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vodomira [7]
2 years ago
7

An ear squeeze occurs when: Select one: The pressure inside the middle ear space is greater than ambient (surrounding) pressure.

The pressure inside the middle ear space is less than ambient (surrounding) pressure. Divers forget to wear ear plugs. None of the above.
Physics
1 answer:
matrenka [14]2 years ago
6 0

B. An ear squeeze occurs when the pressure inside the middle ear space is less than ambient (surrounding) pressure.

<h3>When external ear squeeze occurs</h3>

External ear squeeze occurs when gas trapped in the external ear canal remains at atmospheric pressure while the external water pressure increases during descent.

Thus, we can conclude that, an ear squeeze occurs when the pressure inside the middle ear space is less than ambient (surrounding) pressure.

Learn more about ear squeeze here: brainly.com/question/11430998

#SPJ1

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sineoko [7]

Answer:

I think its B but I may be wrong

3 0
3 years ago
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A light wave moves from diamond (n= 2.4) into water (n= 1.33) at an angleof 24°. what angle does it have in water?
worty [1.4K]

Answer:

n1 sin θ1 = n2 sin θ2      Snell's Law       (θ1 is the angle of incidence)

sin θ2 = n1 / n2 * sin θ1

sin θ2 = 2.4 / 1.33 * sin θ1

sin θ2 = 1.80 * .407 = .734

θ2 = 47.2 deg

3 0
2 years ago
Can someone help me?!!!!!
ladessa [460]

Answer:

magnitude: 21.6; direction: 33.7 degrees

Explanation:

When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have

vector: (-3,-2)

Scalar: -6

so the vector multiplied by the scalar will have components

(-3\cdot (-6), -2 \cdot (-6))=(18,12)

The magnitude is given by Pythagorean's theorem:

m=\sqrt{18^2+12^2}=21.6

and the direction is given by the arctan of the ratio between the y-component and the x-component:

\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}

3 0
3 years ago
A steel ball rolls with a constant velocity on a tabletop 0.950 m high it rolls off and hit the ground 0.352 m from the edge of
sp2606 [1]

Answer:

0.799 m/s if air resistance is negligible.

Explanation:

For how long is the ball in the air?

Acceleration is constant. The change in the ball's height \Delta h depends on the square of the time:

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t,

where

  • \Delta h is the change in the ball's height.
  • g is the acceleration due to gravity.
  • t is the time for which the ball is in the air.
  • v_0 is the initial vertical velocity of the ball.
  • The height of the ball decreases, so this value should be the opposite of the height of the table relative to the ground. \Delta h = -0.950\;\text{m}.
  • Gravity pulls objects toward the earth, so g is also negative. g \approx -9.81\;\text{m}\cdot\text{s}^{-2} near the surface of the earth.
  • Assume that the table is flat. The vertical velocity of the ball will be zero until it falls off the edge. As a result, v_0 = 0.

Solve for t.

\displaystyle \Delta h = \frac{1}{2} \;g\cdot t^{2} + v_0\cdot t;

\displaystyle -0.950 = \frac{1}{2} \times (-9.81) \cdot t^{2};

\displaystyle t^{2} =\frac{-0.950}{1/2 \times (-9.81)};

t \approx 0.440315\;\text{s}.

What's the initial horizontal velocity of the ball?

  • Horizontal displacement of the ball: \Delta x = 0.352\;\text{m};
  • Time taken: \Delta t = 0.440315\;\text{s}

Assume that air resistance is negligible. Only gravity is acting on the ball when it falls from the tabletop. The horizontal velocity of the ball will not change while the ball is in the air. In other words, the ball will move away from the table at the same speed at which it rolls towards the edge.

\begin{aligned}\text{Rolling Velocity}&=\text{Horizontal Velocity} \\&= \text{Average Horizontal Velocity}\\ &=\frac{\Delta x}{\Delta t}=\frac{0.352\;\text{m}}{0.440315\;\text{s}}=0.0799\;\text{m}\cdot\text{s}^{-1}\end{aligned}.

Both values from the question come with 3 significant figures. Keep more significant figures than that during the calculation and round the final result to the same number of significant figures.

3 0
3 years ago
Which expression is equivalent to 5(2+7)? 0 2(5+7) O 2 + 7(5) O 5(2)+7 O 5(2)+5(7)​
professor190 [17]

Answer:

5(2)+5(7)

Explanation:

5(2)+5(7)=10+35=45

6 0
3 years ago
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