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tatyana61 [14]
3 years ago
6

A computer disk drive is turned on starting from rest and has a constant angular acceleration. If it took 0.750 seconds for the

drive to make its second revolution,
1. How long did it take to make the first complete revolution?
2. What is its angular acceleration in rad/s2?
Physics
1 answer:
Irina-Kira [14]3 years ago
6 0

Answer:

a) Θ = ω₀*t + ½αt² To complete first revolution 2π rads = 0*t + ½αt² and to complete the first and second combined 4π rads = 0*t + ½α(t+0.810s)² Divide second by first: 2 = (t + 0.810s)² / t² This is quadratic in t and has roots at t = -0.336 s ← ignore and t = 1.96 s ◄ b) Use either equation from above: 2π rads = 0*t + ½α(1.96s)² α = 3.27 rad/s² ◄ Hope this helps!

Explanation:

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yulyashka [42]

Answer:

Lenz's law, in electromagnetism, statement that an induced electric current flows in a direction such that the current opposes the change that induced it. This law was deduced in 1834 by the Russian physicist Heinrich Friedrich Emil Lenz (1804–65).

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2 years ago
Які лампочки у схемі з'єднані<br> паралельно?
alekssr [168]

Answer:

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5 0
3 years ago
(TCO 5) The relationship between Celsius (º C) and Fahrenheit (º F) degree of measuring temperature is linear. Find the linear e
gtnhenbr [62]

Answer:

C=-\dfrac{10}{17}(F-32)

Explanation:

Given that

32° F corresponds to 0 °C.  ---Point 1

212° F corresponds to 100 °C.----Point 2

We know that if two point is given that equation of line can be found as

y-y_1=\dfrac{y_2-y_1}{x_2-x_2}(x-x_1)

Lets C in y- direction and F in x- direction,so we can say that

C-C_1=\dfrac{C_2-C_1}{F_2-F_2}(F-F_1)

C-0=\dfrac{100-0}{32-212}(F-32)

C=-\dfrac{10}{17}(F-32)

So the linear relationship is

C=-\dfrac{10}{17}(F-32)

8 0
3 years ago
How far will a car travel in 30 min at 40 m/s? (kilometres)
kondor19780726 [428]

Answer:

72km

Explanation:

30 mins --> 30 x 60 s = 1800 s

Distance --> Speed x Time

= 40m/s x 1800s

= 72 000 m

= 72 km (1km is 1000m)

6 0
3 years ago
Ben rushin is waiting at a stoplight. when it finally turns green, ben accelerated from rest at a rate of a 6.00 m/s2 for a time
jasenka [17]

In the 4.10 seconds that elapsed, Ben reaches a velocity of

v_f=v_0+at\implies v_f=0\,\dfrac{\mathrm m}{\mathrm s}+\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)(4.10\,\mathrm s)

\implies v_f=24.6\,\dfrac{\mathrm m}{\mathrm s}

In this time, his displacement \Delta x satisfies

{v_f}^2-{v_0}^2=2a\Delta x\implies\left(24.6\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(6.00\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x

\implies\Delta x=50.4\,\mathrm m

4 0
3 years ago
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