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julsineya [31]
3 years ago
15

An optical disk drive in a computer can spin a disk at up to 10,000 rpm. If a particular disk is spun at 7570 rpm while it is be

ing read, and then is allowed to come to rest over 0.435 s, what is the magnitude of the average angular acceleration of the disk?
Physics
2 answers:
blagie [28]3 years ago
7 0

Answer:

1822.36\ rad/s^2

Explanation:

given,

speed of the disk = 7570 rpm

time of rest of the disk = 0.435 s

average angular acceleration = ?

initial speed of disk = 7570 rpm

                          = 7570 \times \dfrac{2\pi}{60}

                          = 792.73\ rad/s

final angular velocity = 0 rad/s

average angular acceleration = \dfrac{\omega_f-\omega_i}{t}

                                                 = \dfrac{0 -792.73}{0.435}

                                                 = 1822.36\ rad/s^2

the average angular acceleration = 1822.36\ rad/s^2

anastassius [24]3 years ago
5 0

Answer:

The magnitude of the average angular acceleration is calculated as 1822.36\ rad/s^{2}

Explanation:

Maximum speed that can be attained by the disk, N_{m} = 10,000 rpm

Speed of spinning of the disk, N = 7570 rpm

Time taken to come to rest, t = 0.435 s

Now,

The initial angular velocity is given by:

\omega = \frac{2\pi N}{60} = 792.73\ rads^{-1}

Final angular velocity, \omega' = 0\ rads^{- 1}

The average angular acceleration of the disk can be computed by using the kinematic eqn:

\omega' = \omega + \alpha t

0 = 792.73 + 0.435\alpha

\alpha = - 1822.36\ rads^{- 2}

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konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0
3 years ago
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Alik [6]

Answer:

Momentum, p = 23250 kg m/s

Explanation:

Given that

Mass of a car, m = 1550 kg

Speed pf car, v = 15 m/s

We need to find the momentum of the car. The formula for the momentum of an object is given by :

p = mv

Substituting all the values in the above formula

p = 1550 kg × 15 m/s

p = 23250 kg m/s

So, the momentum of the car is 23250 kg m/s.

3 0
3 years ago
A bowling ball with a momentum of 18kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13k
Veronika [31]

Answer:

14.98\ \text{kg m/s}

45.26^{\circ}

Explanation:

P_1 = Initial momentum of the pin = 13 kg m/s

P_i = Initial momentum of the ball = 18 kg m/s

P_2 = Momentum of the ball after hit

55^{\circ} = Angle ball makes with the horizontal after hitting the pin

\theta = Angle the pin makes with the horizotal after getting hit by the ball

Momentum in the x direction

P_i=P_1\cos55^{\circ}+P_2\cos\theta\\\Rightarrow P_2\cos\theta=P_i-P_1\cos55^{\circ}\\\Rightarrow P_2\cos\theta=18-13\cos55^{\circ}\\\Rightarrow P_2\cos\theta=10.54\ \text{kg m/s}

Momentum in the y direction

P_1\sin55=P_2\sin\theta\\\Rightarrow P_2\sin\theta=13\sin55^{\circ}\\\Rightarrow P_2\sin\theta=10.64\ \text{kg m/s}

(P_2\cos\theta)^2+(P_2\sin\theta)^2=P_2^2\\\Rightarrow P_2=\sqrt{10.54^2+10.64^2}\\\Rightarrow P_2=14.98\ \text{kg m/s}

The pin's resultant velocity is 14.98\ \text{kg m/s}

P_2\sin\theta=10.64\\\Rightarrow \theta=sin^{-1}\dfrac{10.64}{14.98}\\\Rightarrow \theta=45.26^{\circ}

The pin's resultant direction is 45.26^{\circ} below the horizontal or to the right.

4 0
3 years ago
How does a wind turbine transform mechanical energy into electrical energy?
Alex777 [14]

Answer:

"A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator."

4 0
3 years ago
A Carnot engine's operating temperatures are 240 ∘C and 20 ∘C. The engine's power output is 910 W . Part A Calculate the rate of
scoray [572]

Answer:1200

Explanation:

Given data

Upper Temprature\left ( T_H\right )=240^{\circ}\approx 513

Lower Temprature \left ( T_L\right )=20^{\circ}\approx 293

Engine power ouput\left ( W\right )=910 W

Efficiency of carnot cycle is given by

\eta =1-\frac{T_L}{T_H}

\eta =\frac{W_s}{Q_s}

1-\frac{293}{513}=\frac{910}{Q_s}

Q_s=2121.954 W

Q_r=1211.954 W

rounding off to two significant figures

Q_r=1200 W

5 0
3 years ago
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