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WINSTONCH [101]
3 years ago
10

a charge of 6.4x10^-7 C experiences an electric force of 1.8x10^-1 N what is the electric field strength

Physics
1 answer:
Andreyy893 years ago
5 0
Field strength = force ÷ charge (E=F/Q)

E= 1.8×10^-1 ÷ 6.4×10^-7... to obtain ur answer

the formula is derived from the definition
Electric field : A region in space where a charged body experiences a force.
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• 2. A 70 kg man on a 100 kg boat throws a ball. The boat moves backwards 5 meters in 10 seconds. What is
Kipish [7]

Answer:

0.5 m/sec

Explanation:

v=S÷t

 =5÷10

v=0.5 m/sec

6 0
3 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

F= ma

a=\dfrac{F}{m}

Put the value of F

a=\dfrac{kq^2}{mr^2}

Put the value into the formula

414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}

q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}

q^2=1.84\times10^{-14}

q=0.135\times10^{-6}\ C

q=0.135\ \mu C

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0
3 years ago
Read 2 more answers
What of the following does NOT influence resistance? *
Marina86 [1]

Answer:

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Explanation:

4 0
3 years ago
A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold
lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

       fr - Wₓ = m    a                      (1)

Y axis  

       N- W_{y} = 0

       N = W_{y}

let's use trigonometry to find the components of the weight

        sin θ = Wₓ / W

        cos θ = W_{y} / W

        Wₓ = W sin θ

        W_{y} = W cos θ

the friction force has the formula

         fr = μ N

         fr = μ Wy

         fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

           F = fr

           F = (μ  mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0
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