answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Maslowich
2 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]2 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

You might be interested in
Complete the passage to summarize factors affecting the speed of a wave. The material or substance that a wave moves through is
pantera1 [17]

the first one is medium, the second one is type, and the third one is temperature . if i gave the correct answer, please give best answer x

6 0
2 years ago
Read 2 more answers
A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f
konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

(c) The maximum speed attained by the object during its motion is 0.577 m/s

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m

(c) the maximum speed attained by the object during its motion

E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s

8 0
2 years ago
A pair of glasses is dropped from the top of a 32.0m stadium. A pen is dropped 2.Os later. How high above the ground is the pen
Svetllana [295]

Answer:

h_p = 30.46\ m

Explanation:

<u>Free Fall Motion</u>

A free-falling object refers to an object that is falling under the sole influence of gravity. If the object is dropped from a certain height h, it moves downwards until it reaches ground level.

The speed vf of the object when a time t has passed is given by:

v_f=g\cdot t

Where g = 9.8 m/s^2

Similarly, the distance y the object has traveled is calculated as follows:

\displaystyle y=\frac{g\cdot t^2}{2}

If we know the height h from which the object was dropped, we can solve the above equation for t:

\displaystyle t=\sqrt{\frac{2\cdot y}{g}}

The stadium is h=32 m high. A pair of glasses is dropped from the top and reaches the ground at a time:

\displaystyle t_1=\sqrt{\frac{2\cdot 32}{9.8}}=2.56\ sec

The pen is dropped 2 seconds after the glasses. When the glasses hit the ground, the pen has been falling for:

t_2=2.56 - 2 = 0.56\ sec

Therefore, it has traveled down a distance:

\displaystyle y=\frac{9.8\cdot 0.56^2}{2} = 1.54\ m

Thus, the height of the pen is:

h_p = 32 - 1.54\Rightarrow h_p=30.46\ m

8 0
2 years ago
A 75-hp (shaft output) motor that has an efficiency of 91.0 percent is worn out and is to be replaced by a high- efficiency moto
daser333 [38]

Convert the shaft ouput from HP to kW

Shaft output = 75HP = 55.93kW

 

1st: Finding for the power consumption based on 55.93kW output

Power consumption (Old) = 55.93kW / .910 = 61.46kW

Power consumption (New) = 55.93kW / .954 = 58.63kW

 

2nd: Total power used in kWh:

Power Used = Power consumption * load factor * Hours:

Power (Old) = 61.46kW * .75 * 4368 = 201343 kWh

Power (New) = 58.63kW * .75 * 4368 = 192072 kWh

Energy saved = 201343 kWh - 192072 kWh = 9,271 kWh

 

3rd: Calculating for the price:

Price = kW-Hr * $/kWh

Price (Old) = 201343kWh * $0.08/kWh = $16107.44

Price (New) = 192072 kWh * $0.08/kWh = $15365.76

Cost saved = $16107.44 - $15365.76 =  $741.68/yr

 

4th: Setting up the cost equation:

Cost over time, F(t) = Motor_Cost + (Price * Number of Years, t)

Cost (Old) = 5449 + 16107.44*t

Cost (New) = 5520 + 15365.76*t

Equate the two to find for t when they cost equally:

5449 + 16107.44*t = 5520 + 15365.76*t

16107.44*t = 15365.76*t +71

16107.44*t - 15365.76*t = 71

741.68*t = 71

t = 71 / 741.68 = .095 years = 35 days

So the payback period is after 35 days.

6 0
2 years ago
Your town is installing a fountain in the main square. If the water is to rise 26.0 m (85.3 feet) above the fountain, how much p
Brums [2.3K]

Answer:

P = 3.55 \times 10^5 Pa

Explanation:

As we know that water from the fountain will raise to maximum height

H = 26.0 m

now by energy conservation we can say that initial speed of the water just after it moves out will be

\frac{1}{2}mv^2 = mgH

v = \sqrt{2gH}

v = \sqrt{2(9.81)(26)}

v = 22.6 m/s

Now we can use Bernuolli's theorem to find the initial pressure inside the pipe

P = P_0 + \frac{1}{2}\rho v^2

P = 10^5 + \frac{1}{2}(1000)(22.6^2)

P = 3.55 \times 10^5 Pa

6 0
2 years ago
Other questions:
  • Scientists in a test lab are testing the hardness of a surface before constructing a building. Calculations indicate that the en
    15·1 answer
  • What body process converts physical energy to electrical energy?
    9·1 answer
  • Which statements describe the book and the forces acting on it? Check all that apply. The forces are balanced. The forces are un
    5·2 answers
  • A turtle takes 3.5 minutes to walk 18 m toward the south along a deserted highway. A truck driver stops and picks up the turtle.
    11·1 answer
  • A ball is kicked horizontally at 8.0 m/s from a cliff 80m high. What is the acceleration of the ball in the vertical
    13·1 answer
  • To a cyclist riding due west with a speed of 4 m/s a wind appears to
    9·1 answer
  • Prof. Swarzenneger picks up his 45 kg brief case on the 17th floor of his office building (85 m above the ground) and prepares t
    15·1 answer
  • A student wants to verify a claim that for a given change in momentum, the average applied force decreases as the time interval
    12·1 answer
  • A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
    11·1 answer
  • Modern wind turbines generate electricity from wind power. The large, massive blades have a large moment of inertia and carry a
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!