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pantera1 [17]
2 years ago
15

A hollow cylinder that is rolling without slipping is given a velocity of 5.0 m/s and rolls up an incline to a vertical height o

f 1.0 m. If a hollow sphere of the same mass and radius is given the same initial velocity, how high vertically does it roll up the incline
Physics
1 answer:
inysia [295]2 years ago
5 0

Answer:

The hollow cylinder rolled up the inclined plane by 1.91 m

Explanation:

From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

M.E_T = \frac{1}{2}mv^2 + \frac{1}{2} I \omega^2 + mgh

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

\frac{1}{2}mv_i^2 + \frac{1}{2} I \omega_i^2 + mg(0) =  \frac{1}{2}mv_f^2 + \frac{1}{2} I \omega_f^2 + mgh

moment of inertia, I, of a hollow cylinder = ¹/₂mr²

substitute for I in the equation above;

\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_i^2) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}mr^2  \omega_f^2) + mgh\\\\ but \ v = r \omega\\\\\frac{1}{2}mv_i^2 + \frac{1}{2} (\frac{1}{2}m v_i^2  ) =  \frac{1}{2}mv_f^2 + \frac{1}{2} (\frac{1}{2}m v_f^2) + mgh\\\\\frac{1}{2}mv_i^2 +\frac{1}{4}mv_i^2 = \frac{1}{2}mv_f^2 +\frac{1}{4}mv_f^2 +mgh\\\\\frac{3}{4}mv_i^2 = \frac{3}{4}mv_f^2 +mgh\\\\mgh = \frac{3}{4}mv_i^2 -  \frac{3}{4}mv_f^2\\\\gh = \frac{3}{4}v_i^2 -  \frac{3}{4}v_f^2\\\\

h = \frac{3}{4g}(v_1^2 -v_f^2)

given;

v₁ = 5.0 m/s

vf = 0

g = 9.8 m/s²

h = \frac{3}{4g}(v_1^2 -v_f^2) =\frac{3}{4*9.8}(5^2 -0) = 1.91 \ m

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m

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Answer:

option C

Explanation:

Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.

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What is relative density?​
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2 years ago
In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it
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Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

                          = 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

5 0
3 years ago
A fisherman is fishing from a bridge and is using a "42.0-N test line." In other words, the line will sustain a maximum force of
lara31 [8.8K]

Answer:

(a) 42 N

(b)36.7 N

Explanation:

Nomenclature

F= force test line (N)

W : fish weight  (N)

Problem development

(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed

We apply Newton's first law of equlibrio because the system moves at constant speed:

∑Fy =0

F-W= 0

42N -W =0  

W = 42N

(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²

We apply Newton's second law because the system moves at constant acceleration:

 m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

∑Fy =m*a

m= W/g , m= W/9.8 ,  m:fish mass , W: fish weight g:acceleration due to gravity

F-W= ( W/9.8 )*a

42-W=  ( W/9.8 )*1.41

42= W+0.1439W

42=1.1439W

W= 42/1.1439

W= 36.7  N

8 0
3 years ago
A 57 kg wagon is pulled with a constant net force of 38 N. Calculate the acceleration of the wagon. F = ma
Dahasolnce [82]

Answer:

<h2>0.67 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{38}{57} =  \frac{2}{3}   \\  = 0.666666...

We have the final answer as

<h3>0.67 m/s²</h3>

Hope this helps you

4 0
2 years ago
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