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pochemuha
2 years ago
10

A block oscillating on a spring has an amplitude of 20 cmcm. Part A Part complete What will the block's amplitude be if its tota

l energy is doubled
Physics
1 answer:
Gnom [1K]2 years ago
7 0

Hi there!

Recall that the amplitude of an oscillating system occurs at a point of maximum potential energy.

For a spring system, potential energy is given as:
U = \frac{1}{2}kx^2

U = Potential Energy (J)

k = Spring constant (N/m)
x = amplitude (m)

We can rearrange the equation to solve for amplitude more easily.

U = \frac{1}{2}kx^2\\\\2U = kx^2\\\\x^2 = \frac{2U}{k}\\\\x = \sqrt{\frac{2U}{k}}

If we double 'U':
x' = \sqrt{\frac{2(2U)}{k}} = \sqrt{2}*\sqrt{\frac{2U}{k}}\\\\x' = \sqrt{2} * x

Thus, the new amplitude would be √2 times greater, so:
20 cm \cdot \sqrt{2} = \boxed{28.284 cm}

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Answer:

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Explanation:

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7 0
3 years ago
The tires on your truck have 0.35 m radius. In a straight line, you drive 2600 m. What is the angular displacement of the tire,
Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
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2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
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3 years ago
Newtons first law 1 to 5. <br>What is each of the net force for all of the 5 questions? ​
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Answer:

1. 65 N.

2. 160 N.

3. 0 N.

4. 210 N.

5. 90 N.

Explanation:

1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

Thus, the net force is 65 N

2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 35 + 125

Fₙ = 160 N

Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

Fₙ = 90 N

Thus, the net force is 90 N

8 0
3 years ago
9. (02.04 LC)
ycow [4]

Answer:

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7 0
3 years ago
An electric iron connected to a 110 V source draws 9 A of current. How much heat (in joules) does it generate in a minute?
meriva
Q = 110V x 9A x 60s = J
7 0
3 years ago
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