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hammer [34]
3 years ago
6

A pitot tube indicates a pressure of 155 kPa when placed in an air stream in which the temperature is 15°C and the Mach number i

s 0.7. Find the static pressure in the flow. Also, find the stagnation temperature in the flow.
Physics
1 answer:
mel-nik [20]3 years ago
5 0

To solve this problem we will start by defining the relationship of the pitot tube and the value indicated in the pressure. We will take advantage of this relationship to obtain a ratio between the defined pressures and proceed to calculate the total and static pressure of the system through the correlation of pressure in isotropic systems. Finally we will perform this same procedure for pressure. Pitot tube measure the total pressure minus static pressure, then

p_0 - p = 155kPa

\frac{p_0}{p} -1 = \frac{155}{p}

\frac{p_0}{p} = \frac{155}{p} +1

Express the total pressure and static pressure ratio

\frac{p_0}{p} = [1+\frac{\gamma-1}{2}M^2]^{\frac{\gamma}{\gamma-1}}

\frac{155}{p} +1 = [1+\frac{1.4-1}{2}(0.7)^2]^{\frac{1.4}{1.4-1}}

\frac{155}{p} +1=1.3871

p = 400.4kPa

Hence the static pressure is 400.4kPa

Express stagnation to static temperature ratio,

\frac{T_0}{T} = 1+\frac{\gamma-1}{2}M^2

\frac{T_0}{(15+273)K} = 1+\frac{1.4-1}{2}(0.7)^2

T_0 = 316.224K

Therefore the stagnation temperature is 316.224K

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As a new electrical technician, you are designing a large solenoid to produce a uniform 0.170 T magnetic field near its center.
lord [1]

Answer:

Explanation:

Magnetic field = permeability x turn density x current

Magnetic field = 0.22T

turn density = 4150 /1.6 = 2593.75 t/m

permeability : µ = k µ°

µ°= 4 π 10^-7

k = 1

I = 0.22 / 4 π 10^-7 * 2593.75 = 0.22 10^7 /32594 = 67.497 A

7 0
3 years ago
A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t
White raven [17]

Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

        v = w r

in the case of the spool the angular velocity for the whole system is constant,

They indicate the linear velocity v₀ = 25.0 cm / s for a radius of r₀ = 1.00 cm,

         w = v₀ /r₀

for the outside of the spool r₁ = 1.5 cm

         w = v₁ / r₁1

since the angular velocity is the same we set the two expressions equal

        \frac{v_o}{r_o} = \frac{v_1}{r_1}

        v1 = \frac{r_1}{r_o} \ \ v_o

let's calculate

       v₁ = \frac{1.50}{1.00} \ \ 25.0

       v₁ = 37.5 cm / s

4 0
3 years ago
Please help. I don’t understand this
skad [1K]

The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is

1/2 • (10.0 m/s) • (4.0 s) = 20.00 m

Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as

<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>

and under constant acceleration,

<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2

According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so

∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2

∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2

∆<em>x</em> = 20.00 m

5 0
3 years ago
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa
Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

5 0
3 years ago
Two observers in different inertial reference frames moving relative to each other at nearly the speed of light see the same two
lions [1.4K]

Answer:

The correct answer is d Both the observer's are correct

Explanation:

We know by postulates of relativity that laws of physics are same in different inertial frames.

Thus for each of the frames they make observations related to their frames and since the observations are true for their individual frames they both are correct. But when we compare the two frames we need to use transformation equations to compare both the results.

3 0
3 years ago
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