Could you please explain the step by step process of setting up this problem?

A flow is isentropically expanded to supersonic speeds in a convergent-divergent nozzle. The reservoir and exit pressures are 1.

Kamila [148]

Answer:

Ae/A* = 1.115

Explanation:

Let the reservoir pressure be $p_0$

Let the exit pressure be $p_e$

Ratio of reservoir pressure and exit pressure

$\frac{p_o}{p_e} = \frac{1}{0.3143}$

= 3.182

For the above value of pressure ratio

Obtain the area ratio from the isentropic flow table

Ae/A* = 1.115

The value of pressure ratio is Ae/A* = 1.115

6 0

3 years ago

For each situation, identify when sound would travel faster and why?

Rudiy27

A. Outside on a summer day, there are less particles that the sound bounces off of (snow and wind absorb the sound)

B. Water, sound travels faster through denser objects. Water has a higher density than air.

C. High air pressure, sound moves faster through matter that is closer together. This can be said because sound can't travel through space. (There is no atmosphere and consistent particles in space for sound to go through)

D. A piece of steel, It's denser than wood and water.

E. 90% nitrogen and 10% helium, nitrogen is denser than helium so it will move faster

8 0

3 years ago

Emily takes a trip, driving with a constant velocity of 87.5 km/h to the north except for a 22 min rest stop. If Emily's average

rjkz [21]

<h2>Answer:</h2>

2.63 hours

<h2>Explanation:</h2>

First we have to read carefully the problem, It start telling us that Emily was driving with average velocity of 87.5 km/h to the north. Velocity is a physical variable that is definded in terms of :

Speed of the object (87.5 km/h)

Direction of the object (north)

Them the problem says that Emily's trip has an average velocity of 76.8 km/h to the north because she made a rest stop for 22 minutes.

Now you can ask yourself: If Emily was driving at 87.5 km/h, why the average speed of the trip was 76.8 km/h?

we need to see how speed is defined:

$Speed=\frac{distance}{time} =\frac{x}{y}$

***Note 1:

Distance will be called as "x" and units will be kilometers(km)

Time will be called as "y" and units will be hours(h)

So, 87.5 km/h would be the average speed for a distance "x" and a time "y" without stops. But Emily made a stop that took 22 minutes. For the same distance Emily's trip took more time and time is in the denominator. If our numerator is constant and our denominator gets higher, the final result will be lower

Now, we have the follow expressions:

$87.5=\frac{x}{y}~~(1) \\76.8=\frac{x}{y+\frac{22}{60}}~~(2)$

***Note 2:

we need to convert 22 minutes to hours. 1 hour= 60 minutes, so we need to apply the follow covert factor:

$y(hours)=y(minutes)*\frac{1(hour)}{60(minutes)}$

The problem aks us about how long does the trip take, it means that we have to find y.

We have two variables (x and y), and we have two equationts. We know that x have the same value for both problems (Because both average speeds have the same distance), so we can solve both equations for x and made equal each other

$87.5*y=x~(3)\\76.8*(y+\frac{22}{60})=x~(4)$

$87.5*y=76.8*(y+\frac{22}{60})~(5)$

We have to expand (5) and then we have to solve for y

$87.5*y=76.8*y+76.8*\frac{22}{60}~~(6)\\87.5*y-76.8*y=76.8*\frac{22}{60}~~(7)\\10.7*y=76.8*\frac{22}{60}~~(8)\\y=\frac{76.8}{10.7}*\frac{22}{60}(hours)~~(9)\\y=2.63(hours)~~(10)$

3 0

3 years ago

Two small plastic spheres are given positive electrical charges. When they are 15.0 cm apart, the repulsive force between them h

Tems11 [23]

Answer:

a) q_1=q_2= 7.42*10^-7 C

b) q_2= 3.7102*10^-7 C , q_1 = 14.8*10^-7 C

Explanation:

Given:

F_e = 0.220 N

separation between spheres r = 0.15 m

Electrostatic constant k = 8.99*10^9

Find: charge on each sphere

part a

q_1 = q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_1^2 = F_e*r^2/k

q_1=q_2= sqrt (F_e*r^2/k)

Plug in the values and evaluate:

q_1=q_2= sqrt (0.22*0.15^2/8.99*10^9)

q_1=q_2= 7.42*10^-7 C

part b

q_1 = 4*q_2

Using coulomb's law:

F_e = k*q_1*q_2 / r^2

q_2^2 = F_e*r^2/4*k

q_2= sqrt (F_e*r^2/4*k)

Plug in the values and evaluate:

q_2= sqrt (0.22*0.15^2/4*8.99*10^9)

q_2= 3.7102*10^-7 C

q_1 = 14.8*10^-7 C

4 0

3 years ago

Read 2 more answers

If the value of the resistor r2 were doubled, how would the value of the resistor r3 have to change in order to keep the current

Reika [66]

This is a Wheatstone bridge, and the ratio of R2 to R1 equals the ratio of Rx to R3. As a result, if R2 is increased, R3 should be reduced by a factor of two.

<h3>Explain Wheatstone bridge?</h3>

A Wheatstone bridge is a type of electrical circuit that is used to measure an unknown electrical resistance by balancing two legs of a bridge circuit, one of which contains the unknown component.

The Wheatstone bridge circuit can be used to compare an unknown resistance RX to others of known value, such as R1 and R2, which have constant values and R3 which can be variable.

If we connected a voltmeter, ammeter, or galvanometer between points C and D, and then changed resistor R3 until the meters read zero, the two arms would be balanced, and the value of RX (substituting R4) would be known as indicated.

To learn more about Wheatstone bridge refer to :

brainly.com/question/15225070

#SPJ4

5 0

1 year ago