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3 years ago

What is the average access time for a hard disk spinning at 360 revolutions per second with a seek time of 10 milliseconds?

Physics

1 answer:

antoniya [11.8K]3 years ago

3 0

As the hard disk makes 360 revolutions per second, so one full rotation requires 1/360 sec = 0.002778 sec = 2.778 ms.

On average, however, we expect to require half a rotation. Therefore, we need 1.389 ms. The second part of the movement depends on the seek time which is 10 ms. Therefore, the total time is 1.389+10=11.389 ms on average.

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13. Direct instruction takes place when

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A skill set is explicitly taught or an activity is completed

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4 years ago

Please help me , thank you

mash [69]

Answer:

Since the reading wasn't specified, it would be most likely A

Explanation:

A is the most similar to a protoplanetary disk, so it'd be A most likely

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3 years ago

A nerve impulse travels along a myelinated neuron at 90.1 m/s.<br> What is this speed in mi/h?

algol [13]

Answer:

201.5537 mph

Explanation:

Given the following data;

Speed = 90.1 m/s

Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.

Mathematically, speed is given by the formula;

Speed = distance/time

To convert this value into miles per hour;

Conversion;

1 meter = 0.000621 mile

90.1 meters = 90.1 * 0.000621 = 0.05595 miles

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90.1 m/s = 201.5537 mph

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3 years ago

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Arte-miy333 [17]

Answer:

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3 years ago

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o

Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

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2 years ago