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Lapatulllka [165]
3 years ago
6

Change 6 revolutions per minute to a linear speed measured in m/s. The radius is 3 m.

Physics
1 answer:
satela [25.4K]3 years ago
7 0

Answer:

Explanation:you know

circumference of a circle =2*pi*r=?

r=radius of a circle =3m

total distance of 6 revolution =s=6*2*pi*r

again, linear speed =v=s/t=?

t=time =60sec

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The law of reflection states that the angle of reflection is equal to the angle of Incidence .

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A car slows down at -5.00 m/s^2 until it comes to a stop after traveling 15.0 m. What was the initial speed of the car?
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8.66mls = v1

vf2 = vi2 + a(d)
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0mls squared = vi2 - 75m squared / s squared
\sqrt{x} 75 m squared/s squared = vi

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The relative hardness of a mineral can best be tested by
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6 0
3 years ago
A bouncy ball is being thrown upwards with a velocity of 34 meters per second if you caught the ball at the same height you rele
Serga [27]

It took the ball 6.94 seconds to make the trip

Explanation:

A bouncy ball is being thrown upwards with a velocity of 34 meters per

second and you caught the ball at the same height you released it

1. The initial velocity of the ball is 34 m/s upward

2. The acceleration of gravity is -9.8 m/s²

3. You caught the ball at the same height you released it

We need to find how long it takes the ball to make the trip

You caught the ball at the same height you released it, then

→ The displacement of the trip s =  zero meter

→ The ball thrown upward with initial velocity u = 34 m/s

→ The acceleration of gravity g = -9.8 m/s²

→ s = u t + \frac{1}{2} g t²

Substitute the values of u , g , s in the rules

→ 0 = 34 t + \frac{1}{2} (-9.8) t²

→ 0 = 34 t - 4.9 t²

Multiply both sides by -1

→ 4.9 t² - 34 t = 0

Take t as common factor

→ t(4.9 t - 34) = 0

Equate each factor by 0

→ t = 0 ⇒ initial position

→ 4.9 t - 34 = 0

Add 34 to both sides

→ 4.9 t = 34

Divide both sides by 4.9

→ t = 6.94 seconds ⇒ final position

It took the ball 6.94 seconds to make the trip

Learn more:

You can learn more about free fall in brainly.com/question/5531630

#LearnwithBrainly

5 0
3 years ago
Why don’t we feel the gravitational force of a large object such as a skyscraper semi-truck?
Kobotan [32]

Answer:

Se the explanation below

Explanation:

We do not feel these forces of these bodies, because they are very small compared to the force of Earth's attraction. Although its mass is greater than that of a human being, its mass is not compared to the Earth's mass. In order to understand this problem we will use numerical data and the universal gravitation formula, to give validity to the explanation.

<u>Force exerted by the Earth on a human being</u>

<u />

F=G*\frac{m_{1}*m_{2}}{r^2}

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 5.97*10^24[kg]

r = distance from the center of the earth to the surface or earth radius = 6371 *10^3 [m]

<u />

Now replacing we have

F = 6.673*10^{-11} *\frac{80*5.97*10^{24}}{(6371*10^{3})^{2}  } \\F = 785[N]

<u>Force exerted by a building on a human being</u>

<u />

Where:

G = universal gravitation constant = 6.673*10^-11 [N*m^2/kg^2]

m1 = mass of the person = 80 [kg]

m2 = mass of the earth 300000 [ton] = 300 *10^6[kg]

r = distance from the building to the person = 2[m]

F = 6.673*10^{-11}*\frac{80*300*10^6}{2^{2} }  \\F= 0.4 [N]

As we can see the force exerted by the Earth is 2000 times greater than that exerted by a building with the proposed data.

8 0
3 years ago
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