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Lorico [155]
3 years ago
5

Why polarization does not occur in dry cell?​

Physics
1 answer:
Arisa [49]3 years ago
7 0

Answer:

Manganese oxide prevents polarisation in dry cells. - Polarization is a defect that occurs in simple electric cells due to the accumulation of hydrogen gas around the positive electrode. ... - MnO2 reacts with H2 and forms water as byproduct, so depolarization doesn't occur.

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Why are water vapor and aerosols important constituents of earth's atmosphere? choose all that apply.
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They are important because numerous serve as areas where water vapour can condense.

<h3></h3><h3>What is condensation?</h3>

The transformation of water vapour into liquid is known as condensation. The process is the opposite of evaporation, in which liquid water turns into a vapour. Either the air is chilled to its dew point or it gets too saturated with water vapour to retain any more water, causing condensation to occur.

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the reason the moon does not crash into the earth is that the group of answer choices moon has less mass than the earth. moon ha
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Two sound waves of equal amplitude interference so that compression of one wave falls on the rarefaction of the other. Which sta
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<span>A. No sound is heard 
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A ball is thrown at 25 m/s at an angle of 55 degrees.
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Explanation:just had the quiz

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1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of
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(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

  • constant velocity of the elevator, u₁ = 10 m/s
  • initial velocity of the ball, u₂ = 20 m/s
  • height of the boy above the elevator floor, h₁ = 2 m
  • height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\  gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s

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3 years ago
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