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1 year ago

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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.

1 answer:

matrenka [14]1 year 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

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A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

$F_s = \mu N$

where $\mu$ is the coefficient of friction and $N$ is the normal force.

When the student pulls on the block with force $F_p$ at an angle $\theta$ , the normal force on the block becomes

$N = Mg- F_psin(\theta)$

and hence the frictional force becomes

$F_s = \mu (Mg- F_psin(\theta))$ .

Now, as we increase $\theta$ , $sin(\theta)$ increases which as a result decreases the normal force $Mg- F_psin(\theta)$ , which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em> $W=mg$ <em> is independent of the external forces on the block. </em>

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2 years ago

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Which of the following objects have gravitational potential energy?

Brut [27]

Answer:

The apple

Explanation:

The apple has gravitational potential energy because it is just sitting there but nothing is pushing it up, it is not sitting on something. This means that it has a lot of gravitational potential energy as nothing is pushing it up.

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2 years ago

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What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?

PtichkaEL [24]

Answer:

$\tau=3.3*10^{-6}s$

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

$C*\frac{dV}{dt}+\frac{V}{R}=0$

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

$\frac{dV}{V}=-\frac{1}{RC}dt$

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

$\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt$

Evaluating the integrals:

$ln(\frac{V}{v})=e^{\frac{-t}{RC} }$

natural logarithm to both sides in order to isolate V:

$V(t)=ve^{-\frac{t}{RC} }$

Where the term RC is called time constant and is given by:

$\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s$

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2 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

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I hope it helps you!

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Rudiy27

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