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3 years ago

Using no more than 3 sentences, explain how a motor and generator are related.

1 answer:

3 0

The Electric Motor and Generator are differentiated on various factors like the main principle of working or function of the motor and generator. Consumption or production of electricity, its driven element, the existence of the current in the winding. Fleming’s rule followed by the motor and generator.

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Has a definite shape and definite volume. Question 4 options: liquid gas solid

ad-work [718]

Answer:

solid

Explanation:

solid

3 0

3 years ago

14*3(36/2(1*3)/6<br><img src="https://tex.z-dn.net/?f=14%20%5Ctimes%203%2836%20%5Cdiv%202%281%20%5Ctimes%203%29%20%5Cdiv%206" id

GarryVolchara [31]

<h2>10.5</h2><h3>remember pemdas</h3>

parentheses
exponents <em>excluded for this problem</em>
multiplication
division
addition
subtraction

<h3>step 1. start with what is at the top of the list.</h3>

parentheses

<h3>step 2. do 1 times 3 since it comes before division, and is in parentheses.</h3>

1 × 3 = 3

<h3>step 3. find 36 divided by 2</h3>

36 ÷ 2 = 18

<h3>step 4. add the values together</h3>

18 + 3 = 21

<h3>step 5. find 14 times 3</h3>

14 × 3 = 42 <em>you can also do 7 × 6 and will get the same result because 7 is half of 14 and 3 is half of 6</em>

<h3>step 6. add what is outside the parentheses</h3>

21 + 42 = 63

<h3>step 7. divide by 6</h3>

63 ÷ 6 = 10.5 $-- >$ 10 remainder of 3 <em>remainder means left over</em>

<em />

7 0

2 years ago

A distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 n

maksim [4K]

Answer:

A) receding from the earth

B) $3.078x10^6m/s$

Explanation:

A) receding from the earth

The wavelength went from 434.1nm to 438.6nm, there was an increase in wavelength (also knowecn as redshift due to the doppler efft), this increase is due to the fact that the source that emits the radiation (the distant galaxy) is moving away and therefore the light waves it emits are "stretched", causing us to see a wavelength greater than the original.

B) $3.078x10^6m/s$

to calculate the relative speed we use the following formula:

$v_{rel}=c(1-\frac{\lambda_{1}}{\lambda_{2}} )$

where $c$ is the speed of light: $c=3x10^8m/s$

$\lambda_{1}$ is the wavelength emited by the source, and

$\lambda_{2}$ is the wavelength measured on earth.

we substitute all the values and do the calculations:

$v_{rel}=(3x10^8m/s)(1-\frac{434.1nm}{438.6nm} )\\\\v_{rel}=(3x10^8m/s)(1-0.98974)\\\\v_{rel}=(3x10^8m/s)(0.01026)\\\\v_{rel}=3.078x10^6m/s$

the relative speed is: $3.078x10^6m/s$

5 0

3 years ago

How does moon changes position every day?

photoshop1234 [79]

It moves relative to the stars

3 0

4 years ago

Please help!!

pantera1 [17]

Answer:

(a) The time the ball stays in the air is approximately 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters

Explanation:

The given parameters of the question are;

The initial velocity at which a baseball cannon fires balls, u = 29 m/s

The angle at which the cannon is tilted, θ = 52°

(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;

$2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}$

Where;

t = The time it takes the baseball to reach maximum height

2·t = The total time of flight = The time the ball stays in the air

θ = The angle at which the ball is tilted = 52°

g = The acceleration due to gravity ≈ 9.81 m/s²

u = The initial velocity = 29 m/s

Therefore, we have;

$2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s$

The time the ball stays in the air, 2·t ≈ 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;

$Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}$

∴ The distance from the cannon at which the ball will hit the ground = R

$Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m$

The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters

$H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}$

Therefore, we have;

$H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m$

The maximum altitude of the ball ≈ 26.62 m

3 0

3 years ago