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2 years ago

5

____________ is an out made when a base runner, forced to run because another teammate must run to the base being occupied, cann

ot reach the next base safely *

1 answer:

maxonik [38]2 years ago

5 0

If I knew the answer I would help but I don’t know sorry

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A car goes North a distance of 120 kilometers during a time period of 3.0 hours. What is the average

DIA [1.3K]

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the average speed of the car during this time interval is 40 m/s i thinkk..

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3 years ago

If the distance between the two mass double what happens to the gravitational force

Kryger [21]

If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

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3 years ago

What are simple machines please?

lubasha [3.4K]

A basic mechanical device that is used for applying a force.

Examples are: inclined plane, lever, wedge, wheel and axle, pulley, and screw...Hope this helps, have a BLESSED and wonderful day!

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3 years ago

Read 2 more answers

A tire measures 20 inches in diameter. How many revolutions will the tire have completed when it has traveled 100 yards? Express

Soloha48 [4]

<h2>Number of revolutions required to travel 100 yards is 57.</h2>

Explanation:

Diameter of tire,D = 20 inches

Perimeter of tire = πD

Perimeter of tire = π x 20 = 62.8 inches

Distance traveled = 100 yards

1 yard = 36 inches

Distance traveled = 100 x 36 = 3600 inches

In one revolution it travels 62.8 inches.

$\texttt{No of revolutions = }\frac{3600}{62.8}\\\\\texttt{No of revolutions = }57.32\\\\\texttt{No of revolutions = }57$

Number of revolutions required to travel 100 yards is 57.

8 0

3 years ago

plz help me with hw A bus of mass 1000 kg moving with a speed of 90km/hr stops after 6 sec by applying brakes then calculate the

Lelechka [254]

Answer:

Mass, M = 1000 kg

Speed, v = 90 km/h = 25 m/s

time, t = 6 sec.

Distance:

${ \tt{distance = speed \times time }} \\ { \tt{distance = 25 \times 6}} \\ { \tt{distance = 150 \: m}}$

Force:

${ \tt{force = mass \times acceleration}} \\ { \bf{but \: for \: acceleration : }} \\ from \: second \: equation \: of \: motion : \\ { \bf{s = ut + \frac{1}{2} {at}^{2} }} \\ \\ { \tt{150 = (0 \times 6) + ( \frac{1}{2} \times a \times {6}^{2} ) }} \\ \\ { \tt{acceleration = 8.33 \: {ms}^{ - 2} }} \\ \\ { \tt{force = 1000 \times 8.33}} \\ { \tt{force = 8333.3 \: newtons}}$

5 0

3 years ago