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3 years ago

A 45 kg man is running 22 m/s. Calculate the kinetic energy

Physics

1 answer:

Eddi Din [679]3 years ago

6 0

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 45 \times {22}^{2} \\ = \frac{1}{2} \times 45 \times 484 \\ = 45 \times 242 \\ = 10890$

We have the final answer as

Hope this helps you

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3) A 15.0kg cannonball is ascending after being launched. Neglect drag.

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Answer:

Explanation:

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3 years ago

Which one of the following temperatures is equal to 5°C?

natali 33 [55]

Answer : The correct option is, (D) 278 K

Explanation :

We are given temperature $5^oC$ .

Now the conversion factor used for the temperature is,

$K=^oC+273$

where, K is kelvin and $^oC$ is Celsius.

Now put the value of temperature, we get

$K=5^oC+273=278K$

Therefore, the temperature 278 K is equal to the $5^oC$

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3 years ago

How much time would it take for the sound of thunder to travel 1500 meters if sound travels at a speed of 330 m/s

Elis [28]

Data given:

Δx=1500m

v=330m/s

t=?

Formula:

V=Δx/t

Solution:

t=1500m/330m/s

t=4.5s

7 0

3 years ago

A 0.0663 kg ingot of metal is heated to 241◦C

Westkost [7]

Answer:280.216j/kg°C

Explanation:

Mass of metal=0.0663kg

mass of water=0.395kg

Final temperature=27.4°C

Temperature of metal=241°C

Temperature of water=25°C

specific heat capacity of water=4186j/kg°C

0.0663xax(241-27.4)=0.395x4186x(27.4-25)

0.0663xax213.6=0.395x4186x2.4

14.16168a=3968.328

a=3968.328 ➗ 14.16168

a=280.216j/kg°C

4 0

3 years ago

How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33Ã—Ã—10âˆ

AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, $F=3.33\times 10^{-21}\ N$

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

$F=k\dfrac{q^2}{r^2}$

$q=\sqrt{\dfrac{Fr^2}{k}}$

$q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}$

$q=1.21\times 10^{-16}\ C$

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

$n=\dfrac{q}{e}$

$n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}$

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0

3 years ago