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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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An airplane touches down on the runway with a speed of 70 m/s2. Determine the airplane after each second of its deceleration.

ivann1987 [24]

<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters

</span>

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3 years ago

Efficiency is the ratio of output work to input work, expressed as a percentage. Light bulbs put out less light energy than the

Ivanshal [37]

The answer is 10%. just took the test. hope it helps!!

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3 years ago

Read 2 more answers

A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.

cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

$a=1.5 m/s^2$

So we can re-arrange the equation above to find their combined mass:

$m=\frac{F}{a}=\frac{75}{1.5}=50 kg$

3 0

3 years ago

An aeroplane, flying in a straight line at a constant

dedylja [7]

Answer:

200

Explanation:

3 0

3 years ago

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

Mamont248 [21]

Answer:

74.86°C

Explanation:

P₂ = Vapour pressure of water at sea level = 760 mmHg

P₁ = Pressure at base camp = 296 mmHg

T₂ = Temperature of water = 373 K

ΔH°vap for H2O = 40.7 kJ/mol = 40700 J/mol

R = Gas constant = 8.314 J/mol K

From Claussius Clapeyron equation

$ln\frac{P_2}{P_1}=\frac{\Delta H}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\\\Rightarrow ln\frac{760}{296}=\frac{40700}{8.314}\left(\frac{1}{T_1}-\frac{1}{373}\right)\\\Rightarrow ln\frac{760}{296}\times \frac{8.314}{40700}+\frac{1}{373}=\frac{1}{T_1}\\\Rightarrow 0.0028735=\frac{1}{T_1}\\\Rightarrow T_1=347.996\ K$

T₁ = 347.996 K = 74.86°C

∴Water will boil at 74.86°C

4 0

3 years ago