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3 years ago

A baseball is hit into the air with a velocity of 27 m/s. How high does it go?

Physics

1 answer:

Marizza181 [45]3 years ago

8 0

Answer:

Explanation:

If a baseball is hit into the air with a velocity of 27 m/s, we want to determine the maximum height of the ball. Using the projecile formula;

Max height H = u²/2g

u is the initial velocity of the body = 27m/s

g is the acceleration due to gravity = 9.81m/s²

H = 27²/2(9.81)

H = 729/19.62

H = 37.16m

Hence the ball went 37.16m high

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Does most psychologists believe that ESP exist

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3 years ago

The weight of a basketball with a mass of 0.5 kg is _N.

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Answer:

4.9N

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3 years ago

Scientists want to place a telescope on the moon to improve their view of distant planets. The telescope weighs 200 pounds on Ea

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33,02 lb

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3 years ago

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The rope of a swing is 3.10 m long. Calculate the angle from the vertical at which a 76.0 kg man must begin to swing in order to

kakasveta [241]

Answer:

The angle from the vertical is 48.72°

Explanation:

Given :

Length of rope $l = 3.10$ m

Mass of man $m = 76$ Kg

Mass of car $M = 1520$ Kg

Velocity of car $v = 1.01$ $\frac{m}{s}$

According to conservation law,

Potential energy of man is converted to kinetic energy of car moving,

$\frac{1}{2} M v^{2} = mgh$

We calculate height,

$h = \frac{M v^{2} }{2mg}$

$h = \frac{1520(1.01)^{2} }{2\times 75 \times 9.8}$ ( ∵ $g = 9.8 \frac{m}{s^{2} }$ )

$h = 1.05$ m

This is the distance of the rope at the bottom,

So we take difference of it.

⇒ $3.10 - 1.05 = 2.05$

We can calculate angle between them,

$\cos \theta = \frac{2.05}{3.10} = 0.6597$

$\theta =$ 48.72°

Therefore, the angle from the vertical is 48.72°

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3 years ago