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seropon [69]
3 years ago
14

Suppose a spring compresses by 5cm if you put a 250 gram weight on top of it. How

Physics
1 answer:
Leokris [45]3 years ago
8 0

Answer:

Option D. 0.061 J

Explanation:

We'll begin by converting 5 cm to m. This can be obtained as follow:

100 cm = 1 m

Therefore,

5 cm = 5 cm × 1 m / 100 cm

5 cm = 0.05 m

Next, we shall convert 250 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

250 g = 250 × 1 Kg / 1000

250 g = 0.25 Kg

Next, we shall determine the weight of the object. This can be obtained as follow:

Mass (m) = 0.25 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

W = m × g

W = 0.25 × 9.8

W = 2.45 N

Next, we shall determine the spring constant. This can be obtained as follow:

Compression (e) = 0.05 m

Weight = Force (F) = 2.45 N

Spring constant (K) =?

F = Ke

2.45 = K × 0.05

Divide both side by 0.05

K = 2.45 / 0.05

K = 49 N/m

Finally, we shall determine the potential energy contained in the compressed spring. This can be obtained as follow:

Compression (e) = 0.05 m

Spring constant (K) = 49 N/m

Energy (E) =?

E = ½Ke²

E = ½ × 49 × (0.05)²

E = 24.5 × 0.0025

E = 0.061 J

Therefore, the potential energy contained in the compressed spring is 0.061 J

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3 years ago
It takes 50 seconds for Tyler to do 450 Joules of work. How much power did Tyler use?
Tomtit [17]

Answer:

<h2>The answer is 9 W</h2>

Explanation:

The power used by an object can be found by using the formula

p =  \frac{w}{t}  \\

w is the workdone

t is time taken

From the question we have

p =  \frac{450}{50}  =  \frac{45}{5}  \\

We have the final answer as

<h3>9 W</h3>

Hope this helps you

8 0
3 years ago
If spring has a spring constant of 500 N/m and is stretched .50 meters,how much energy is stored in the spring ((show work for f
Free_Kalibri [48]

Answer:

The energy stored is: 62.5 Joules

Explanation:

Given

k = 500N/m --- spring constant

x = 0.5m --- stretch

Required

The amount of energy

This is calculated as:

U = \frac{1}{2} kx^2

U = \frac{1}{2} * 500N/m * (0.5m)^2

U = 250N/m * (0.5m)^2

U = 62.5\ J

7 0
3 years ago
Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a
Helen [10]

Complete Question

Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be added to the gas in the form of heat by applying a flame to the outside of the container. Conversely, energy can also be removed from the gas in the form of heat by immersing the container in ice water. Energy can be added to the system in the form of work by pushing the piston in, thereby compressing the gas. Conversely, if the gas pushes the piston out, thereby pushing some atmosphere aside, the internal energy of the gas is reduced by the amount of work done.

          pV=nRT

so the absolute temperature T is directly proportional to the product of the absolute pressure p and the volume V,Here n denotes the amount of gas moles,which is a constant because the gas is confined and R is the universal constant

What is the \triangle U as the system of ideal gas goes from point A to point B on the graph recall u is proportional to T

Answer:

\triangle T=0

\triangle V=0

The gas A and B have same internal energy

Explanation:

From the question we are told that

Pa=u atm\\Va=1m^3\\Pb=1 atm\\Vb=4m^3

Generally the equation of temperature is mathematically given as

Ta=\frac{Pv}{nR}

Ta=\frac{u*1}{nR}

And

Tb=\frac{PbVb}{nR}

Tb=\frac{u*1}{nR}

Generally the change in temperature \triangle T is mathematically given as

\triangle T=Tb-Ta=Tb=\frac{u*1}{nR}-\frac{u*1}{nR}

\triangle T=0

Generally the change in internal energy \triangle V

\triangle V=nC_v \triangle T\\

\triangle V=0

Therefore with

\triangle T=0

\triangle v=0

The gas A and B have same internal energy

6 0
3 years ago
Encontrar la distancia y desplazamiento de las dos trayectorias si se mueve el móvil Desde A hasta B
Lerok [7]

Answer:

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Explanation:

6 0
3 years ago
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