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Sloan [31]
2 years ago
12

If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

lar speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the CD, as it spins to a stop?
Physics
2 answers:
gavmur [86]2 years ago
8 0

Answer:

Explanation:

Given that,

Initial Angular velocity w=500rpm

Converting from rpm to rad/s

1rev =2πrad

1minutes =60secs

500rpm=500rev/mins

w = 500×2π/60

wi=52.36rad/s

The final angular velocity wf=0rad/s

Time to stop is t=2.6sec

We want to find angular acceleration α

Using the equation of angular motion

wf = wi + αt.

0 = 52.36 + 2.6α

-52.36=2.6α

α = -52.36/2.6

α = -20.14rad/s²

The angular acceleration is negative because it is decelerating.

Then, α=20.14rad/s²

Tems11 [23]2 years ago
8 0

Answer:

Explanation:

Given:

Initial angular velocity, wi = 500 rpm

Converting from rom to rad/s,

500 rev/min × 2pi rad/1 rev × 1 min/60s

= 52.36 rad/s

Final angular velocity, wf = 0 rad/s

Time, t = 2.6 s

Using equations of angular motion,

wf = wi + αt

α = (52.36 - 0)/2.6

= 20.138 rad/s^2

= 20.14 rad/s^2

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A worker is thinking about two ways to get a box up 1.2 m onto a loading dock. He can use a force of 250 N to lift it straight u
harina [27]

Answer:

<em>The second option has a lower power output. P=30 W</em>

Explanation:

<u>Mechanical Power </u>

It is a physical magnitude that measures the rate a work W is done over time t.

\displaystyle P=\frac{W}{t}

Since W=F.d

\displaystyle P=\frac{F.d}{t}

The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of

\displaystyle P=\frac{(250). (1.2)}{3}=100\ Watt

The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is

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The second option has a lower power output

7 0
3 years ago
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

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3 years ago
Joe Burrow of the Cincinnati Bengals ate a turkey sandwich and an apple for lunch. Later that day, he spent 2 hours on the footb
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Answer:

A Thermal energy was converted to kinetic energy

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2 years ago
Please help
Snezhnost [94]

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

Range=v_i^2\frac{sin(2\theta)}{g}

which for our case renders:

Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m

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2 years ago
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