All categories

3 years ago

What is the Kinetic Energy of a 1200 kg object that is moving with a speed of 24 m/s

Physics

1 answer:

SVEN [57.7K]3 years ago

8 0

Kinetic Energy = 1/2mv^2

m= 1200kg

v= 24 m/s

KE = 1/2 (1200kg)(24m/s)^2 = 345,600 N

You might be interested in

A man walks 30 m to the west, then 5 m to the east in 45 seconds.

katen-ka-za [31]

Answer:

a. Displacement=30²+5²=925= 30.4m

b. Total distance=30m+5m=35m

c. V=s/t. = 30.4/45=0.6m/s

8 0

2 years ago

if the flower pot in problem 3 falls off the windowsill and falls 20 meters downwards(i.e., is 10 meters from hitting the ground

morpeh [17]

Answer:

te energy is

Explanation:

8 0

3 years ago

Why is it possible for a gas to take the shape of its container?

Sedaia [141]

I believe the answer is A

8 0

3 years ago

Read 2 more answers

A baseball leaves a bat with a horizontal velocity of 20 m/s. In a time of 0.25 s, How far will it have traveled horizontally?

Maurinko [17]

Distance traveled by the ball is given by

$distance = speed \times time$

here we know that

speed = 20 m/s

times = 0.25 s

now we have

$distance = 20 \times 0.25$

$distance = 5 m$

so ball will travel 5 m distance in the given interval of time

6 0

3 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

4 0

3 years ago