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otez555 [7]
2 years ago
13

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

0 rpm over an interval of 5.0 s. How many revolutions does the ball make during these 5.0 s?
Physics
1 answer:
liraira [26]2 years ago
7 0

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

Therefore, the ball made 75 revolutions during the 5 seconds.

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Ahat [919]
There is no acceleration of g in the x direction because the gravitational acceleration points downward. Also, on most studies we ignore the tidal forces since we are dealing with small bodies compared to the size of the earth.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
5 0
3 years ago
Explain why it takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron fr
slava [35]
It takes more energy to remove the second electron from a lithium atom than it does to remove the fourth electron from a carbon atom because its inner core e, not valence e. C's 4th removed e is still a valence e. And also <span>because more nuclear charge acting on the second electron, it is more close to the nucleus, thus the the protons attract it more than the 4th electron.</span>
8 0
3 years ago
Read 2 more answers
Given that an electric field of 3×106V/m3×106V/m is required to produce an electrical spark within a volume of air, estimate the
Andre45 [30]

Answer:

Length, l = 33.4 m

Explanation:

Given that,

Electrical field, E=3\times 10^6\ V/m

Let the electrical potential is, V=10^8\ V

We need to find the length of a thundercloud lightning bolt. The relation between electric field and the electric potential is given by :

V=E\times d\\\\d=\dfrac{V}{E}\\\\d=\dfrac{10^8}{3\times 10^6}\\\\d=33.4\ m

So, the length of a thundercloud lightning bolt is 33.4 meters. Hence, this is the required solution.

5 0
3 years ago
URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally i
STatiana [176]

Answer:

Explanation:

heat lost by water will be used to increase the temperature of  ice

heat gained by ice

= mass x specific heat  x rise in temperature

1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt  where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186  

8372

heat lost = heat gained

1 x 2090 x t  = 8372

t = 4°C

There will be a rise of  4 degree in the temperature of ice.  

 

5 0
3 years ago
A student is swimming south applying a force of 256 N. The water exerts a westward force of 104 N. If the student has a mass of
grigory [225]

Answer:

a=2.9\ m/sec^2

Explanation:

<u>Net Forces and Acceleration</u>

The second Newton's Law relates the net force F_r acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.

\vec F_r=m\vec a

According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.

The magnitude of the resulting force F_r is computed as the hypotenuse of a right triangle

|F_r|=\sqrt{256^2+104^2}

|F_r|=276.32\ Nw

The acceleration can be obtained from the formula

F_r=ma

Note we are using only magnitudes here

\displaystyle a=\frac{F_r}{m}

\displaystyle a=\frac{276.32Nw}{95.3Kg}

\boxed{a=2.9\ m/sec^2}

7 0
3 years ago
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