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ahrayia [7]
3 years ago
11

The maximum pressure variations the human ear can withstand are about how much above and below atmospheric pressure?

Physics
1 answer:
Crank3 years ago
4 0

The maximum pressure variations the human ear can withstand  above and below atmospheric pressure is around 30 pa. the normal atmospheric pressure is around 101325 pa. hence the variation in the maximum pressure for human ear is very small as compared to the atmospheric pressure. if the ear is exposed to a pressure greater than this , it can cause permanent damage to the ear.

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What is the weight of a dog with a mass of 12 kg? Explain how you obtained your answer.
butalik [34]

Answer: 117.60N

Explanation:

Weight is a force. Therefore, we can use the force formula to find weight.

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A 0.500-kg block slides up a plane inclined at a 30° angle. If it slides 1.50 m before coming to rest while encountering a frict
julia-pushkina [17]
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3 years ago
Help me find the acceleration
ANEK [815]

a = 3.09 m/s²

<h3>Explanation</h3>

This question doesn't tell anything about how long it took for the car to go through 105 meters. As a result, the <em>timeless </em>suvat equation is likely what you need for this question.

In the <em>timeless</em> suvat equation,

a = \dfrac{v^2 - u^2}{2\; x}

where

  • a is the acceleration of the car;
  • v is the <em>final</em> velocity of the car;
  • u is the <em>initial</em> velocity of the car; and
  • x is the displacement of the car.

Note that <em>v</em> and <em>u</em> are velocities. Make sure that you include their signs in the calculation.

In this question,

  • a is the unknown;
  • v = -10.9 \; \text{m} \cdot \text{s}^{-2};
  • u = -27.7 \; \text{m} \cdot \text{s}^{-2}; and
  • x = - 105 \; \text{m}.

Apply the <em>timeless</em> suvat equation:

a = \dfrac{v^{2} - u^{2}}{2\; x}\\\phantom{a} = \dfrac{(-10.9)^{2} - (-27.7)^{2}}{2 \times (-105)}\\\phantom{a} = 3.09 \; \text{m} \cdot \text{s}^{-2}.

The value of a is greater than zero, which is reasonable. Velocity of the car is negative, meaning that the car is moving backward. The car now moves to the back at a slower speed. Effectively it accelerates to the front. Its acceleration shall thus be positive.

7 0
3 years ago
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