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3 years ago

Calculate the kinetic energy of a 5.0kg object moving at 4.0 m/s

Physics

1 answer:

Nana76 [90]3 years ago

3 0

Answer:

The answer is

Explanation:

The kinetic energy of an object given it's mass and velocity can be found by using the formula

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

where

m is the mass

v is the velocity

From the question

m = 5 kg

v = 4 m/s

The kinetic energy is

$KE = \frac{1}{2} \times 5 \times {4}^{2} \\ = \frac{1}{2} \times 5 \times 16 \\ = 5 \times 8$

We have the final answer as

Hope this helps you

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A 100 N force is applied to move an object a horizontal distance of 5 meters at constant speed in 10 seconds. How much power is

Tpy6a [65]

Answer:

50 W

Explanation:

<h3><u>Given :</u></h3>

Force applied = 100 N
Distance covered = 5 metres
Time = 10 seconds

Power

<h3><u>Solution :</u></h3>

For calculating power, we first need to know about the work done.

$\bf \boxed{Work = Force \times displacement}$

Now, substituting values in the above formula;

Work = 100 × 5

= 500 Nm or 500 J

We know that,

$\bf \boxed{Power=\dfrac{Work\:done}{Time\: taken}}$

Substituting values in above formula;

Power = 500/ 10

= 50 Nm/s or 50 W

Hence, power = 50 W .

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3 years ago

At which temperature does the motion of atoms and molecules stop?

Scorpion4ik [409]

Answer:

0 Kelvin

Explanation:

Atoms in absolute temperature get approximatelly motionless since 0 Kelvin is -273 degrees Celcius. The kinetic energy of atoms/particles in matter has the possible lowest value ( almost zero), so that there is nothing colder than 0 Kelvin.

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3 years ago

Why do all freely falling objects have the same acceleration

iren2701 [21]

They are falling under the sole influence of gravity all objects<span> will </span>fall<span> with the </span>same<span> rate of </span><span>acceleration needless of there size</span>

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3 years ago

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A 500-n crate needs to be lifted 1 meter vertically in order to get it into the back of a pickup truck. what gives the crate a g

worty [1.4K]

The weight an weight of the truck

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3 years ago

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A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is

pochemuha

Answer:

2856.96 J

$\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

6.78822 m/s

Explanation:

$v_i$ = Initial velocity = 9.6 m/s

g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J$

At height y = 0 the potential energy is 0 as

$P=mgy\\\Rightarrow P=mg0=0$

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

$K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f$

Half of maximum height

$\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}$

$h_i=\frac{v_i^2}{2g}$

$v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s$

The velocity of the athlete at half the maximum height is 6.78822 m/s

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2 years ago