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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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An electron has a negative electric charge.

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Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
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We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
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By rearranging the equation, we find d:
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3 0
3 years ago
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

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2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

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80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

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