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jolli1 [7]
3 years ago
5

On a vacation flight, you look out the window of the jet and wonder about the forces exerted on the window. Suppose the air outs

ide the window moves with a speed of approximately 150m/s shortly after takeoff, and that the air inside the plane is at atmospheric pressure.(A) Find the pressure difference between the inside and outside of the window.(B) If the window is 25cm by 45cm, find the force exerted on the window by air pressure.
Physics
1 answer:
user100 [1]3 years ago
5 0

Answer:

A) \Delta P =  14512.5 Pa = 14.512 kPa

B) F = 1632.65 N

Explanation:

Given details

outside air speed is given as v_2 = 150 m/s

since inside air is atmospheric , v_1 = 0 m/s

a) By using bernoulli equation between outside and inside of flight

P_1 + \frac{1}{2} \rho v_1^2 + \rho gh = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh

\Delta P = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2

\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]

\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]

\Delta P =  14512.5 Pa = 14.512 kPa

b) force exerted on window

Area of window  = 25\times 45 = 1125 cm^2 = 0.1125 m^2

We know that force is given as

F = P\times A

F = 14512.5 \times 0.1125

F = 1632.65 N

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Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta

\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h

\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h

h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g} (1)

Where:

h - Maximum height of the wood block, in meters.

v - Initial speed of the block, in meters per second.

\mu_{k} - Kinetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

m - Mass, in kilograms.

s - Distance travelled by the wood block along the wooden ramp, in meters.

\theta - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that v = 10\,\frac{m}{s}, \mu_{k} = 0.20 and g = 9.807\,\frac{m}{s^{2}}, then the height reached by the block above its starting point is:

h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

h = 4.249\,m

The wood block reaches a height of 4.249 meters above its starting point.

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C. is responsible for sliding friction and contact forces

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Sorry don't know

Explanation:

Ask google

8 0
2 years ago
Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc
snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

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Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0
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A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
sergij07 [2.7K]

Answer:

0.98kW

Explanation:

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\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

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\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

\dot{W} = \frac{(5)(9.81)(0-20)}{1000}

\dot{W} = -0.98kW

8 0
3 years ago
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