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nikklg [1K]
3 years ago
14

An airplane is cruising at an altitude of several kilometers. The pressure outside the craft is 0.258 atm within the passenger c

ompartment the pressure is 1.00 atm and the temperature is 20∘C. The density of air is 1.20 kg/m3 at 20∘C and 1 atm of pressure. A small leak occurs in one of the window seals in the passenger compartment. Model the air as an ideal fluid to find the speed of the stream of air flowing through the leak.

Physics
1 answer:
garik1379 [7]3 years ago
8 0

Answer:

<em>The speed of the stream of air flowing through the leak is 340.754 m/sec</em>

Explanation:

Carefully applying the Bernoulli's equation the speed of the leak can be obtained. The attached images show step by step explanation of the question, while applying the Bernoulli's equation;

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So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

We have an applied force of 400 N to the right, this means that:
<span>The magnitude of the net force is 400, directed to the right.</span>
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3 years ago
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A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?
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Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19×\frac{1}{2\pi} *60 revolution per min

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Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

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Juli2301 [7.4K]

Answer:

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Explanation:

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For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

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Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

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Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

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Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

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