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elena-14-01-66 [18.8K]

3 years ago

12

A windmill on a farm rotates at a constant speed and completes one-half of a rotation in 0. 5 seconds. What is its rotation spee

d?

1 answer:

vesna_86 [32]3 years ago

6 0

Answer:

The angular velocity is the rotation speed, which is the angle of rotation

of the windmill per second, which is 2·π radians.

Explanation:

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Finding dead insects or mouse droppings in the kitchen indicates which of the following.

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Finding dead insects or mouse droppings in the kitchen is an indication of pest infestation.

Pest infestation is the occurrence of one or more pest species in an area or location where their numbers and impact are currently or potentially at intolerable levels. It can also be defined as a sudden increase in the population numbers of a pest species in a given area.

Finding dead insects or mouse droppings in the kitchen is an indication of pest infestation.

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A common motif in the poetry of this unit is death

Viefleur [7K]

Various poets address the element of death in their poems by referring to it as something dark.

<h3>How poets address the element of death</h3>

Various poets address the element of death in their poems by mentioning something that relates to dark.

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Thus, poets address the element of death in their poems by referring to it as something dark.

The complete question is below:

A common motif in the poetry of this unit is death. How do the various poets address the element of death?

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2 years ago

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of len

Citrus2011 [14]

Answer:

The additional words is $2^n$

Explanation:

Given

$B(n) = 2^n$

Required

Determine the additional words; i.e. $B(n + 1) - B(n)$

From the given parameters, we have that;

B is a function of n

Such that;

$B(n) = 2^n$

To calculate $B(n+1)$ , we simply substitute n + 1 for n

$B(n) = 2^n$

$B(n + 1) = 2^{n + 1}$

Applying laws of indices

$B(n + 1) = 2^{n} * 2^1$

$B(n + 1) = 2^{n} * 2$

$B(n + 1) = 2(2^{n})$

Calculating Additional Binary Code;

$B(n + 1) - B(n)$

Substitute values for B(n + 1) and B(n)

$B(n + 1) - B(n) = 2(2^n) - 2^n$

Express $2^n$ as $2^ n * 1$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 1$

Express 1 as $2^0$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 2^0$

Factorize

$B(n + 1) - B(n) = 2^n(2 - 2^0)$

$B(n + 1) - B(n) = 2^n(2 - 1)$

$B(n + 1) - B(n) = 2^n(1)$

$B(n + 1) - B(n) = 2^n$

Hence, the additional words is $2^n$

8 0

3 years ago