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3 years ago

Use the figure below to answe

Physics

1 answer:

vovikov84 [41]3 years ago

7 0

Answer:

B. Q

A. P

Explanation:

I assume the figure is the one included in my answer.

The pivot (also called the fulcrum) is point Q.

The effort needed to keep the bar balanced is force P.

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______ can occur when water-saturated soil turns from a solid to a liquid as a result of an earthquake.

jek_recluse [69]

The answer is liquefaction

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2 years ago

What does a lunar eclipse and a solar eclipse have in common

liraira [26]

During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".

3 0

3 years ago

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

A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?

Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

$W=(mg)h$