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Maslowich
4 years ago
5

At a given location the airspeed is 22 m/s and the pressure gradient along the streamline is 105 N/m3. Estimate the airspeed at

a point 0.5 m farther along the streamline. First write the equation for the pressure gradient along the streamline. If neglest gravity,
Physics
1 answer:
scoray [572]4 years ago
8 0

Answer

given,

Air speed, v = 20 m/s

Pressure gradient, = 105 N/m³

distance,dl = 0.5 m

density of air = 1.23 kg/m³

Airspeed at 0.5 m = ?

relation between pressure gradient and speed changed

-\frac{\partial p}{\partial l}-\gamma sin\theta = \rho_{air}v\frac{\partial v}{\partial l}

neglecting the gravity

-\frac{\partial p}{\partial l}-0 = \rho_{air}v\frac{\partial v}{\partial l}

-105= 1.23\times 20\times \frac{\partial v}{\partial l}

\frac{\partial v}{\partial l} = -4.26

\partial v = -4.26 \times \partial l

\partial v = -4.26 \times 0.5

\partial v = -2.13\ m/s

Speed of the air at 0.5 m

V = 20 - 2.13

V = 17.87 m/s

Hence, the speed of air is equal to V = 17.87 m/s

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pickupchik [31]

Answer:

T = 0.607 seconds

Explanation:

Given:

Mass, M = 1.50 × 10⁻² kg

Radius, R = 5.50 × 10⁻² m

Now,

the time period in terms of moment of inertia is given as:

T = 2\pi\sqrt\frac{I}{mgR}    .....................1

where, T is the time period

g is the acceleration due to gravity

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Now,

Moment of inertia, I is given as:

I = \frac{5mR^{2}}{3}

on substituting the moment of inertia in the equation 1, we get

T = 2\pi\sqrt\frac{\frac{5mR^{2}}{3}}{mgR}

or

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on substituting the valeus, we get

T = 2\pi\sqrt\frac{{5\times5.50\times10^{-2}}}{3\times9.8}

or

T = 0.607 seconds

Hence, the time period is 0.607 seconds

5 0
3 years ago
Only living things have energy T or F?
Angelina_Jolie [31]

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. living things only have energy : true.

8 0
3 years ago
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A 26-cm-long wire with a linear density of 20 g/m passes across the open end of an 86-cm-long open-closed tube of air. If the wi
damaskus [11]

Answer: T = 472.71 N

Explanation: The wire vibrates thus making sound waves in the tube.

The frequency of sound wave on the string equals frequency of sound wave in the tube.

L= Length of wire = 26cm = 0.26m

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Length of open close tube = 86cm = 0.86m

Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as

L = 3λ/4

0.86 = 3λ/4

3λ = 4 * 0.86

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λ = 3.44/3 = 1.15m.

Speed of sound in the tube = 340 m/s

Hence to get frequency of sound, we use the formulae below.

v = fλ

340 = f * 1.15

f = 340/ 1.15

f = 295.65Hz.

f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.

The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as

L = λ/2

0.26 = λ/2

λ = 0.52m

The speed of sound in string is given as v = fλ

Where λ = 0.52m f = 295.65 Hz

v = 295.65 * 0.52

v = 153.738 m/s.

The velocity of sound in the string is related to tension, linear density and tension is given below as

v = √(T/u)

153.738 = √T/ 0.02

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153.738² = T / 0.02

T = 153.738² * 0.02

T = 23,635.372 * 0.02

T= 472.71 N

3 0
3 years ago
A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .
Usimov [2.4K]

The body moves at a velocity of 1.62m/s after the bullet emerges.

<h3>Given:</h3>

Mass of bullet, m_1 = 22g

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Mass of the block, m_2 = 1.9 kg

Velocity of bullet , v_1 = 265 m/s

v_2 = 0

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

v^{'}_1 =125 m/s

v^{'}_2=?

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

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Therefore, It moves with a velocity of 1.62 m/s.

Learn more about momentum here:

brainly.com/question/25121535

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inn [45]

Answer:

False

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Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

8 0
3 years ago
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