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3 years ago

Select all the facts about a motor

1 answer:

vladimir1956 [14]3 years ago

4 0

Well, it would do you well to number the answers, but Electric input, Mechanical output, and Magnetic forces turn the loop.

You'll have to trust me on this one.

You might be interested in

What is the magnitude of a point charge that produces a potential of -200V at a distance of 1.00 mm?

schepotkina [342]

Answer:

$q=-2.22*10^{-11}C$

Explanation:

The potential produces by a point charge is given by:

$V=\frac{kq}{r}$

Here, k is the Coulomb constant, q is the signed magnitude of the point charge and r is the distance between the charge and the point at which the electric potential is measured. Solving for q:

$q=\frac{rV}{k}\\q=\frac{1*10^{-3}m(-200V)}{8.99*10^9\frac{V\cdot m}{C}}\\q=-2.22*10^{-11}C$

5 0

3 years ago

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

5 0

4 years ago

A sound wave has a speed of 345 m/s and a wavelength of 4.15 meters. What is the frequency of this wave?

Flauer [41]

D i believe. bdbdbdbdbdbndndd

6 0

2 years ago

An archer pulls back the string of a bow to release an arrow at a target. Which kind of potential energy is transformed to cause

Ugo [173]

Elastic Potential Energy because the elasticity in the string stores up the energy.

3 0

4 years ago

Read 2 more answers

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago