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Art [367]
3 years ago
9

What is voltage and current?

Physics
2 answers:
MrMuchimi3 years ago
4 0

Voltage - An electromotive force or potential difference expressed in volts.

Current - A flow of electricity which results from the ordered directional movement of electrically charged particles.

vladimir1956 [14]3 years ago
4 0
Current is a rate at which electric charge flows past a point in a circuit. Voltage is the potential difference in charge between two points in an electrical field.
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At a location near the equator, the earth’s magnetic field is horizontal and points north. An electron is moving vertically upwa
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Answer:

(b) EAST

Explanation:

you can assume that the magnetic field points rightward, that is, in the positive x direction (NORTH). Furthermore, you can assume that the direction of the motion of the electron is in the positive y direction. Hence, you have:

\vec{B}=B_o\hat{i}\\\\\vec{v}=v_o\hat{j}

You use the Lorentz formula to known which is the direction of the magnetic force over the electron:

F=qv\ X\ B

which implies the cross product between the unitary vecors j and i, that is

\hat{i} \ X\ \hat{j} = -\hat{k}  (WEST)

However, the minus sign of the charge of the electron changes the direction 180°. Hence, the direction is k. That is, to the EAST

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The Beams And Joints That Hold It .
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Carol has several test tubes of different liquids. If Carol removes heat from the liquid substances, which of the following is m
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d

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What causes competition among organisms?
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Both organisms attempt to use the same limited sources
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In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
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