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2 years ago

A baseball is thrown a distance of 17.2 m What is its speed if it takes 0.56 seconds to cover the distance?

Physics

1 answer:

lubasha [3.4K]2 years ago

7 0

Answer:

v = (S2 - S1) / t = 17.2 m / .56 s = 30.7 m/s

This is about 100 ft/sec or 69 mph

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A ball is throw vertically upwards with a speed of 20ms,calculate the maximum height reach(g=10)

Goshia [24]

Please check the attached picture

4 0

2 years ago

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp

sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s$

So, the speed of the box is 0.489 m/s.

3 0

3 years ago

A baseball leaves a bat with a horizontal velocity of 20.0 m/s. When it has left the bat for a time of 0.250 s (Assume air resis

miss Akunina [59]

Answer:

The distance is 5 m.

Explanation:

Given that,

Horizontal velocity = 20.0 m/s

Time t = 0.250 s

We need to calculate the horizontal distance

The distance traveled by the baseball with horizontal velocity in time t is given by

Using formula for horizontal distance

$d = v\times t$

$d = 20.0\times0.250$

$d=5\ m$

Hence, The distance is 5 m.

5 0

3 years ago

If a light is moved twice (2x) as far from a surface, the area the light covers is ___ as big.

Ksenya-84 [330]

Answer:

twice

Explanation:

From magnification = height of image / height of object

Distance of image/ distance of object = magnification

If the distance and height of the object represents the initial light distance and the exposed surface respectively.

And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.

Hence the new image exposure would be twice as large.

If we use the formula our point of investigation is Height of image,

H2= D2/D1× H1

H2 = 2D2/D1 × H1

H2 = 2H1

6 0

3 years ago

Which of the following forces can change the motion of an object? A. push B. shove C. pull D. all of these

Goshia [24]

Answer: The answer is D all of these sorry if i am wrong

Explanation:

3 0

2 years ago