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2 years ago

How fast must a 70 kg student be running to have a kinetic energy of 568 J?!

Physics

1 answer:

Dmitry [639]2 years ago

4 0

Answer: 4.03 m/s

Explanation: Kinetic energy is

KE = 1/2 m v²

So we derive the equation to have v.

v = √ 2KE / m

= 2 x 568 J / 70 kg

= 16.23

= √ 16.23

= 4.03 m/s

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A positive point charge (q = +8.65 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 2.97 m. A p

ella [17]

Answer:

$r_B = 1.88 m$

Explanation:

As we know that work done by electric force is given as

$W_e = -q\Delta V$

so here we know that charge is moving from

$r_A = 2.97 m$

to another position

so we will have

$W_{AB} = \frac{kq_1q_2}{r_A} - \frac{kq_1q_2}{r_B}$

$-6.95\times 10^{-9} = (9\times 10^9)(8.65\times 10^{-8})(4.56\times 10^{-11})(\frac{1}{2.97} - \frac{1}{r_B})$

$-0.196 = (\frac{1}{2.97} - \frac{1}{r_B})$

$r_B = 1.88 m$

7 0

3 years ago

WHAT HAPPENS TO THE VELOCITY AND ACCELERATION OF A BODY FALLING FREELY IN A VACUUM?

mixas84 [53]

both remains constant

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7 0

3 years ago

Question 5 of 32

harkovskaia [24]

The answer is D
Neon has 20.0026 element more then the other

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2 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

Oksana_A [137]

Answer:

a.Attractive

$2.31531\times 10^{-16}\ N$

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

$q_1=q_2$ = Charge of electron and proton = $1.6\times 10^{-19}\ C$

r = Distance between them = 997 nm

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N$

The force of attraction between the particles will be $2.31531\times 10^{-16}\ N$

8 0

2 years ago

Problem: Suppose that the rate of flow of water through a fire hose is 21.4 kg/s and the stream of water from the hose moves at

Sloan [31]

Answer:

The force exerted is 318.86 N

Explanation:

The force exerted by such a stream is calculated by multiplying the mass flow rate of water by the velocity of the stream of water.

mass flow rate = 21.4 kg/s

velocity = 14.9 m/s

Force exerted = 21.4 kg/s × 14.9 m/s = 318.86 kgm/s^2 = 318.86 N

6 0

3 years ago

Read 2 more answers