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4 years ago

10

10. Define and give 2 examples of kinetic energy. Be sure to include the two factors that make an object’s kinetic energy increa

se.

1 answer:

Mademuasel [1]4 years ago

7 0

You can describe kinetic energy and the potential energy of motion to get your answer.

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If the temperature of a volume of ideal gas increases for 100°C to 200°C, what happens to the average kinetic energy of the mole

mina [271]

Answer: $\Delta E=2.0715\times 10^{-21} J$

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

$E=\frac{3}{2}\cdot \frac{R}{N_A}T$

Where

R=Gas constant

$N_A$ =Avogadro's number

T=Temperature in kelvin

$\frac{R}{N_A}=1.381\times 10^{-23}$

So kinetic energy increases by

$\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )$

$\Delta E=2.0715\times 10^{-21} J$

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3 years ago

What is the velocity of Car B after collision?

zubka84 [21]

Answer:

30m/s

Explanation:

m1v1=m2v2

12 * 10=4* v2

120/4 =v2

V2=30m/s

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3 years ago

A coworker did not clean his work area before going home this could cause an accident so you quickly clean up the next day you s

defon

Answer:

THE FIRST ONE YOU SHOULD TELL HIM AND THE LAST ONE YOU SHOUDENT DO BECAUSE HE WILL DO IT AGAIN AND EXPECT OTHERS TO CLEAN UP AFTER HIM

Explanation:

5 0

3 years ago

Read 2 more answers

A horse has an acceleration of 2 m/s2. If it starts from rest, how fast is it going after 1.7 seconds?

fgiga [73]

(2 m/s²) · (1.7 s) = 3.4 m/s

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3 years ago

Read 2 more answers

A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1

aivan3 [116]

Answer:

Part a)

$F_n = 306 N$

Part b)

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

$mg - F_n = ma_c$

now we will have

$mg - F_n = \frac{mv^2}{R}$

$F_n = mg - \frac{mv^2}{R}$

now we have

$F_n = 100(9.81) - \frac{100(9^2)}{12}$

$F_n = 981 - 675$

$F_n = 306 N$

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

$F_n + mg = ma_c$

$F_n = 0$ at minimum speed

$mg = \frac{mv^2}{R}$

$v = \sqrt{Rg}$

$v = \sqrt{15 \times 9.81}$

$v = 12.1 m/s$

So this speed is independent of the mass of the rider

5 0

4 years ago