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lord [1]
2 years ago
11

A small charge q is placed near a large spherical charge Q. The force experienced by both charges is F. The electric eld created

by Q at the position of q is:
Physics
1 answer:
anastassius [24]2 years ago
7 0

The electric field created by Q at the position of q is \frac{F}{Q}.

The given parameters:

  • <em>Magnitude of charge, = q</em>
  • <em>Spherical charge, = Q</em>
  • <em>Force experienced by both charges, =  F</em>

The electric field created by Q at the position of q is calculated as follows;

E = \frac{F}{Q} \\\\

where;

  • <em>E is the magnitude of electric field strength </em>
  • <em>F is the force experienced by both charges</em>
  • <em>Q is the charge</em>

Thus, the electric field created by Q at the position of q is \frac{F}{Q}.

Learn more about electric field here: brainly.com/question/14372078

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d=vi*t+(1/2)gt²

d=11 m
g=9.8 m/s²
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11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
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t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s

Answer: the time the sone is in flight is 1.5 s
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A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts
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A small boy is playing with a ball on a stationary train. If he places the ball on the floor of the train, when the train starts moving the ball moves toward the back of the train. This happened due to inertia

An object at rest remains at rest, or if in motion, remains in motion unless a net external force acts on it .  

When a train starts moving forward, the ball placed on the floor  tends to fall backward is an example of inertia of rest. Due to the reason that the lower part of the ball is in contact with the surface and rest of the part is not . As the train starts moving, its lower part gets the motion as the floor starts moving but the upper part will remain as it is as it is not in contact with the floor , hence do not attain any motion due to the inertia of rest simultaneously i.e. it tends to remain at the same place.

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8 0
1 year ago
Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren
nikklg [1K]

Answer:

4.77\ \text{A}

Explanation:

F = Magnetic force = 4.11 N

I_n = Net current

I_2 = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

\theta = Angle between current and magnetic field = 65^{\circ}

l = Length of wires = 2.64 m

I = Current in the other wire

Magnetic force is given by

F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}

Net current is given by

I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}

The current I is 4.77\ \text{A}.

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