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adoni [48]
3 years ago
13

A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

how much elastic potential energy is stored in the spring? 700 J 1050 J 1575 J 4375 J
Physics
2 answers:
Vlad [161]3 years ago
6 0

Answer:

1575 J

Explanation:

The elastic potential energy of a string is given by:

E=\frac{1}{2}k \Delta x^2

where

k is the spring constant

\Delta x is the stretching/compression of the string with respect to its natural length

For the spring in the problem, we have:

- Spring constant: k=1400 N/m

- Stretching: \Delta x=2.5 m -1.0 m=1.5 m

Therefore, the elastic potential energy stored in the spring is

E=\frac{1}{2}(1400 N/m)(1.5 m)^2=1575 J

Elena L [17]3 years ago
5 0

The answer is C. 1575 J. I just took this review.

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Answer:

Answers A and D are the correct solution.

Explanation:

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2 years ago
A small 20-kg canoe is floating downriver at a speed of 2 m/s. What is the canoe’s kinetic energy? A. 40 J B. 80 J C. 18 J
rusak2 [61]

A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.

Answer: Option A

<u>Explanation:</u>

The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,

Given that mass of the canoe = 20 kg and its speed =1 m/s

As we know that the Kinetic energy has the formula,

\text {Kinetic energy}=\frac{1}{2} \boldsymbol{m} \boldsymbol{v}^{2}

Therefore, substituting the value into the equation, we get,  

K . E .=\frac{1}{2} \times 20 \times 2^{2} = 40 J

4 0
3 years ago
Read 2 more answers
The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n
aleksandr82 [10.1K]
You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.

T - F = ma
T = ma + F
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T = 355.06 N

5 0
3 years ago
In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin
Natasha2012 [34]

Answer:

y = 0.0233 m

Explanation:

In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:

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λ = wavelength of light

L = Distance between screen and slits

d = Slit Separation

Now, for initial case:

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8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d

L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)

L/d = 2.08 x 10⁴

using this for λ = 560 nm = 5.6 x 10⁻⁷ m:

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3 0
3 years ago
What part of the electromagnetic spectrum can be seen by humans
Amanda [17]

Answer:

Visible light

Explanation:

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All these waves are invisible to human eye, except for the part referred as 'visible light'. The electromagnetic waves of this part of the spectrum are visible to human eye, and they appear as a different color depending on their wavelength. In particular, we have:

Violet: 380-450 nm

Blue: 450-495 nm

Green: 495-570 nm

Yellow: 570-590 nm

Orange: 590-620 nm

Red: 620-750 nm

4 0
3 years ago
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