division of time on the geologic time scale is the shortest?
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Explanation:

In geology, there are 5 divisions of time on the time scale. They are:

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Eons are the largest time period, while ages are the shortest time period. The rest of the above listed are in between the two.

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The capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separ
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Answer:

U₁ = (ϵAV²)/6d

This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

Explanation:

The energy stored in a capacitor is given by (1/2) (CV²)

Energy in the capacitor initially

U = CV²/2

V = voltage across the plates of the capacitor

C = capacitance of the capacitor

But the capacitance of a capacitor depends on the geometry of the capacitor is given by

C = ϵA/d

ϵ = Absolute permissivity of the dielectric material

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d = separation between the capacitor

So,

U = CV²/2

Substituting for C

U = ϵAV²/2d

Now, for U₁, the new distance between plates, d₁ = 3d

U₁ = ϵAV²/2d₁

U₁ = ϵAV²/(2(3d))

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This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.

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As a roller coaster car crosses the top of a 50-m-diameter loop-the-loop, its apparent weight is the same as its true weight.
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R=W\quad \quad [\text{apparent weight =Actual weight}]

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