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Dmitrij [34]
2 years ago
9

500 kg car takes a car of radius 50m with a velocity of 36km/hr.the centripetal force is​

Physics
1 answer:
mylen [45]2 years ago
3 0

\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>a </u><u>car </u><u>weighing </u><u>5</u><u>0</u><u>0</u><u> </u><u>kg </u><u>takes </u><u>a </u><u>turn </u><u>of </u><u>radius </u><u>5</u><u>0</u><u>m</u><u> </u><u>with </u><u>a </u><u>velocity</u><u> </u><u>of </u><u>3</u><u>6</u><u> </u><u>km/</u><u>hr</u>

  • To calculate - <u>the </u><u>centripetal </u><u>force</u>

  • Formula used -

\bold{F =  \frac{mv {}^{2} }{r}}  \\

where ,

<u>r </u><u>=</u><u> </u><u>radius</u>

<u>m </u><u>=</u><u> </u><u>mass </u><u>of </u><u>the </u><u>object</u>

<u>v </u><u>=</u><u> </u><u>velocity</u><u> </u><u>with </u><u>which </u><u>the </u><u>object </u><u>moves</u>

According to Question ,

<u>Mass </u><u>of </u><u>car </u><u>=</u><u> </u><u>5</u><u>0</u><u>0</u><u> </u><u>kg</u>

<u>Radius </u><u>of </u><u>the </u><u>round </u><u>=</u><u> </u><u>5</u><u>0</u><u> </u><u>m </u>

<u>velocity </u><u>of </u><u>car </u><u>=</u><u> </u><u>3</u><u>6</u><u> </u><u>km/</u><u>hr</u>

velocity = \cancel{36 }\times  \frac{5}{\cancel{18}}  \\  \\ \implies \: velocity = 10 \: ms {}^{ - 1}

<u>Substituting</u><u> </u><u>the </u><u>values</u><u> </u><u>in </u><u>the </u><u>formula</u><u> </u><u>,</u>

F =  \frac{\cancel{500} \times 10 {}^{2} }{\cancel{50}}  \\  \\ \dashrightarrow \: F = 10 \times 10 {}^{2}  \\  \\ \dashrightarrow \: F = 1000 \: N

hope helpful ~

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An insulating sphere is 8.00 cm in diameter and carries a 6.50 µC charge uniformly distributed throughout its interior volume.
Kobotan [32]

Explanation:

(a)   Formula to calculate the density is as follows.

            \rho = \frac{Q}{\frac{4}{3}\pi a^{3}}

                       = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.04)^{3}}

                     = 2.42 \times 10^{-2} C/m^{3}

Now, calculate the charge as follows.

            q_{in} = \rho(\frac{4}{3} \pi r^{3})

                      = 2.42 \times 10^{-2} C/m^{3} \times 4.1762 \times (0.01)^{3}

                      = 10.106 \times 10^{-8} C

or,                   = 101.06 nC

(b)  For r = 6.50 cm, the value of charge will be calculated as follows.

                q_{in} = \frac{Q}{\frac{4}{3}\pi a^{3}}

                          = \frac{6.50 \times 10^{-6}}{\frac{4}{3} \times 3.14 \times (0.065)^{3}}

                          = 7.454 \mu C

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Which process do single-celled organisms go through to reproduce asexually?
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The answer is A-Mitosis
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How does upwelling affect the weather of a coastal region?
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c.The warm surface water results in moist air and more rainfall.

Explanation:

  • During upwelling, cold water in the ocean is stirred up and brought to the surface.
  • The warmer surface water is then taken into deeper parts of the ocean.
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  • The warm surface water causes the air to be moisty.
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  • The air then becomes cold and dry and it rises up.
  • Therefore, warm surface water results in moist air and more rainfall.

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Ocean current brainly.com/question/4117397

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Read 2 more answers
Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is t
den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

F\propto \dfrac{1}{r^2}

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N    

We need to find the electric force between charges if the new separation of 0.25 cm. So,

\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N

So, the new force is 64 N if the separation between charges is 64 N.

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3 years ago
What evidence supports the ages of our planets, moon, sun, and asteroids
Pie

It is indeed true that scientists have known about the background radiation (commonly known as the Cosmic Microwave Background) since the early 60s. It was first discovered quite by accident by Penzias and Wilson working at Bell Labs, who detected it as an unexplainable interference in their precision radio equipment. When people finally figured out exactly what it was they were seeing, they won the Nobel Prize for their discovery. Only a few years before, George Gamow had predicted that if the Big Bang theory were correct, we should observe just such a background radiation. The CMB is not the only evidence in favor of the Big Bang, but it is one of the most important. It is a natural consequence of the theory, and is pretty unexplainable in steady-state cosmology.

The 15-20 billion year number comes not from the CMB, but rather predominantly from measurements of nearby and distant galaxies, particularly their rates of expansion away from us. We find that the distance to a galaxy is proportional to its recessional velocity. The constant of proportionality is the Hubble Constant, H, which turns out to be (approximately) the reciprocal of the age of the universe. So we measure the age by measuring recessional velocities. T = 1/H is only true, however, if the universe is not significantly accelerating or decelerating its expansion rate. If the rate of expansion is rapidly accelerating, the universe may be older than 1/H = 15 billion years, give or take. Such an acceleration would be caused by a large value of the Cosmological Constant, a sort of anti-gravity force predicted by General Relativity. There is some evidence that this might be the case.

So finally, yes, the age of the universe, being based on the empirical determination of H, is based on the observed evidence.

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