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Natalka [10]
3 years ago
12

What two things surround every moving charge?

Physics
1 answer:
Marina86 [1]3 years ago
3 0
A magnetic field and an electric field surround every moving electric charge.

Hope this helps. - M
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The most likely answer will be B. Overfishing is still a large issue today, and it can really ruin a habitat. =)
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When a circuit is arranged in parallel
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Answer:

A. There are multiple paths that electrons can take through the circuit, and it is possible for the electron to pass through one circuit component but not another.

Explanation:

Parallel arrangement of components in an electric circuit puts different parts of the circuit on different branches. In a parallel connection, there are multiple paths for the electrons to take, and it is possible for electrons to pass through on circuit component without going through another. This is the reason why If there is a break in one branch of the circuit, electrons can still flow in other branches, and the same reason why one bulb going off in your home does prevent the other components in your home from coming on (your home is wired in a parallel electric circuit).

5 0
3 years ago
1. A small block of mass m slides in a vertical circle of radius R on the inside of a circular track. There is no friction betwe
Nookie1986 [14]

Answer:

a) v_{b} = \sqrt{\frac{R}{m}( n_{b} - mg)}

b) v_{t} = \sqrt{\frac{R}{m}( n_{b} - mg) - 4gR}

c) N_{t} = = n_{b} - 6mg

Explanation:

NOTE: No values are provided in this exercise, any values of the given parameters can be substituted into the answers provided

Let the mass of the block = m

Radius  = R

Normal force acting on the block, N =  n_{b}

a) Calculate the velocity of the block at the bottom of its path, v_{b}

According to Newton's second law of motion,

\sum F = ma\\\sum F = N - mg

N - mg = ma

Since the block clides in a vertical circle, the acceleration, a, is a centripetal acceleration

a = \frac{v_{b} ^{2} }{R}

n_{b} - mg = \frac{mv_{b} ^{2} }{R}

\frac{R}{m}( n_{b} - mg) = v_{b} ^{2} \\ v_{b} = \sqrt{\frac{R}{m}( n_{b} - mg)}

b) Use conservation of energy to find the velocity of the block at the top of its path.

According to the law of conservation of energy

K_{b} = K_{t} + PE_{t}.................(1)

PE_{t} = mgh

The height of the block's path at the top, h = twice the radius

h = 2R

PE_{t} = mg(2R)\\PE_{t} = 2mgR

From equation (1)

\frac{1}{2} mv_{b} ^{2} = \frac{1}{2} mv_{t} ^{2} + 2mgR\\\frac{1}{2} v_{b} ^{2} = \frac{1}{2} v_{t} ^{2} + 2gR\\\frac{1}{2} v_{b} ^{2} -  2gR= \frac{1}{2} v_{t} ^{2}

Substituting v_{b} into the equation above

\frac{1}{2} (\sqrt{\frac{R}{m}( n_{b} - mg)})^{2} - 2gR = \frac{1}{2} v_{t} ^{2} \\\frac{R}{m}( n_{b} - mg) - 4gR = v_{t} ^{2}\\v_{t} = \sqrt{\frac{R}{m}( n_{b} - mg) - 4gR}

c) Find the normal force that the track exerts on the block at the top of its path.

From the Newton's law of motion applied in part a

N_{t} + mg = ma

a = \frac{v_{t} ^{2} }{R}

N_{t} = \frac{mv_{t}^{2}} {R} - mg

N_{t} = \frac{m(\frac{R}{m}( n_{b} - mg) - 4gR) }{R} - mg\\N_{t} = (n_{b} - mg) - 4mg - mg\\N_{t} = = n_{b} - 6mg

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3 years ago
An object accelerates at 3 meters per second when a 10-newton (N) force is applied to it. Which force would cause this object to
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Answer: Net force

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The masses of two metal blocks are measured. Block A has a mass of 8.45 g and Block B has a mass of 45.87 g. How many significan
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3 sig figs total mass is 54.32g

Explanation:

It is 3 sig figs because that is the accuracy that Block A was measured to

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