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djyliett [7]
3 years ago
11

How much gravitational force do two lead balls with a mass of 8 kilograms, the centers of mass of which are 17 cm apart, affect

each other?
Physics
1 answer:
Arisa [49]3 years ago
7 0

Answer:

1.48×10⁻⁷ Newtons

Explanation:

From the question,

According to newton's law of universal gravitation.

F = Gmm'/r²........................ Equation 1

F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.

Given: m = m' = 8 kg, r = 17 cm = 0.17 m,

Constant : G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F =  (6.67×10⁻¹¹×8×8)/(0.17²)

F = 1.48×10⁻⁷ N

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2. Neutrons have a ____<br> charge.<br> a. positive<br> b. negative<br> c. neutral
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b. negative

Explanation:

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A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

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7 0
2 years ago
In magazine car tests an important indicator of performance is the zero to 60 mph (0 to 96.6 km/h) acceleration time. A time bel
slavikrds [6]

Answer:

2.73414 seconds

467622.66798 J

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

v=60\times \dfrac{1609.34}{3600}=26.822\ m/s

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s

or

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s

The time taken is 2.73414 seconds

The potential energy is given by

U=mgh\\\Rightarrow U=1300\times 9.81\times 36.66766\\\Rightarrow U=467622.66798\ J

The change in potential energy is 467622.66798 J

5 0
3 years ago
Suppose that you are holding a pencil balanced on its point. If you release the pencil and it begins to fall, what will be the a
Naily [24]

Answer:

The angular acceleration of the pencil<em> α  = 17 rad·s⁻²</em>

Explanation:

Using Newton's second angular law or torque to find angular acceleration, we get the following expressions:

    τ = I α                              (1)

    W r = I α                          (2)

The weight is that the pencil has is,

   sin 10 = r / (L/2)

   r = L/2(sin(10))

 

The shape of the pencil can be approximated to be a cylinder that rotates on one end and therefore its moment of inertia will be:

    I = 1/3 M L²

Thus,

   mg(L / 2)sin(10) = (1/3 m L²)(α) 

   α(f) = 3/2(g) / Lsin(10)

   α  = 3/2(9.8) / 0.150sin(10)

  <em> α  = 17 rad·s⁻²</em>

Therefore, the angular acceleration of the pencil<em> </em>is<em> 17 rad·s⁻²</em>

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3 years ago
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