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jasenka [17]
3 years ago
5

Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i

n newtons) that holds the spring stretched at the same distance (11 cm)
Physics
1 answer:
Sidana [21]3 years ago
5 0
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where 
xn is the natural length
x₁ is the stretched length 
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
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Answer:

yes because physics want that

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2 years ago
Ballon volume of 3200ml of xenon gas is at a gauge pressure of 122kPa and a temperature of 27c. What is the volume when the ball
N76 [4]

Given:

The initial volume of the gas, V₁=3200 ml=3.2×10⁻³ m³

The initial pressure of the gas, P₁=122 kPa

The initial temperature of the gas, T₁=27 °C=300 K

The final temperature, T₂=65 °C=338 K

The final pressure, P₂=112 kPa

To find:

The final volume of xenon gas.

Explanation:

From the combined gas law,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

Where V₂ is the volume after it is heated.

On rearranging the above equation,

V_2=\frac{T_2P_1V_1}{T_1P_2}

On substituting the known values,

\begin{gathered} V_2=\frac{338\times112\times10^3\times3.2\times10^{-3}}{300\times112\times10^3} \\ =3.61\text{ m}^3 \end{gathered}

Final answer:

The volume of the balloon when it is heated is 3.61 m³

4 0
1 year ago
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Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

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The single interface equation is \frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}

Substituting the quantities given in the problem,

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The image distance is then y^i=-\frac{18}{4}in =-4.5in

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x = Distance between center of mass of the objects.

From this equation we can observe that the Force is inversely proportional to the squared distance between the two objects. The greater the distance, the lower the force of gravity and vice versa.

F \propto \frac{1}{R^2}

If you want to increase the force of gravity, you need to reduce the distance of the two. Therefore the correct option is B. Talk to the distance between them.

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Efficincy of simple machine is 60% what does it mean?<br>​
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