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nignag [31]
4 years ago
9

A bug is on the rim of a 78 rev/min, 12 in. diameter record. The record moves from rest to its final angular speed in 2.67 s. Fi

nd the bug’s centripetal acceleration 1.5 s after the bug starts from rest. (1 in = 2.54 cm).
Physics
1 answer:
viva [34]4 years ago
3 0

Answer:

6.41m/s2

Explanation:

unit conversion:

78 rev/min = 78rev/min * 2π rad/rev * 1/60 min/sec = 8.168 rad/s

12 in = 12 in * 2.54cm/in = 30.48 cm = 0.3048 m

So the record accelerate from rest to 8.168 rad/s, its (assuming constant) angular acceleration would be

\alpha = \Delta \omega / t = (8.168 - 0) / 2.67 = 3.06 rad/s^2

From here we can calculate the angular velocity of the bug at t2 = 1.5s

\omega_2= \alpha*t_2 = 3.06 * 1.5 = 4.59 rad/s

Its centripetal acceleration would then be

a = \omega_2^2*R = 4.59^2*0.3048 = 6.41 m/s^2

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x ≈ 56 m

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t- time is found solving quadratic equation.

horizontal velocity = v_{0x}=25m/s*cos(30^{o})=21.65 m/s

Horizontal velocity is constant, so distance x=v_{0x}*t =21.65 m/s *2.584 s=55.9 = 56 m

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4 years ago
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3 years ago
The sun appears to move through the background stars. This apparent motion would not exist if:_________.
otez555 [7]

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b) the earth did not orbit the sun

Explanation:

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A billiard ball is dropped from a height of 64 feet. Use the position function s(t) = –16???? 2 + ????0???? + ????0 to answer th
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Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

a(t) = -32 ft/s^2

v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

Find:-

a. Determine the position function s(t),

b. the velocity function v(t),

c. the acceleration function a(t).

d. What is the velocity of the ball at impact?

Solution:-

- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

                                  a(t) = d v(t) / dt

                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

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