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
Tom [10]
3 years ago
6

Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applie

d to raise the water. (a) Calculate the pressure needed to raise the water. (b) What is unreasonable about this pressure? (c) What is unreasonable about the premise?
Physics
1 answer:
WITCHER [35]3 years ago
4 0

Answer:

Explanation:

Given

Pipe is lowered to the water h=90\ m

Negative Pressure is applied to raise the water

Pressure is given by

P=\rho gh

where P=pressure

\rho =Density

h=depth

P=10^3\times 9.8\times 90

P=8.82\times 10^{5}\ N/m^2\approx 8.82\ atm

(b)8.82 atm is much lower than the vapor pressure of water

(c)The fact of applying a negative pressure of 8.74 below the vapor pressure of water

You might be interested in
Who invented the telephone?
krok68 [10]
<h2>Answer:</h2><h3><em><u>Alexander Graham Bell</u></em></h3><h2>Explanation:</h2>

Alexander Graham Bell is often credited as the inventor of the telephone since he was awarded the first successful patent.

5 0
2 years ago
Read 2 more answers
A rock has a specific gravity of 2.32 and a volume of 8.64 in3 how much does it weigh
lianna [129]

Answer: 3.21 N

Specific\hspace{1mm} gravity = \frac {Density\hspace{1mm}of\hspace{1mm}substance}{Density\hspace{1mm} of \hspace{1mm}water}

\Rightarrow Density \hspace{1mm}of\hspace{1mm}substance= 2.32\times 1000\hspace{1mm} kg/m^3 = 2320\hspace{1mm}kg/m^3\\ Mass =Density\times volume\\ \Rightarrow 2320 \hspace{1mm} kg/m^3\times 8.64 \hspace{1mm}in^3\times \frac {1.64\times10^{-5} m^3}{1\hspace{1mm}in^3}=0.328 kg

For weight, we will multiply by g=9.8 m/s^{-2}

weight= 0.328\times9.8=3.21\hspace{1mm}N

Hence, the rock would weigh 3.21 N.

3 0
3 years ago
A 2-Newton upward net force is being applied to a 10-kg object. What is the magnitude of the upward
Colt1911 [192]
Acceleration in m/s^2 = 2/10 = 0.2 m/s^2
7 0
3 years ago
Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by
Ulleksa [173]

A) v=\sqrt{\frac{2qV}{m}}

B) r=\frac{mv}{qB}

C) T=\frac{2\pi m}{qB}

D) \omega=\frac{qB}{m}

E) r=\frac{\sqrt{2mK}}{qB}

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

\Delta U = qV

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

\Delta K=\frac{1}{2}mv^2

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

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

And solving for v, we find the speed v at which the particle enters the cyclotron:

v=\sqrt{\frac{2qV}{m}}

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

F=qvB

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

F=m\frac{v^2}{r}

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

qvB=m\frac{v^2}{r}

And solving for r, we find the radius of the orbit:

r=\frac{mv}{qB} (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference (2\pi r) and the velocity of the particle (v):

T=\frac{2\pi r}{v} (2)

From eq.(1), we can rewrite the velocity of the particle as

v=\frac{qBr}{m}

Substituting into(2), we can rewrite the period of revolution of the particle as:

T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

\omega=\frac{2\pi}{T} (3)

where T is the period.

The period has been found in part C:

T=\frac{2\pi m}{qB}

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

v=\frac{qBr}{m}

which can be rewritten as

r=\frac{mv}{qB} (4)

The kinetic energy of the particle is written as

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

So, from this we can find another expression for the velocity:

v=\sqrt{\frac{2K}{m}}

And substitutin into (4), we find:

r=\frac{\sqrt{2mK}}{qB}

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0
3 years ago
A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh
Greeley [361]

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

4 0
3 years ago
Read 2 more answers
Other questions:
  • What voltage would be measured across the 45 ohm resistor?
    9·2 answers
  • How many coulombs of charge are needed to produce 29.1 mol of solid zinc?
    10·1 answer
  • A parallel plate capacitor is charged up by a battery. The battery is then disconnected, but the charge remains on the plates. T
    9·1 answer
  • What is a population?
    13·1 answer
  • A 306 g cart moves on a horizontal, frictionless surface with a constant speed of 14.2 cm/s. A 76.3 g piece of modeling clay is
    7·1 answer
  • You are driving a car at the 25-mi/h speed limit when you observe the light at the intersection 65 m in front of you turn yellow
    7·1 answer
  • Small pockets of synovial fluid that reduce friction and act as a shock absorber where ligaments and tendons rub against other t
    12·1 answer
  • How far can you get away from your little brother with a paintball marker if you can travel at 3 m/s and you have 15s before he
    13·1 answer
  • What is the difference between temperature and heat???
    15·1 answer
  • In a simple 2-bulb series circuit, why does the bulb light when you close the switch?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!