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il63 [147K]
3 years ago
15

A spring with a force constant of 5.3 n/m has a relaxed length of 2.60 m. when a mass is attached to the end of the spring and a

llowed to come to rest, the vertical length of the shaik (rs53237) – work, energy, power – theisen – (20167) 2 spring is 3.62 m. calculate the elastic potential energy stored in the spring. answer in units of j
Physics
1 answer:
olganol [36]3 years ago
6 0
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant. 
<span> U = (1/2)kx^2 
</span><span> U = (1/2)(5.3)(3.62-2.60)^2 
</span> U = <span> <span>2.75706 </span></span>J
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Ms. Mary Mack walked around her block from her house for 200 meters. She arrived back at her house in 15 minutes. What was her d
inysia [295]

Answer:

d = 0 [m]

Explanation:

Displacement is understood as the length and direction that a body travels to move from an initial point to an endpoint.

This displacement is represented with a vector or straight line that indicates the distance of the displacement and its length.

This displacement in an easier way to understand. It is the distance between the start point and the endpoint of the journey. Since the second point is equal to the first point, since Mary returns to the same place, there is no difference between the displacement.

Therefore the displacement is zero.

5 0
3 years ago
The lower the__ the frequency, the __the pitch <br><br> -lower <br> -higher
rjkz [21]

Answer:

umm the lower the frequency the higher the pitch

Explanation:

6 0
2 years ago
What is an independent or manipulated variable?
Paha777 [63]

Is the variable you change, independent, I, something I change.

5 0
3 years ago
A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th
Stells [14]

Answer:

The relationship between the initial stored energy PE_{i} and the stored energy after the dielectric is inserted PE_{f} is:

c) PE_{f} =0.5\ PE_{i}

Explanation:

A parallel plate capacitor with C_{o} that is connected to a voltage source V_{o} holds a charge of Q_{o} =C_{o} V_{o}. Then we disconnect the voltage source and keep the charge Q_{o} constant . If we insert a dielectric of \kappa=2 between the plates while we keep the charge constant, we found that the potential decreases as:

                                                     V=\frac{V_{o}}{\kappa}

The capacitance is modified as:

                                              C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}

The stored energy without the dielectric is

                                               PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}

The stored energy after the dielectric is inserted is:

                                               PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}

If we replace in the above equation the values of V and C we get that

                                         PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})

                                                   PE_{f} =\frac{PE_{i}}{\kappa}

Finally

                                                  PE_{f} =0.5\ PE_{i}

                                               

                                     

5 0
3 years ago
A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m
Marrrta [24]

Answer:

-1.47 m/s^2

Explanation:

We can use the following SUVAT equation to solve the problem:

v^2 - u^2 = 2ad

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2

4 0
2 years ago
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