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Sergeeva-Olga [200]
3 years ago
13

A jet travels 1000 kilometers in 5 hours. What is its average speed?

Physics
1 answer:
nlexa [21]3 years ago
7 0

Given:

Total distance = 1000 kilometer

Total time = 5 hours

To find:

Average speed = ?

Formula used:

v_{avg} = \frac{s}{t}

Where v_{avg} = average speed

s = total distance

t = total time

Solution:

Average speed of the jet is given by,

v_{avg} = \frac{s}{t}

Where v_{avg} = average speed

s = total distance

t = total time

v_{avg} = \frac{1000}{5}

v_{avg} = 200 km/ h

Thus, average speed of the jet is 200 km/h.

Hence, Option (A) is correct.


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If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is
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Answer:

A.) 42.7 m/s

B.) 0.33 m/s^2

C.) 90 kg

Explanation:

A.) If Justin races his Chevy S-10 down highway 37 north for 2,560 meters in 60 seconds, what is his velocity? 

Velocity = displacement/time

Velocity = 2560/60

Velocity = 42.67 m/s

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Acceleration = 10/30

Acceleration = 0.33 m/s^2

C.) The S-10 has a force of 30 N. What is the mass of the car?

Force = mass × acceleration

30 = mass × 0.33

Mass = 30/ 0.33

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What do we know that might help here ?

-- Temperature of a gas is actually the average kinetic energy of its molecules.

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Knowing just these little factoids, we realize that as a gas gets hotter, the average speed of its molecules increases.

That's exactly what Graph #1 shows.

How about the other graphs ?

-- Graph #3 says that as the temperature goes up, the molecules' speed DEcreases.  That can't be right.

-- Graph #4 says that as the temperature goes up, the molecules' speed doesn't change at all.  That can't be right.

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A 0.2 kg block sliding on a horizontal table slows down from 25 m/s to 20 m/s. How much energy does the block lose due to fricti
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Answer:

the kinetic energy lost due to friction is 22.5 J

Explanation:

Given;

mass of the block, m = 0.2 kg

initial velocity of the block, u = 25 m/s

final velocity of the block, v = 20 m/s

The kinetic energy lost due to friction is calculated as;

\Delta K.E= K.E_f - K.E_i\\\\\Delta K.E= \frac{1}{2}mv^2 -  \frac{1}{2}mu^2\\\\\Delta K.E= \frac{1}{2}m(v^2 -u^2)\\\\\Delta K.E= \frac{1}{2} \times 0.2 (20^2 - 25^2)\\\\\Delta K.E= -22.5 \ J

Therefore, the kinetic energy lost due to friction is 22.5 J

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Answer:

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