on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit
2 answers:
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Answer:
<em>a) 105935.7 Pa</em>
<em>b) 103630.35 Pa</em>
Explanation:
The volume of the container = 0.025 m^3
The radius of the container = 13 cm = 0.13 m
We have to find the height of the tank
From the equation for finding the volume of the cylinder,
V =
where
V is the volume of the cylinder
h is the height of the cylinder
substituting values, we have
0.025 = 3.142 x x h
0.025 = 0.0531h
h = 0.025/0.0531 = 0.47 m
Pressure at the bottom of the tank P = ρgh
where
ρ is the density of water = 1000 kg/m^3
g is the acceleration due to gravity = 9.81 m/s^2
h is the depth of water which is equal to the height of the tank
substituting values, we have
P = 1000 x 9.81 x 0.47 = 4610.7 Pa
atmospheric pressure = 101325 Pa
therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>
b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m
pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa
This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>
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:
(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):
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:
The energy density is 17230.6 J/m³
c. The total energy contained in the solenoid is:
(2)
V is the volume of the solenoid and is calculated by assuming the solenoid as a perfect cylinder:
A: cross-sectional area of the solenoid = 0.550 cm^2 = 5.5*10^-5m^2
Then, the energy contained in the solenoid is:
The energy contained is 0.236J
d. The inductance of the solenoid is calculated as follow:
The inductance of the solenoid is 5.84*10^-5 H