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Olin [163]
3 years ago
7

What parts are found in an electric generator? Check all that apply.

Physics
2 answers:
Blizzard [7]3 years ago
6 0

Answer:

an armature a permanent magnet brushes slip rings

Explanation:

Snowcat [4.5K]3 years ago
4 0

Answer:

armature, permanent magnet, brushes, slip rings on edge:-)

Explanation:

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Which describes the electric field at the center of a ring of uniform negative charge?.
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A negatively charged particle -q is placed at the center of a uniformly charged ring, where the ring has a total positive charge Q as shown in the following figure. The particle, confined to move along the x axis, is moved a small distance x along the axis ( where x << a) and released. Show that the particle oscillates in simple harmonic motion with a frequency given by,

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4 years ago
The box shown on the rough ramp above is sliding up the ramp.calculate the force of friction on the box
Anna [14]

We are given a box that slides up a ramp. To determine the force of friction we will use the following relationship:

F_f=\mu N

Where.

\begin{gathered} N=\text{ normal force} \\ \mu=\text{ coefficient of friction} \end{gathered}

To determine the Normal force we will add the forces in the direction perpendicular to the ramp, we will call this direction the y-direction as shown in the following diagram:

In the diagram we have:

\begin{gathered} m=\text{ mass}_{} \\ g=\text{ acceleration of gravity} \\ mg=\text{ weight} \\ mg_y=y-\text{component of the weight. } \end{gathered}

Adding the forces in the y-direction we get:

\Sigma F_y=N-mg_y

Since there is no movement in the y-direction the sum of forces must be equal to zero:

N-mg_y=0

Now we solve for the normal force:

N=mg_y

To determine the y-component of the weight we will use the trigonometric function cosine:

\cos 40=\frac{mg_y}{mg}

Now we multiply both sides by "mg":

mg\cos 40=mg_y

Now we substitute this value in the expression for the normal force:

N=mg\cos 40

Now we substitute this in the expression for the friction force:

F_f=\mu mg\cos 40

Now we substitute the given values:

F_f=(0.2)(10\operatorname{kg})(9.8\frac{m}{s^2})\cos 40

Solving the operations:

F_f=15.01N

Therefore, the force of friction is 15.01 Newtons.

3 0
1 year ago
Which geologic feature would most likely be represented by contour lines drawn far apart from one another?
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