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3 years ago

When throwing curveballs in baseball, is it important to have air hit the ______ part of the ball.

Physics

2 answers:

Licemer1 [7]3 years ago

7 0

Answer: A (smooth)

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Ede4ka [16]3 years ago

5 0

Answer:

Rough

Explanation:

You might be interested in

according to newton's first law of motion what is the reason for a ball throwing up in the airfall back to earth

skelet666 [1.2K]

The force of earth's gravitational field is always directed downwards (towards the center of the earth. When the ball is thrown up, it is going against the earth's gravitational field and so, the earth's gravitational force pulls it back down, accelerating it downwards.

5 0

3 years ago

A 200kg ball on the end of string is swung in horizontal circle with radius of 0.5m . The ball makes revolution every 2second th

ANEK [815]

Answer: $\dfrac{\pi}{2} ms^{-1}$

Explanation:

Let $T$ be the time required to make one revolution.

Let $r$ be the radius of the circular path.

Let $d$ be the distance travelled by ball in one revolution.

As we know,the distance travelled in one revolution is the circumference of the circle.

So, $d=2\pi r$

Given, $d=0.5m\\T=2sec$

$d=2\times \pi \times 0.5=\pi m$

Speed of an object moving is circular path is define as the ratio of distance travelled in one revolution to the time taken by the object to complete one revolution.

Let $s$ be the speed of the ball.

$s=\frac{d}{T}=\frac{\pi }{2}ms^{-1}$

So,the speed of the ball is $\frac{\pi }{2}ms^{-1}$

5 0

3 years ago

1. Calculati greutatea unui sac cu 5 kg de cartofi într-o zonă in care acceleratia gravitatională

harkovskaia [24]

Answer:

the weight is 49.1 N

Explanation:

The computation of the weight is shown below:

As we know that

= 5kg of potatoes × gravitational acceleration

= 5kg of potatoes × 9.82 m/s

= 49.1 N

Hence, the weight is 49.1 N

We simply applied the above formula in order to determine the weight

6 0

3 years ago

A 3874-kg rollercoaster is brought to the top of a 42m hill in 40 seconds,then drops 28m before the next hill.

Ede4ka [16]

(a) The work required to get the coaster to the top of the first hill is 1,594,538.4 J.

(b) The power required to bring the train to the top of the first hill is 39,863.46 W.

(d) The left at the bottom is determined as 1,063,025.6 J.

<h3>Work done to bring the rollercoaster top of the hill</h3>

W = Fn x d = mgh

W = 3874 x 9.8 x 42

W = 1,594,538.4 J

<h3>Power dissipated in bringing the rollercoaster on top hill</h3>

P = Fv

P = Fd/t

P = W/t

P = 1,594,538.4 /40 = 39,863.46 W

<h3>Energy lost when the coaster drops</h3>

E = 1,594,538.4 - (3874 x 9.8 x 28)

E = 531,512.8 J

<h3>Energy left at the bottom</h3>

E = 3874 x 9.8 x 28 = 1,063,025.6 J

Learn more about energy here: brainly.com/question/13881533

#SPJ1

6 0

2 years ago

An object with a charge of 0.9 x 10^-5 C is separated from a second object with a chare of 2.5 x 10^-4 C by a distance of 0.5 m.

meriva

Answer: Force = 81 N

Explanation:

from Columbs law,

F = k(q1*q2)/r²

k = 9 x 10^9 Nm²/C²

F = (9 x 10^9)x (0.9x10^-5 x 2.5x10^-4)/(0.5)²

F = 81 Newtons

4 0

3 years ago