1answer.
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
ArbitrLikvidat [17]
3 years ago
8

At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is

150 ft/s. determine the pressure at a stagnation
Physics
1 answer:
Nastasia [14]3 years ago
5 0
At a point on the streamline, Bernoulli's equation is
p/ρ + v²/(2g) = constant
where
p = pressure
v = velocity
ρ = density of air, 0.075 lb/ft³ (standard conditions)
g = 32 ft/s²

Point 1:
p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft²
v₁ = 150 ft/s

Point 2 (stagnation):
At the stagnation point, the velocity is zero.

The density remains constant.
Let p₂ = pressure at the stagnation point.
Then,
p₂ = ρ(p₁/ρ + v₁²/(2g))
p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²)
     = 314.37 lb/ft²
     = 314.37/144 = 2.18 lb/in²

Answer: 2.2 psi

You might be interested in
These two magnets are held close together, as shown. What directions will they move once released?
frozen [14]
I think the magnets will move towards each other due to the opposite poles being attracted to each other
3 0
2 years ago
An electric dipole is formed from ± 5.0 nC point charges spaced 3.0 mm apart. The dipole is centered at the origin, oriented alo
Ymorist [56]

Answer:

The electric field strength at point (x,y) = ( 20 mm ,0cm) is =<u>16321.0769 N/C</u>

The electric field strength at point (x,y) = (0cm, 20 mm) is =<u>35321.58999 N/C</u>

Explanation:

Question: What is the electric field strength at point (x,y) = ( 20 mm ,0cm)?

Answer:

The electric field at any given point of the dipole is given as:

E= (KP) ÷ (r^2 + a^2)^3/2

Where:

K = 9x10^9 Nm^2/c^2 (coloumb constant)

P = (0.003) (5x10^-9c) which is the movement of the dipole

(0.003) is arrived at when mm is converted to m. 3.0 mm space apart was converted to a meter.

r= the point, in the question above is 20mm = 0.02m

Now, the electric field, E can be calculated by putting the values in the formula above:

E = (KP) ÷ (r^2 + a^2)^3/2

= (9x10^9 Nm^2/c^2) (0.003 m) (5x10^-9c) ÷ [ (0.02m)^2 + (0.003)^2]^3/2

= 0.135 ÷ (8.271513x10^-6)

=<u>16321.0769 N/C</u>

 Question: What is the electric field strength at point (x,y) = (0cm, 20 mm )?

Answer:

Here, the electric field, E= 2krp ÷ (r^2 - a^2)^2

E= 2 (9x10^9 Nm^2/c^2) (0.02m) (0.003 m) (5x10^-9c) ÷ [(0.02m)^2 - (0.003)^2]^2

= 0.0054 ÷  0.000000152881

=<u>35321.58999 N/C</u>

8 0
2 years ago
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
6. Which of these contain muscles that are not under the mind's control? (Select all that apply.)
Arada [10]

Answer:

a   c

Explanation:

edu 2021

3 0
3 years ago
A 1.5 kg ball pushed with a force of 13.5 N accelerates to the left. What is the acceleration of the ball?​
GarryVolchara [31]

Explanation:

F= ma

13.5 =1.5a

1.5a/1.5 =13.5/1.5

a= 9 m/s^2

4 0
3 years ago
Other questions:
  • A race-car drives around a circular track of radius RRR. The race-car speeds around its first lap at linear speed v_iv i ​ v, st
    15·1 answer
  • The image above shows a block pulled across a table. The spring scale which measures in Newtons, reads the force resisting the m
    5·1 answer
  • A wooden log of density 2g/cm3 and volume 50 cm3 fall on the surface of water. Calculate its buoyant force.
    14·1 answer
  • How do you find weight =mass×gravitational acceleration
    5·2 answers
  • A car with a mass of 860 kg sits at the top of a 32 meter high hill. Describe the transformations between potential and kinetic
    11·1 answer
  • If you measure a change in temperature, then the _____.
    7·1 answer
  • 1. Calculate the work done by a 47 N force pushing a pencil 0.26 m.<br> 47x .26 = 12J<br> Solve it
    5·1 answer
  • Why do you think lightning is so dangerous if it strikes a person?
    8·1 answer
  • What element conducts heat and electricity:
    6·2 answers
  • What is the displacement if a person traveled from point B to E?
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!