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DaniilM [7]
2 years ago
7

Keeping the applied voltage constant at approximately 0.70 volts, measure the electric field strength at different points betwee

n the parallel plates. To do this, you have to move the handle of the electric field detector. What do you observe about the electric field strength and the direction of the electric field
Physics
1 answer:
Vitek1552 [10]2 years ago
3 0

Answer: I observed that the electric field strength is the same at all points between the plates. The value of the field is 70 volts per meter. This is exactly 100 times the applied voltage. The electric field lines point from the positive plate to the negative plate, as the downward arrow on the detector shows.

Explanation:

the sample answer, don't directly copy it!

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Vectors are arrows that tell us two things about force...WEIGHT and DIRECTION
nexus9112 [7]

Answer:

False only direction not weight.

Explanation:

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How much charge passes through a wire in 4.0 s if the current is 3.0 A?
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12 coloumbs

Explanation:

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How many times higher could an astronaut jump on the Moon than on Earth if his takeoff speed is the same in both locations (grav
JulsSmile [24]

Answer:

maximum height on moon is 6 times more than the maximum height on Earth

Explanation:

Let the Astronaut has its maximum speed by which he can jump is "v"

now for the maximum height that it can jump is given as

v_f^2 - v_i^2 = 2 aH

now from above equation we will have

0 - v^2 = 2(-g)H

now we have

H = \frac{v^2}{2g}

now if Astronaut jump on the surface of moon with same speed

then we know that the acceleration of gravity on surface of moon is 1/6 times the gravity on earth

so at surface of moon we have

0 - v^2 = 2(-g/6) H

now we have

H = \frac{6v^2}{2g}

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8 0
3 years ago
. A mass m is traveling at an initial speed of 25.0 m/s. It is brought to rest in a distance of 62.5 m by a net force of 15.0 N.
harkovskaia [24]

Answer:

m = 3 kg

The mass m is 3 kg

Explanation:

From the equations of motion;

s = 0.5(u+v)t

Making t thr subject of formula;

t = 2s/(u+v)

t = time taken

s = distance travelled during deceleration = 62.5 m

u = initial speed = 25 m/s

v = final velocity = 0

Substituting the given values;

t = (2×62.5)/(25+0)

t = 5

Since, t = 5 the acceleration during this period is;

acceleration a = ∆v/t = (v-u)/t

a = (25)/5

a = 5 m/s^2

Force F = mass × acceleration

F = ma

Making m the subject of formula;

m = F/a

net force F = 15.0N

Substituting the values

m = 15/5

m = 3 kg

The mass m is 3 kg

7 0
3 years ago
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