If a bullet travels 2500 meters at a speed of 5 m/s, how long did it take?

Physics

1 answer:

MatroZZZ [7]2 years ago

3 0

It would take 500 seconds, because 2500 divided by 5 is 500

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They would feel that the water is cold.

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2 years ago

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A ball droped from a building. How fast is it traveling after falling 3.55s

algol [13]

Answer:

d = 61.75 m

Explanation:

Given that,

A ball droped from a building.

We need to find how fast is it traveling after falling 3.55 s.

As it is dropped, its initial velocity is equal to 0.

Let d is the distance it covers after falling 3.55 s.

We can use second equation of motion to find d.

$d=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a =g

$d=\dfrac{1}{2}gt^2\\\\d=\dfrac{1}{2}\times 9.8\times (3.55)^2\\\\d=61.75\ m$

So, it will cover 61.75 m after falling 3.55 seconds.

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3 years ago

The instant before a batter hits a 0.14-kilogram baseball, the velocity of the ball is 45 meters per second west. The instant af

NemiM [27]

(1) The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is 8,000 m/s² east.

(2) The magnitude of the average force the Bat exerts on the ball while they are in contact is 1,120 N.

The given parameters:

Mass of the baseball, m = 0.14 kg
Initial velocity of the baseball, u = 45 m/s west (negative direction)
Final velocity of the baseball, v = 35 m/s east (positive direction)
Time of contact, t = 0.01 s

The magnitude and direction of the average acceleration of the baseball while it is in contact with the bat is calculated as follows:

$a = \frac{v- u}{t} \\\\a = \frac{35 - (-45)}{0.01} \\\\a = \frac{80}{0.01} \\\\a = 8,000 \ m/s^2$