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brilliants [131]

3 years ago

13

A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction

less table at 2m/s2, then what is the magnitude of the repulsive force?

1 answer:

makkiz [27]3 years ago

5 0

Answer:

0.1 N

Explanation:

using F = ma yields 0.1 N.

net

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A solar-powered car is traveling at constant speed around a circular track. what happens to the centripetal acceleration of the

anyanavicka [17]

Centripetal acceleration is the acceleration of an object acted upon a force that pulls it toward the center of origin. For constant speed, there is no acceleration, since acceleration is defined as the difference of velocities at an elapsed time. Hence, if speed doubles, then the centripetal acceleration would increase as well.

5 0

3 years ago

You observe a distant galaxy. You find that a spectral line, resulting from an electron transition in hydrogen, is shifted from

alexandr1967 [171]

Answer:

The galaxy is moving away from the observer

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So when we see the light has shifted to the infrarred part of the spectrum, it means the source is traveling away from us, making the light waves we percieve streched and move from visible light to infrarred.

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2 years ago

If we interpret the large, angular rocks to have originated from the outcrop at the top of the hill, we are using __________ rea

IrinaK [193]

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brainly.com/question/2631497

3 0

2 years ago

What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?

iren [92.7K]

Answer:

$\lambda=1090nm$

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

$\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where $R_H$ is the Rydberg constant for hydrogen and $n_1$ , $n_2$ are the lower energy state and the higher energy state, respectively.

$\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm$

4 0

3 years ago

A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper nee

baherus [9]

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095 and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

4 0

3 years ago