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3 years ago

9

Briefly explain how an electric motor works.

1 answer:

NeTakaya3 years ago

3 0

You put electricity<span> into it at one end and an </span>axle<span> (metal rod) rotates at the other end giving you the power to drive a machine of some kind.

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Can y’all help me with 5 plsssss

Alona [7]

Answer:

Bar graph

Explanation:

each day collects data so a bar graph would work.

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3 years ago

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The components of vector A are Ax = +2.2 and Ay = -6.9 , and the components of vector B are given are Bx = -6.1 and By = -2.2. W

Zina [86]

For simplicity, let's call vector B-A vector C Then C is
Cx = (-6.1 - 2.2)
Cy = (-2.2 - (-6.9)) Or,
Cx = -8.3 Cy = 4.7
The magnitude is found with the Pythagorean theorem
||C|| = √(-8.3² + 4.7²) = 9.538

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3 years ago

How does the suns energy contribute to the carbon cycle

netineya [11]

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Plants are a good starting point when looking at the carbon cycle on Earth. Plants have a process called photosynthesis that enables them to take carbon dioxide out of the atmosphere and combine it with water. Using the energy of the Sun, plants make sugars and oxygen molecules.

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3 years ago

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

3 years ago

A 0.700-kg ball is on the end of a rope that is 2.30 m in length. The ball and rope are attached to a pole and the entire appara

otez555 [7]

Answer:

The tangential speed of the ball is 11.213 m/s

Explanation:

The radius is equal:

$r=2.3*sin70=2.161m$ (ball rotates in a circle)

If the system is in equilibrium, the tension is:

$Tcos70=mg\\Tsin70=\frac{mv^{2} }{r}$

Replacing:

$\frac{mg}{cos70} sin70=\frac{mv^{2} }{r} \\Clearing-v:\\v=\sqrt{rgtan70}$

Replacing:

$v=\sqrt{2.161x^{2}*9.8*tan70 } =11.213m/s$

7 0

3 years ago