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Arturiano [62]
2 years ago
9

Electric potential is associated with both electric fields due to static charges and induced electric fields. Electric potential

is associated with magnetic fields but not with electric fields due to static charges. Electric potential is associated with both electric fields due to static charges and magnetic fields. Electric potential is associated with electric fields due to static charges but not with induced electric fields. Electric potential is associated with induced electric fields but not with electric fields due to static charges.
Physics
1 answer:
maksim [4K]2 years ago
4 0

Answer:

Electric potential is associated with electric fields due to static charges but not with induced electric fields.

Explanation:

An electric potential is the amount of work needed to move a unit charge from a reference point to a specific point inside the field without producing an acceleration. Typically, the reference point is the Earth or a point at infinity. Induced electricity is as a result of changing magnetic flux linkage (no charge is involved)

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Answer:

The natural frequency = 50 rad/s = 7.96 Hz

Damping ratio = 0.5

Explanation:

The natural frequency is calculated in this manner

w = √(k/m)

k = spring constant = 5 N/m

m = mass = 2 g = 0.002 kg

w = √(5/0.002) = 50 rad/s

w = 2πf

50 = 2πf

f = 50/(2π) = 7.96 Hz

Damping ratio = c/[2√(mk)] = 0.1/(2 × √(5 × 0.002)) = 0.5

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A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
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Answer:

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v(t) = (-9.8m/s^2)*t + v0

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v(t) = (-9.8m/s^2)*t

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p(t) = (-4.9m/s^2)*t^2+ 20m

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The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) =  (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

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