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yarga [219]
3 years ago
12

You’ve set up a frictionless car racetrack, with a loop of radius 0.5 m. Your toy car of mass 2 kg starts at rest at the top of

a ramp and is released so that it slides down and approaches the loop. What condition must be present so that the car just completes the loop? How high must you make the ramp in order to complete the loop?
Physics
1 answer:
Anna007 [38]3 years ago
4 0

Answer

h = 1.12 m

Explanation:

given,

mass of the car = 2 kg

radius = 0.5 m

let h be the height of the ramp

when the car reaches at the top point

gravity = centripetal force  

mg =\dfrac{mv^2}{r}

g =\dfrac{v^2}{r}

v = \sqrt{gr}

using conservation of energy

\PE_i = PE_f + KE_f

m g h = m g (2r)+\dfrac{1}{2}mv^2

g h =g (2r)+\dfrac{1}{2}(\sqrt{gr})^2

g h =2 gr +\dfrac{gr}{2}

h =\dfrac{5r}{2}

h =\dfrac{5\times 0.5}{2}

h = 1.12 m

hence, height of the ramp should be greater than  1.12 m so that it can complete the loop.

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A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

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Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

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8 0
2 years ago
How long will it take a 2190 W motor to lift a 1.47 x 104 g box, 6.34 x 104 mm vertically.​
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Answer:

t = 4.17 [s]

Explanation:

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where:

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In order to find the force, we must determine the weight of the box, the weight can be determined by means of the product of mass by gravitational acceleration.

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