4 years ago

II

Physics

1 answer:

Kamila [148]4 years ago

3 0

Answer:

wait 1 2 3 and then i jump and gravity falls

Explanation:

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What is the centripetal acceleration of the tip of the fan blade?

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4 years ago

An object with a mass of 78 kg is lifted through a height of 6 meters how much work is done

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We got the following:
m = 78kg
h = 6m
g = 9.8 m/ $sec^{2}$ ≈ 10 m/ $sec^{2}$
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A = ?

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3 years ago

A bob of mass of 0.18 kilograms is released from a height of 45 meters above the ground level. What is the value of the kinetic

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4 years ago

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A 50.0 g toy car is released from rest on a frictionless track with a vertical loop of radius R (loop-the-loop). The initial hei

Mariana [72]

Answer:

the speed of the car at the top of the vertical loop $v_{top} = 2.0 \sqrt{gR \ \ }$

the magnitude of the normal force acting on the car at the top of the vertical loop $F_{N} = 1.47 \ \ N$

Explanation:

Using the law of conservation of energy ;

$mgh = mg (2R) + \frac{1}{2}mv^2_{top}\\\\mg ( 4.00 \ R) = mg (2R) + \frac{1}{2}mv^2_{top}\\\\g(4.00 \ R) = g (2R) + \frac{1}{2}v^2 _{top}\\\\v_{top} = \sqrt{2g(4.00R - 2R)}\\\\v_{top} = \sqrt{2g(4.00-2)R$

$v_{top} = 2.0 \sqrt{gR \ \ }$

The magnitude of the normal force acting on the car at the top of the vertical loop can be calculated as:

$F_{N} = \frac{mv^2_{top}}{R} \ - mg\\\\F_{N} = \frac{m(2.0 \sqrt{gR})^2}{R} \ - mg\\\\F_{N} = [(2.0^2-1]mg\\\\F_{N} = [(2.0)^2 -1) (50*10^{-3} \ kg)(9.8 \ m/s^2]\\\\$