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geniusboy [140]

1 year ago

12

A hollow cylinder is given a velocity of 5.3 m/s and rolls up an incline to a height of 2.87 m. If a hollow sphere of the same m

ass and radius is given the same initial velocity, how high (in m) does it roll up the incline

1 answer:

Valentin [98]1 year ago

5 0

Answer:

E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2

E = 1/2 M V^2 (1 + I / (M R^2))

For a cylinder I = M R^2

For a sphere I = 2/3 M R^2

E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2

E(sphere) = 1 + 2/3 = 1.67

E(cylinder) / E(sphere) = 2 / 1.67 = 1.2

The cylinder initially has 1.20 the energy of the sphere

The PE attained is proportional to the initial KE

H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE

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Answer:

Part a)

$H = 44.1 m$

Part b)

$y = 13.48 m$

Part c)

$d = 8.86 m$

Explanation:

Part a)

As we know that ball will reach at maximum height at

t = 3 s

now we will have

$t = \frac{v sin\theta}{g}$

now we have

$3 = \frac{vsin\theta}{9.8}$

$v sin\theta = 29.4 m/s$

Now maximum height above ground is given as

$H = \frac{v^2sin^2\theta}{2g}$

$H = \frac{29.4^2}{2(9.8)}$

$H = 44.1 m$

Part b)

Height of the fence is given as

$y = (vsin\theta) t - \frac{1}{2}gt^2$

$y = (29.4)(5.5) - \frac{1}{2}(9.8)(5.5^2)$

$y = 13.48 m$

Part c)

As we know that its horizontal distance moved by the ball in 5.5 s is given as

$x = v_x t$

$97.5 = v_x (5.5)$

$v_x = 17.72 m/s$

now total time of flight is given as

$T = 3 + 3 = 6 s$

so range is given as

$R = v_x T$

$R = (17.72)(6)$

$R = 106.4 m$

so the distance from the fence is given as

$d = 106.4 - 97.5$

$d = 8.86 m$

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The 4kg head of a sledge hammer is moving at 6m/s when it strikes a chisel, driving it into a log. The duration of the impact (o

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