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Maksim231197 [3]
2 years ago
15

If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?

Physics
1 answer:
Mariana [72]2 years ago
7 0

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

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A closely wound search coil has an area of 3.21 cm2, 120 turns, and a resistance of 58.7 O. It is connected to a charge-measurin
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Answer:

The magnetic field in the System is 0.095T

Explanation:

To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.

By Faraday's law we know that

\epsilon = \frac{NBA}{t}

Where,

\epsilon  =electromotive force

N = Number of loops

B = Magnetic field

A = Area

t= Time

For Ohm's law we now that,

V = IR

Where,

I = Current

R = Resistance

V = Voltage (Same that the electromotive force at this case)

In this system we have that the resistance in series of coil and charge measuring device is given by,

R = R_c + R_d

And that the current can be expressed as function of charge and time, then

I = \frac{q}{t}

Equation Faraday's law and Ohm's law we have,

V = \epsilon

IR = \frac{NBA}{t}

(\frac{q}{t})(R_c+R_d) = \frac{NBA}{t}

Re-arrange for Magnetic Field B, we have

B = \frac{q(R_c+R_d)}{NA}

Our values are given as,

R_c = 58.7\Omega

R_d = 45.5\Omega

N = 120

q = 3.53*10^{-5}C

A = 3.21cm^2 = 3.21*10^{-4}m^2

Replacing,

B = \frac{(3.53*10^{-5})(58.7+45.5)}{120*3.21*10^{-4}}

B = 0.095T

Therefore the magnetic field in the System is 0.095T

3 0
3 years ago
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