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3 years ago

An element with only neutrons?

Physics

1 answer:

nirvana33 [79]3 years ago

3 0

All except for one tom of the element with different number of neutron

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2 years ago

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Is the amplitude just 2? Or do I combine all of them and do 6?

oksian1 [2.3K]

Answer:

combine them all

6 m

Explanation:

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2 years ago

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Roman55 [17]

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3 years ago

A car’s bumper is designed to withstand a 4.0-km/h (1.1-m/s) collision with an immovable object without damage to the body of th

Alex Ar [27]

Answer:

avriage force F = 2722.5 N

Explanation:

For this problem we can use Newton's second law, to calculate the average force and acceleration we can find it by kinematics.

vf² = v₀² - 2 ax

The final carriage speed is zero (vf = 0)

0 = v₀² - 2ax

a = v₀² / 2x

a = 1.1²/(2 0.200)

a = 3.025 m / s²

a = 3.0 m/s²

We calculate the average force

F = ma

F = 900 3,025

F = 2722.5 N

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