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jekas [21]
2 years ago
10

A 1.90-kg mass vibrating up and down on the end of a vertical spring has a maximum speed of 2.30 m/s. What is the total potentia

l energy of the mass on the spring when the mass is at either endpoint of its motion? NOTE: Assume that the potential energy of the mass on the vertical spring is zero when the mass is at the midpoint of its motion. Give your answer in joules.
Physics
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

The answer to the question is;

The total potential energy of the mass on the spring when the mass is at either endpoint of its motion is 5.0255 Joules.

Explanation:

To answer the question, we note that the maximum speed is 2.30 m/s and the mass is 1.90 kg

Therefore the maximum kinetic energy of motion is given by

Kinetic Energy, KE = \frac{1}{2} mv^{2}

Where,

m = Attached vibrating mass = 1.90 kg

v = velocity of the string = 2.3 m/s

Therefore Kinetic Energy, KE = \frac{1}{2}×1.9×2.3² = 5.0255 J

From the law of conservation of energy, we have the kinetic energy, during the cause of the vibration is converted to potential energy when the mass is at either endpoint of its motion

Therefore Potential Energy PE at end point = Kinetic Energy, KE at the middle of the motion

That is the total potential energy of the mass on the spring when the mass is at either endpoint of its motion is equal to the maximum kinetic energy.

Total PE = Maximum KE = 5.0255 J.

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The 5.5 million vinyl long-playing (LP) records sold in the United States per year pales in comparison with the 1.26 billion dig
miv72 [106K]

Answer:

decline

Explanation:

Based on the scenario being described within the question it can be said that these types of firms are in the decline stage of the product life cycle. This stage refers to when a product has already passed it's peak potential and sales begin to decline until production is ultimately halted and the product dies off. Which is exactly what is happening to the LP's since everyone has moved on to digital downloads.

4 0
3 years ago
Jupiter's moon Io has a radius of 1.82 ✕ 106 m and a measured gravitational field of 1.80 m/s2. What is its mass (in kg)?
AveGali [126]

Answer:

8.94*10^22 kg

Explanation:

Given that

Mass of Lo, M = ?

Radius of Lo, r = 1.82*10^6 m

Acceleration on Lo, g = 1.80 m/s²

Gravitational constant, G = 6.67*10^-11

Using the formula

g = GM/r²

Solution is attached below

Answer is 8.94*10^22 kg

7 0
3 years ago
Name three methods that can be used to change the overall density of an object.
Gnesinka [82]
<span> changing its shape
changing its mass
or changing its volume</span>
6 0
3 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
Waves combine to produce a smaller or zero-amplitude wave in a process called
vampirchik [111]
ANSWER: Destructive Interference

-This is the exact definition of the question you have provided, look this term up if you do not believe lol.

Hope this Helps!
6 0
3 years ago
Read 2 more answers
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