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Irina-Kira [14]
3 years ago
11

You compress a spring by a certain distance. then you decide to compress it further so that its elastic potential energy increas

es by another 50%. what was the percent increase in the spring's compression distance?
Physics
1 answer:
Likurg_2 [28]3 years ago
5 0
Elastic potential energy = 1/2 k * change of x^2
k- coefficient
x - change in length.
to increase energy 1.5 times you have to change x (compress) into \sqrt{1.5}  times (it's abot 1.22 or 22%)
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Two carts connected by a 0.05 m spring hit a wall, compressing the spring to 0.02 m. The spring constant k is
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Answer:

0.045 J

Explanation:

From the question,

The elastic potential energy stored in a spring is given as,

E = 1/2ke²...................... Equation 1

Where E = elastic potential energy, k = spring constant, e = compression.

Given: k = 100 N/m, e = 0.05-0.02 = 0.03 m

Substitute these values into equation 1

E = 1/2(100)(0.03²)

E = 50(9×10⁻⁴)

E = 0.045 J

Hence the right option is 0.045 J

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Which moment corresponds to the maximum potential energy of the system?
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A solenoid 25.0 cmcm long and with a cross-sectional area of 0.550 cm^2 contains 460 turns of wire and carries a current of 90.0
ankoles [38]

Answer:

a.  B = 0.20T

b.  u = 17230.6 J/m³

c.  E = 0.236J

d.  L = 5.84*10^-5 H

Explanation:

a. In order to calculate the magnetic field in the solenoid you use the following formula:

B=\frac{\mu_o n i}{L}               (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

n: turns of the solenoid = 460

L: length of the solenoid = 25.0cm = 0.25m

i: current  = 90.0A

You replace the values of the parameters in the equation (1):

B=\frac{(4\pi*10^{-7}T/A)(460)(90.0A)}{0.25m}=0.20T

The magnetic field in the solenoid is 0.20T

b. The magnetic permeability of air is approximately equal to the magnetic permeability of vacuum. To calculate the energy density in the solenoid you use:

u=\frac{B^2}{2\mu_o}=\frac{(0.20T)^2}{2(4\pi*10^{-7}T/A)}=17230.6\frac{J}{m^3}

The energy density is 17230.6 J/m³

c. The total energy contained in the solenoid is:

E=uV           (2)

V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:

V=AL

A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2

V=(5.5*10^{-5}m^2)(0.25m)=1.375*10^{-5}m^3

Then, the energy contained in the solenoid is:

E=(17230.6J/m^3)(1.375*10^{-5}m^3)=0.236J

The energy contained is 0.236J

d. The inductance of the solenoid is calculated as follow:

L=\frac{\mu_o N^2 A}{L}=\frac{(4\pi*10^{-7}T/A)(460)^2(5.5*10^{-5}m^2)}{0.25m}\\\\L=5.84*10^{-5}H

The inductance of the solenoid is 5.84*10^-5 H

3 0
3 years ago
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