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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?

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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3 years ago

A lawyer drives from her home, located 1 mile east and 8 miles north of the town courthouse, to her office, located 4 miles w

givi [52]

Answer:

d = 13 miles

Explanation:

Lets say the position of court house is origin in this case

her office is located at 4 miles west and 4 miles south of court house

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$r_1 = (-4\hat i - 4\hat j)$

now the position of her home is located at 1 miles east and 8 miles north of the court house

so the coordinates of her home is given as

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3 years ago

Differences between freezing point and melting point (Atleast 5 differences)

labwork [276]

Answer:

What is freezing point?

A liquid's freezing point is determined at which it turns into a solid. Corresponding to the melting point, the freezing point often rises with increasing pressure. In the case of combinations and for some organic substances, such as lipids, the freezing point is lower than the melting point. The first solid which develops when a combination freezes often differs in composition from the liquid, and the development of the solid alters the composition of the remaining liquid, typically lowering the freezing point gradually. Utilizing successive melting and freezing to gradually separate the components, this approach is used to purify mixtures.

What is melting point?

The temperature at which a purified substance's solid and liquid phases may coexist in equilibrium is referred to as the melting point. A solid's temperature goes up when heat is added to it until the melting point is achieved. The solid will then turn into a liquid with further heating without changing temperature. Additional heat will raise the temperature of the liquid once all of the solid has melted. It is possible to recognize pure compounds and elements by their distinctive melting temperature, which is a characteristic number.

The difference between freezing point and melting point:

While a substance's melting point develops when it transforms from a solid to a liquid, a substance's freezing point happens when a liquid transforms into a solid when the heat from the substance is removed.
When the temperature rises, the melting point can be seen, and when the temperature falls, the freezing point can be seen.
When a solid reaches its melting point, its volume increases; meanwhile, when a liquid reaches its freezing point, its volume decreases.
While a substance's freezing point is not thought of as a distinctive attribute, its melting point is.
While external pressure is a significant component in freezing point, atmospheric pressure is a significant element in melting point.
Heat must be supplied from an outside source in order to reach the melting point for such a state shift. When a material is at its freezing point, heat is needed to remove it from the substance in order to alter its condition.

<em>Reference: Berry, R. Stephen. "When the melting and freezing points are not the same." Scientific American 263.2 (1990): 68-75.</em>

7 0

1 year ago

Flapping flight is very energy intensive. A wind tunnel test

steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

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1 kg=1000 g

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Speed, v=11 m/s

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We know that

Metabolic power for starting flight= $\frac{P}{m}$

Using the formula

Metabolic power for starting flight= $\frac{12}{0.089}$

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

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3 years ago

A sock stuck to the inside of the clothes dryer spins around the drum once every 2.0 s at a distance of 0.50 m from the center o

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a) 1.57 m/s

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T = 2.0 s

Therefore, the angular velocity of the sock is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2.0}=3.14 rad/s$

The linear speed of the sock is given by

$v=\omega r$

where

$\omega$ is the angular velocity

r = 0.50 m is the radius of the circular path of the sock

Substituting, we find:

$v=(3.14)(0.50)=1.57 m/s$

B) Faster

In this case, the drum is twice as wide, so the new radius of the circular path of the sock is twice the previous one:

$r' = 2r = 1.00 m$

At the same time, the drum spins at the same frequency as before, therefore the angular frequency as not changed:

$\omega' = \omega = 3.14 rad/s$

Therefore, the new linear speed would be:

$v'=\omega' r' = \omega (2r)$

And substituting,

$v'=(3.14)(1.00)=3.14 rad/s = 2v$

So, we see that the linear speed has doubled.

8 0

2 years ago