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scoundrel [369]
3 years ago
5

Calculate the force exerted by a mental ball having a mass of 70kg moving with speed of 20m/s>2

Physics
1 answer:
lord [1]3 years ago
8 0

Answer:

F = 1400 N

Explanation:

It is given that,

Mass of the ball, m = 70 kg

It is moving with an acceleration of 20 m/s². We need to find the force exerted by the ball.

Force is given by the product of mass and acceleration. So,

F = ma

F=70\ kg\times \ 20m/s^2\\\\F=1400\ N

So, the force of 1400 N is exerted by a metal ball.

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Troyanec [42]

Answer:

the current is 15.68 × 10^{5} A

Explanation:

mass = 1000 kg

magnetic fied = 2T

rail separation = 2m

escape velocity is 11.2km/s  = 11.2 × 10^{3} m/s

distance = 10 km = 10^{4} m

to find out

determine the current

solution

we know force F = I×L×B

here I is current and L is rail separation and B is magnetic field

so F = I ×2×2  = 4 I

so

acceleration is a = \frac{F}{mass}

a =  \frac{4I}{1000} m/s²

so equation of motion

v²-u² = 2 a S

here u is initial velocity and S is distance and a is acceleration and v is final velocity

11.2 × 10^{3} - 0 = 2×  \frac{4I}{1000} × 10^{4}

solve we get I

I = 15.68 × 10^{5} A

so the current is 15.68 × 10^{5} A

4 0
3 years ago
A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum
uysha [10]

To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor

C = \frac{Q}{V}

Here,

C = Capacitance

V = Potential difference between the plates

Q = Charge between the capacitor plates

At the same time the energy stored in the capacitor can be defined as,

U = \frac{1}{2} CV^2

We will start by finding the value of the capacitance, so we will have to,

C = \frac{2\mu C}{12.0V}

C = 0.166\mu F

Finally using the expression for the energy we have that,

U = \frac{1}{2} CV^2

U = \frac{1}{2} (0.166\muF)(12.0V)^2

U = 11.0*10^{-6} J

Therefore the minimum amount of work that must be done in charging this capacitor is 11.0*10^{-6} J

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