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3 years ago

Which is the average kinetic energy of particles in an object?

Physics

1 answer:

alina1380 [7]3 years ago

5 0

The amount of heat in the body in joule

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a ball is shot from the ground straight up into the air with initial velocity of 50 50 ft/sec. assuming that the air resistance

Anna [14]

The height attained by the ball is 11.86m

a ball is shot from the ground straight up into the air its initial and final velocity is

initial velocity, u = 50 ft/s = 50×0.305 = 15.25m/s

final velocity ,v = 0 m/s

gravity =-9.8 m/s²

( negative sign shows acceleration in opposite direction)

height =?

using the newton motion of equation

v² = u² + 2as

where

a= acceleration due to gravity(g)

s = height

v² = u² + 2gs

(0)² = (15.25)² + 2×(-9.8)×s

0 = (15.25)² - 19.6 × s

s= - (15.25)²/ 19.6

s = 11.86m

after ignoring the air resistance the maximum height of the ball is 11.86m

To learn more about motion under gravity -

brainly.com/question/27962354

#SPJ4

3 0

1 year ago

What force is needed to propel a 421-kg car with an acceleration of 3.77 m/s2?

nekit [7.7K]

brainly.com/question/11542618?answering=true&answeringSource=greatJob%2FquestionPage

7 0

3 years ago

You make a straight line when an object is

swat32

Answer:

can you explain more or..,.

Explanation:

8 0

3 years ago

Read 2 more answers

When two pool balls collide, what happens to the momentum of each one?

Schach [20]

Answer:

They slow down.

Explanation:

The collided so the slow down till they stop.

7 0

3 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

maria [59]

Answer:

W = 0.678 rad/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

Roll up and hill without slipping is the sumatory of two energys, rotational and translational, so:

$\frac{1}{2}IW^2+ \frac{1}{2}mV^2 = mgh$

where I is the moment of inertia, W the angular velocity at the base of the hill, m the mass of the ball, V the velocity at the base of the hill, g the gravity and h the altitude.

First, we will find the moment of inertia as:

I = $\frac{2}{3}mR^2$

where m is the mass and R the radius, so:

I = $\frac{2}{3}(0.426kg)(11.3m)^2$

I = 36.26 Kg*m^2

Then, replacing values on the initial equation, we get:

$\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)V^2 = (0.426kg)(9.8)(5m)$

also we know that:

V =WR

so:

$\frac{1}{2}(36.26)W^2+ \frac{1}{2}(0.426kg)W^2R^2 = (0.426kg)(9.8)(5m)$

Finally, solving for W, we get:

$W^2(\frac{1}{2}(36.26)+ \frac{1}{2}(0.426kg)(11.3m)^2) = (0.426kg)(9.8)(5m)$

W = 0.678 rad/s

8 0

3 years ago