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ankoles [38]
4 years ago
10

Why would a soccer ball keep going forever without physics? Pls help and explain why

Physics
2 answers:
Pani-rosa [81]4 years ago
7 0
A soccer ball would keep moving forever without physics, because without force to act upon the soccer ball, it could, or will not be able to stop the acceleration. And force is a factor in physics.
WITCHER [35]4 years ago
7 0
If physics did not exist, I guess we are stating that there is no friction. Without friction, the ball will never slow down, unless it runs into something else.
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Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an
Paladinen [302]

Answer:

\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}

Explanation:

We need to find the acceleration of the 2 kilogram object. Let's complete this in 2 steps.

<h3>1. Force of 1st Object </h3>

First, we can find the force of the first, 8 kilogram object.

According to Newton's Second Law of Motion, force is the product of mass and acceleration.

F=m \times a

The mass of the object is 8 kilograms and the acceleration is 1.6 meters per square second.

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Substitute these values into the formula.

F= 8 \ kg * 1.6 \ m/s^2

Multiply.

F= 12.8 \ kg*m/s^2

<h3>2. Acceleration of the 2nd Object </h3>

Now,  use the force we just calculated to complete the second part of the problem. We use the same formula:

F= m \times a

This time, we know the force is 12.8 kilograms meters per square second and the mass is 2 kilograms.

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Substitute the values into the formula.

12.8 \ kg*m/s^2= 2 \ kg *a

Since we are solving for the acceleration, we must isolate the variable (a). It is being multiplied by 2 kg. The inverse of multiplication is division. Divide both sides of the equation by 2 kg.

\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}

\frac {12.8 \ kg*m/s^2}{2 \ kg}=a

The units of kilograms cancel.

\frac {12.8}{2}\ m/s^2=a

6.4 \ m/s^2=a

The acceleration is 6.4 meters per square second.

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Explanation:

v=s/t

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Amplitude: Meters

Frequency: Hertz

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Explanation:

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Now, we will calculate maximum load from allowable stress as follows.

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Hence, P_{max} will be calculated as follows.

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