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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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A box is placed on a table. If an upward pulling force on the box is slightly less than the weight of the box, what will happen
aniked [119]

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The rate (in cubic feet per hour) that a spherical snowball melts is proportional to the snowball's volume raised to the 2/3 pow
Darina [25.2K]

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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

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Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

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g = G \cdot M/ r^{2} <em>(equation 4)</em>

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Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

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Answer:

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b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

3 0
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