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muminat
2 years ago
11

We wrap a light, nonstretching cable around a 10.0 kg kg solid cylinder with diameter of 38.0 cm cm . The cylinder rotates with

negligible friction about a stationary horizontal axis. We tie the free end of the cable to a 14.0 kg kg block and release the block from rest. As the block falls, the cable unwinds without stretching or slipping.How far will the mass have to descend to give the cylinder 510 J of kinetic energy?
Physics
1 answer:
amm18122 years ago
6 0

Answer: 14.16

Explanation:

Given

d = 38cm

r = d/2 = 38/2 = 19cm = 0.19m

K.E = 510J

m = 10kg

I = 1/2mr²

I = 1/2*10*0.19²

I = 0.18kgm²

When it has 510J of Kinetic Energy then,

510J = 1/2Iω²

ω² = 1020/I

ω² = 1020/0.18

ω² = 5666.67

ω = √5666.67 = 75.28 rad/s

Velocity is the block, v = ωr

V = 75.28 * 0.19

V = 14.30m/s

The "effective mass" M of the system is

M = (14.0 + ½*10.0) kg = 19.0 kg

The motive force would be

F = ma

F = 14 * 9.8

F = 137.2N

so that the acceleration would be

a = F/m

a = 137.2/19

a = 7.22m/s²

Finally, using equation of motion.

V² = u² + 2as

14.3² = 0 + 2*7.22*s

204.49 = 14.44s

s = 204.49/14.44

s = 14.16m

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Answer:

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Explanation:

(a) Just as one-dimensional numbers add on a number line by putting them end-to-end, so two-dimensional numbers add on a coordinate plane the same way.

Here, we choose to let the positive y-axis represent North, and the positive x-axis, East. This is the way a map is conventionally oriented. The velocity of the plane is represented by a vector pointing north (up). Its length represents the magnitude of the velocity. Likewise, the wind is represented by a vector of length 15 pointing east (right). The sum of these is the hypotenuse of the triangle they form.

The magnitude of the sum can be found here using the Pythagorean theorem, but for the purpose of this question, you're not asked to find that.

Instead, you're asked to estimate whether it is more or less than 25 (m/s).

Your knowledge of the triangle inequality will tell you that the hypotenuse (resultant) must be shorter than the sum of the lengths of the sides of the triangle, hence must be less than 10+15 = 25.

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(b) The triangle inequality says the resultant is less than the sum of the other two sides of the triangle.

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4 0
3 years ago
What was the direction of the ball’s velocity
Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

s=ut+\frac{1}{2}at^2

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

a=g=9.8 m/s^2 is the acceleration of gravity

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

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Answer:

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The tension in the cable is

T = ma_R = (1200\text{ kg})(8.37\text{ m/s}^2) = 10044 \text{ N}

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