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3 years ago

What is the kinetic energy of a 5 kg object moving at 4 m/s

Physics

1 answer:

mixas84 [53]3 years ago

4 0

Answer:

Explanation:

rule: K.E = 1/2 m×v^2

1/2 × 5 × (4)^2= 40 joule

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A cyclist reaches the top of a hill with

rjkz [21]

Answer:Based on the excerpt, which best describes Harburg’s view of the Great Depression?

He has no interest in financial success for himself.

He values artistic success over financial success for himself.

He believes most people benefited from losing their financial stability.

He regrets the fact that he gave away his money to benefit his art.

Explanation:

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2 years ago

At a particular instant, a hot air balloon is 100 m in the air and descending at a constant speed of 2.0 m/s. at this exact inst

rewona [7]

Answer:

86.4 m horizontal from landing spot

Explanation:

Find out how long before the ball hits the ground

vertical speed of ball = -2 m/s gravity = - 9.81 m/s^2

find time to hit ground from 100 m

( height will be<u> zero</u> when it hits the ground)

<u>0 </u>= 100 - 2 t - 1/2 ( 9.81) t^2

use Quadratic Formula to find t = 4.32 seconds

horizontal speed of ball = 20 m/s

in 4.32 seconds it will travel horizontally 20 m/s * 4.32 s = 86.4 m

3 0

1 year ago

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.5 kg mass of the pulley is concentrated on i

alexandr1967 [171]

Explanation:

63 kg ice skater finishes her performance and crossed the finish line with a speed of 10.8 m/s

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3 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0

3 years ago

A small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal

Gemiola [76]

Answer:

Time taken, $T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Explanation:

It is given that, a small metal ball is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone as shown in attached figure.

From the figure,

The sum of forces in y direction is :

$T\ cos\theta-mg=0$

$T=\dfrac{mg}{cos\theta}$

Sum of forces in x direction,

$T\ sin\theta=\dfrac{mv^2}{r}$

$mg\ tan\theta=\dfrac{mv^2}{r}$ .............(1)

Also, $r=l\ sin\theta$

Equation (1) becomes :

$mg\ tan\theta=\dfrac{mv^2}{l\ sin\theta}$

$v=\sqrt{gl\ tan\theta.sin\theta}$ ...............(2)

Let t is the time taken for the ball to rotate once around the axis. It is given by :

$T=\dfrac{2\pi r}{v}$

Put the value of T from equation (2) to the above expression:

$T=\dfrac{2\pi r}{\sqrt{gl\ tan\theta.sin\theta}}$

$T=\dfrac{2\pi l\ sin\theta}{\sqrt{gl\ tan\theta.sin\theta}}$

On solving above equation :

$T=2\pi \sqrt{\dfrac{l\ cos\theta}{g}}$

Hence, this is the required solution.

4 0

2 years ago