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Aleksandr [31]
3 years ago
15

A spring with k = 500 N/m stores 704 J. How far is it extended from the equilibrium position

Physics
1 answer:
kaheart [24]3 years ago
4 0

Given that:

k = 500 n/m,

work (W) = 704 J

spring extension (x) = ?

         we know that,

                      Work = (1/2) k x²

                          704 = (1/2) × 500 × x²

                            x = 1.67 m

A spring stretched for 1.67 m distance.

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The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its
Scilla [17]
<h2>Power of cheetah is 5576.85 W = 7.48 hp</h2>

Explanation:

Power is the ratio of energy to time.

Here we need to consider kinetic energy,

Mass, m = 102 kg

Initial velocity = 0 m/s

Final velocity = 16.2 m/s

Time, t = 2.4 s

Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J

Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J

Change in energy = Final kinetic energy - Initial kinetic energy

Change in energy = 13384.44 - 0

Change in energy = 13384.44 J

Power = 13384.44  ÷ 2.4 = 5576.85 W = 7.48 hp

Power of cheetah is 5576.85 W = 7.48 hp

7 0
3 years ago
What happens to the electrostatic force between two charged particles if the distance between them is doubled?
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According to Coulomb's Law , The size of the force varies inversely as the square of the distance between the two charges. So ,if the distance between the two charges is doubled, the  electrostatic force will become weak by one fourth of the original force.

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Type the correct answer in the box. Round your answer to the nearest whole number. Calculate the man’s mass. (Use PE = m × g × h
Blababa [14]

Answer:

56 kg

Explanation:

The change in potential energy of the man is given by:

\Delta U = mg \Delta h

where

m is the man's mass

g is the gravitational acceleration

\Delta h is the change in height of the man

In this problem, we have:

\Delta U=4620 J is the gain in potential energy

g = 9.8 m/s^2 is the gravitational acceleration

\Delta h=8.4 m is the change in height

Re-arranging the equation and substituting the numbers, we find the mass:

m=\frac{\Delta U}{g\Delta h}=\frac{4620 J}{(9.8 m/s^2)(8.4 m)}=56 kg

6 0
3 years ago
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Identify two structural characteristics specific to muscle tissue.
Semmy [17]

Answer:

Explanation:

Muscle cells are excitable; they respond to a stimulus.meaning they can shorten and generate a pulling force. When attached between two movable objects, such as two bones, contraction of the muscles cause the bones to move.It contains protein fibers which contract to make the cell shorter.

7 0
3 years ago
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What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire
Marianna [84]

Answer:

The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

When two point are at same height so ,

P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

Q=A_{1}v_{1}

v_{1}=\dfrac{Q}{A_{1}}

v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}

v_{1}=56.58\ m/s

For output velocity,

v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}

v_{2}=6.28\ m/s

Put the value into the formula

P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)

\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)

\Delta P=1.58\times10^{6}\ N/m^2

(b). We need to calculate the maximum height

Using formula of height

\Delta P=\rho g h

Put the value into the formula

1.58\times10^{6}=1000\times9.8\times h

h=\dfrac{1.58\times10^{6}}{1000\times9.8}

h=161.22\ m

Hence, The pressure and maximum height are 1.58\times10^{6}\ N/m^2 and 161.22 m respectively.

8 0
3 years ago
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