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zlopas [31]
3 years ago
9

I WILL MARK YOU AS BRAINLIEST IF RIGHT The fastest pitch ever recorded in MLB was thrown by Aroldis Chapman at 5 points 105.1 mp

h . If the 0.145 kg baseball accelerated 96 m/s / s , how much force did Chapman apply to the ball? Round your answer to the nearest whole number
Physics
1 answer:
Goshia [24]3 years ago
5 0

Answer: The force was 13.92 Newtons.

Explanation:

First, let's recall the second Newton's law:

The net force is equal to the mass times the acceleration, or:

F = m*a

where:

F = force

m = mass

a = acceleration.

When the player hits the ball with the bat, he applies a force that accelerates the ball for a small period of time, that increases greatly the speed of the ball.

In this case, we know that:

the mass of the ball is 0.145 kg

The acceleration of the ball is 96m/s^2

Then we can input those values in the above equation to find the force.

F = 0.145kg*96m/s^2 = 13.92 N

The force was 13.92 Newtons.

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A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The moment of inertia of t
emmasim [6.3K]

Answer:

Angular acceleration = 6.37rad/sec²

Approximately, Angular acceleration =

6.4 rad/sec²

Explanation:

Length of the rod = 2.0m long

Inclination of the rod (horizontal) = 30°

Mass of the rod is not given so we would refer to it as = M

Rotational Inertia of the Rod(I) = 1/3ML²

Angular Acceleration = ?

There is an equation that shows us the relationship between Torque and Angular acceleration.

The equation is :

Torque(T) = Inertia × Angular Acceleration

Angular acceleration = Torque ÷ Inertia

Where:

Torque = L/2(MgCosθ)

Where M = Mass

L = Length = 2.0m

θ = Inclination of the rod (horizontal) = 30°

g = Acceleration due to gravity = 9.81m/s²

Inertia = 1/3ML²

Angular Acceleration =  (Mass × g × Cos (30°) × (L÷2)) ÷ 1/3ML²

Angular Acceleration =

(3 × g × cos 30°) ÷ 2× L

Angular Acceleration = (3 × 9.81m/s² × cos 30°) ÷ 2× L

Angular Acceleration = 3 × 9.81m/s² × cos 30°) ÷ 2× 2.0m

Angular Acceleration = 6.37rad/sec²

Approximately Angular Acceleration =

6.4rad/sec²

5 0
3 years ago
A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff
Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0
2 years ago
Please answer D in the image with an explanation
puteri [66]

Answer:

The force is 274 N.

Explanation:

In figure 2:

(d) Let the tension in the string is T.

According to the Newton's second law,

Net force = mass x acceleration

Apply for 200N.

T - 200 sin 35 =\frac{200}{9.8}\times a \\T - 114.7 = 20.4 a..... (1)\\220 - T = \frac{220}{9.8}\times a\\220 - T = 22.45 a..... (2)\\Adding both the equations\\334.7 = 42.85 aa =7.81 m/s^{2}

Now put in (1)

T - 114.7 = 20.4 x 7.81

T = 274 N

4 0
2 years ago
Light from the star Betelgeuse takes 640 years to reach Earth. How far away is Betelgeuse in units of light-years? Name any hist
Eduardwww [97]

Answer:

Betelgeuse is 640 light years away from earth.

Explanation:

A light-year is an astronomical unit to measure the distance the light travels in a calendar year.

If the light from a star takes 640 years to reach us, then that star its 640ly away from us.

Betelgeuse has been labeled as a Variant Star, which means that its brightness can fluctuate over the course of years, this has made difficult for astronomers to measure the exact distance of the star. Right now the star is estimated to be around 613 and 881ly away from earth, although, for the sake of your second question, we will take 640 years as our estimated value.

In 1380 (640 years ago) the Battle of Kulikovo took place. A battle of remarkable importance to Russian history, in which the Russian army, led by Prince Dmitry of Moscow, defeated the Mongol army, defining a turning point in the Mongol dominance, and setting the bases for what Russia is today.

6 0
3 years ago
Read 2 more answers
Which of the following has more energy than visible light?
Tom [10]
The answer is D.


Hope it helps!
5 0
3 years ago
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