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
kipiarov [429]
3 years ago
8

A garden hose is attached to a water faucet on one end and a spray nozzle on the other end. The water faucet is turned on, but t

he nozzle is turned off so that no water flows through the hose. The hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of 0.78 m.
What is the pressure inside the hose?
Physics
1 answer:
GarryVolchara [31]3 years ago
5 0

To solve this problem it is necessary to use the concepts related to pressure depending on the depth (or height) in which the object is on that fluid. By definition this expression is given as

P = P_{atm} +\rho gh

Where,

P_{atm} = Atmospheric Pressure

\rho =Density, water at this case

g = Gravity

h = Height

The equation basically tells us that under a reference pressure, which is terrestrial, as one of the three variables (gravity, density or height) increases the pressure exerted on the body. In this case density and gravity are constant variables. The only variable that changes in the frame of reference is the height.

Our values are given as

P_{atm} = 1.013*10^5Pa

\rho = 1000Kg/m^3

g = 9.8m/s^2

h = 0.78m

Replacing at the equation we have,

P = P_{atm} +\rho gh

P = 1.013*10^5 +(1000)(9.8)(0.78)

P = 108944Pa

P = 0.1089Mpa

Therefore the pressure inside the hose is 0.1089Mpa

You might be interested in
An object is released from rest and falls in free fall motion. The speed v of the object after it has fallen a distance y is giv
konstantin123 [22]

Answer:

8.91 %

Explanation:

Since v² = 2gy

By the relative error formula,

2Δv/v = Δg/g + Δy/y multiplying by 100%, we have

2Δv/v × 100% = Δg/g × 100 % + Δy/y × 100%

2(Δv/v × 100%) = Δg/g × 100 % + Δy/y × 100%

Δg/g × 100 % = 2(Δv/v × 100%) - Δy/y × 100%

Since Δv/v × 100% = 3.69 % and Δy/y × 100% = 5 %

Since we have a difference for the percentage error in g, we square the percentage errors and add them together. So,

[Δg/g × 100 %]² = [2(Δv/v × 100%)]² + [Δy/y × 100%]²

[Δg/g × 100 %]² = [2(3.69)]² + [5%]²

[Δg/g × 100 %]² = [4)(3.69 %)² + [5%]²

[Δg/g × 100 %]² = 54.4644 %² + 25%²

[Δg/g × 100 %]² = 79.4644 %²

taking square-root of both sides, we have

[Δg/g × 100 %] = 8.91 %

So, the percent uncertainty in the calculated value of g is 8.91 %

6 0
3 years ago
What are two parts of an atom
seraphim [82]

Answer:

Nucleus And electron cloud

Explanation:

Hope this helps

7 0
3 years ago
18 points! Brainliest! Physics!
Naddik [55]
You will use the Pythagorean Theorem to solve it.
c^2 = a^2 + b^2
c^2 = (1.5)^2 + (2)^2
c^2 = 6.25
c = square root of 6.25
c = 2.5
I hope this helps!
3 0
3 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
Read 2 more answers
You have a 16-kg suitcase. what work did you do when you have slowly lifted it 0.80 m upward?
Anna71 [15]
The formula is:
Work = Force · Displacement
F = m · g
F = 16 kg · 9.8 m/s²  = 156.8 N
and we know that:
d = 0.8 m
W = 156.8 N · 0.8 m = 125.44 J
Answer:
W = 125.44 J.
8 0
3 years ago
Other questions:
  • What are some ways you can vary your tone of voice to help you communicate effectively with others? A. Speaking aggressively
    11·2 answers
  • A jogger runs at an average speed of 4.20 mi/h. (a) how fast is she running in m/s? (report your answer to the correct number of
    9·1 answer
  • If Earth's obliquity was 157 degrees, would the seasons be more severe, less severe, or about the same?
    15·1 answer
  • How the forces makes effect on the state of motion of an object ? <br><br> Plz ans fast urgent
    14·1 answer
  • A copper rod at 25°C is 2.5 m long. How long would it take a sound to move through the rod from one end to another? how would I
    10·2 answers
  • A seed can be round when parents are crossed, scientists refer to the first generation of offspring as
    8·1 answer
  • A force does work on an object if a component of the force:a. is perpendicular to the displacement of the object b. is parallel
    6·1 answer
  • 1.Convert 340 cm into m *(answer=0.34m)
    12·1 answer
  • A force of 40 N accelerates a 5 block at 6 m/s ^ 2 along a horizontal surface a. What would the block's acceleration be if the s
    10·1 answer
  • The question and answer options are in the photo!
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!