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2 years ago

13

What is the average kinetic energy of an object's particles?

2 answers:

vova2212 [387]2 years ago

7 0

What is the average kinetic energy of an object's particles?

the answer is temperature

Lorico [155]2 years ago

4 0

The answer is Temperature!

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You toss a ball in the air. what is the work done by gravity as the ball goes up?

jolli1 [7]

Answer: Gravity slows the ball down as it goes up and eventually stops it from going up and starts to pull it back down to earth.

Explanation:

7 0

2 years ago

A rifle with a mass of 0.9 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 750 m/s

SCORPION-xisa [38]

Answer:

a ) 4.5 N.s

b) V =5 m/s

Explanation:

given,

mass of rifle(M) = 0.9 kg

mass of bullet(m) = 6 g = 0.006 kg

velocity of the bullet(v) = 750 m/s

a) momentum of bullet = m × v

= 750 × 0.006

= 4.5 N.s

b) recoil velocity

m × u + M × U = m × v + M × V

0 + 0 = 0.006 × 750 - 0.9 × V

V = $\dfrac{4.5}{0.9}$

V =5 m/s

5 0

3 years ago

A weight lifter applies an upward force of 1100 N while lowering a dumbbell

Paul [167]

Answer:

A

Explanation:

$work = force \times distance$

$work = 1100 \times 0.5$

$= 550 \: j$

hope it helped a lot

pls mark brainliest with due respect .

6 0

2 years ago

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

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2 years ago

Which model shows how a comet's tail changes during its orbit?

pentagon [3]

Third model shows how a comet's tail changes during its orbit...

mark brainliest

5 0

2 years ago

Read 2 more answers