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serg [7]
4 years ago
8

When a mass is placed on a spring with a spring constant of 15 newtons per meter, the spring is compressed 0.25 meter. How much

elastic potential energy is stored in the spring?
Physics
1 answer:
Andre45 [30]4 years ago
7 0

Answer:

0.47 J

Explanation:

The elastic potential energy of a spring is given as,

E = 1/2ke²........................ Equation 1

Where E = Elastic potential energy, k = spring constant, e = extension/compression.

Given: k = 15 N/m, e = 0.25 m.

Substitute into equation 1.

E = 1/2(15)(0.25)²

E = 0.46875

E ≈ 0.47 J.

Hence the elastic potential energy stored in the spring = 0.47 J

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B). the atom is made up of smaller particles

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A 0.280-kg volleyball approaches a player horizontally with a speed of 15.0 m/s. The player strikes the ball with her fist and c
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Answer:

Impulse = 10.36 kg m/s

average force = 172.667 N

Explanation:

given data

mass = 0.280 kg

speed = 15.0 m/s

speed = 22.0 m/s

to find out

impulse and magnitude of the average force

solution

we know that Impulse is change in momentum that is

initial momentum = mass × speed    ..........1

initial momentum = 0.28 × (15)

initial momentum = 4.2 kg m/s

Final momentum = mass × speed     ..........2

Final momentum = 0.28 × (-22)

Final momentum = -6.16 kg m/s

so now we get Impulse that is

Impulse = 4.2 - (-6.16)

Impulse = 10.36 kg m/s

and

average force will be

average force = impulse ÷ time

average force = \frac{10.36}{0.060}

average force = 172.667 N

6 0
3 years ago
Noble gases, such as argon and neon, are known for being extremely non-reactive. Neon and argon are non-reactive because they A.
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How long does it take an airplane to fly 1200 miles if it maintains a speed of 300 miles per hour
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Answer:

The answer is

<h2>4 hrs</h2>

Explanation:

To find the time taken we use the formula

t =  \frac{d}{v}  \\

where

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v is the velocity

t is the time

From the question

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v = 300 mi/hr

We have

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5 0
3 years ago
A positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 6.5 μg is floating a dista
fenix001 [56]

Answer:

Q₂ = (9.83 × 10⁻⁹) C = +9.83 nC

Explanation:

The force of attraction/repulsion between the two charges is equal to the force of gravity on the charge 2.

The force of gravity on the charge two = mg

= 6.5 μg × 9.8 m/s² = (6.37 × 10⁻⁵) N

Since the gravity force is directed downwards, the force on charge 2 due to charge 1 has to be directed upwards, hence it is a force of repulsion.

The magnitude of the force between the two charges is given according to the Coulomb's law.

F = kQ₁Q₂/r²

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Q₂ = ?

r = 25 cm = 0.25 m

F = (6.37 × 10⁻⁵) N

(6.37 × 10⁻⁵) = [(9.0 × 10⁹) × (45 × 10⁻⁹) × Q₂] ÷ 0.25²

Q₂ = 0.00000000983 C = (9.83 × 10⁻⁹) C = 9.83 nC

Hope this Helps!!!

7 0
4 years ago
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