Ariana is heating a beaker of water. She notices that the water at the bottom

Determine the kinetic energy of a 575-kg roller coaster car that is moving with a speed of 16.0 m/s. __ J What is the kinetic en

worty [1.4K]

KE at 16m/s
==========
Givens
=====
m = 575 kg
v = 16 m/s

Formula
======
KE = 1/2 m* v^2

KE = 1/2 575 * 16^2
KE = 73600

Second Question
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m = 575
v = 32 m/s

KE = 1/2 mv^2
KE = 1/2 575 * 32^2
KE = 294400

Third Question
===========
Ke2 = KE1 * k where k is the number of times greater the second is than the first.
294400 = k * 73600
k = 294400 / 73600
k = 4 So the KE of the second condition is 4 times the first.

8 0

4 years ago

Which of these models of the solar system first proved that the sun is at one of the foci of Earth's elliptical orbit?

Ivenika [448]

Heliocentric model is the name of the model that shows that earth orbits the sun

5 0

3 years ago

Read 2 more answers

A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. H

lukranit [14]

Answer:

d = 6.32 m

Explanation:

Given that,

The mass of a puck, m = 2 kg

It is pushed straight north with a constant force of 5N for 1.50 s and then let go.

We need to find the distance covered by the puck when move from rest in 2.25 s.

We know that,

F = ma

$a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2$

Let d is the distance moved in 2.25 s. Using second equation of motion,

$d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m$

So, it will move 6.32 m from rest in 2.25 seconds.

4 0

3 years ago

a 68 kg skydiver with a parachute falls at constant velocity for 100 m how much work does the earth do on the skydiver

Ulleksa [173]

<span>The earth does no work on her if she's falling in the direction of frictional force. But suppose it does work on the woman i.e in the opposite direction, N. Then we have F = mg where g is acceleration due to grav. So F = 68 * 9.8 = 666.4 Newton. Then work done = force * distance = 666.4 * 100 = 666400</span>

8 0

4 years ago

Read 2 more answers

In anticipation of a long 10 upgrade, a bus driver accelerates at a constant rate of 5 ft/s2 while still on a level section of a

fgiga [73]

Answer:

$S = 20903.4$ ft

Explanation:

First we will determine the acceleration of the bus while it is moving upward

The equilibrium equation would be

$F - W sin\theta = ma'\\ma - W sin\theta = ma'\\m = \frac{W}{g} \\\frac{W}{g}*a - W sin\theta = \frac{W}{g} * a'\\\frac{a}{g} - sin\theta = \frac{a'}{g}\\\frac{x}{y} \frac{5}{32.2} - sin 10 = \frac{a'}{32.2} \\\a' = -0.59$

Let the displacement be $S_0$

As per newton's third law of motion

$v^2 -u^2 = 2as\\u = 80 \frac{mi}{h} = 117.33\frac{ft}{s}\\v = 50 \frac{mi}{h} = 73.33\frac{ft}{s}\\73.33 ^ 2 - 117.33^2 = 2 * (-0.59) * (S-S_0)\\\\S_0 = 0\\S = 20903.4$

3 0

3 years ago