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
vfiekz [6]
3 years ago
13

You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m

agnetic system by modeling it inthe laboratory. The safety system is a conducting bar that slides on two parallel conducting rails thatrun down the ramp (similar to the one in the previous problem). The bar is perpendicular to the railsand is in contact with them. At the bottom of the ramp, the two rails are connected together. The box slides down the rails through a uniform vertical magnetic field. The magnetic field is supposed to causethe bar to slide down the ramp at a constant velocity even when friction between the bar and the rails negligible.Before setting up the laboratory model, your task is to calculate the constant velocity of the bar (sliding down the ramp on rails in a vertical magnetic field) as a function of the mass of the bar, the strength ofthe magnetic field, the angle of the ramp from the horizontal, the length of the bar (which is the same asthe distance between the rails), and the resistance of the bar. Assume that all of the other conductors inthe system have a much smaller resistance than the bar.a) If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.b)Write out the units for each of your variables and show (by cancellation and substitution) that the units for your veloctiy will be m/s on both left and right side of your equation.
Physics
1 answer:
Fofino [41]3 years ago
8 0

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

You might be interested in
A kayakeris paddling 2.50 m/s at an angle of 45° (northeast) and the current is moving 1.25 m/s at an angle of 315° (southeast).
PIT_PIT [208]

The kayaker has velocity vector

<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )

<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )

and the current has velocity vector

<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )

<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )

The kayaker's total velocity is the sum of these:

<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>

That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that

tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s)   →   <em>θ</em> ≈ 18.4º

or about 18.4º north of east.

5 0
3 years ago
What kind of radiation is emitted in the following nuclear reaction
miv72 [106K]

Beta emission is occurring in the given nuclear reaction.

Answer: Option B

<u>Explanation:</u>

In this equation, the reactant is the Thorium atom, which is reduced to palladium. As the atomic number get decreased by one, so an electron will be emitted. This process of emission of electrons by radiation or decaying the reactant nuclei to form a new product nuclei is termed as beta emission.

So, the electrons are generally termed as beta particles while the positrons are termed as positive beta particles. So this is a kind of radioactive reactions where the reactant changes to new element by releasing an electron and thus there is a change in the atomic number of the product by one.

4 0
3 years ago
Then, complete the riddle below by finding the matching number and writing the letter.
liq [111]

Answer:

I have no clue I'm just trying to get points

Explanation:

:) sorry

3 0
2 years ago
A thin partition divides a thermally insulated vessel into a lower compartment of volume V and an upper compartment of volume 2V
Anni [7]

Answer:

1.89mol

Explanation:

The entropy change during free expansion is express as

S_{f}-S_{i}=nRln(\frac{V_{F}}{V_{I}})\\

Where S is the entropy of the system,

            n is the amount of mole

             R is the gas constant = 8.314 and

           V is the volume occupied at the initial and final stage

since the process is n adiabatic free expansion, the entropy of the system is constant. Hence we can re-write the equation as

S=nRln(\frac{V_{F}}{V_{I}})\\

where the  V_{i}=v\\ and V_{f}=2v+v=3v\\

S=17.28J/k\\ and

R=8.314\\

Now if we substitute in values we arrive at

17.28=(8.314)n*ln(\frac{3v}{v} )\\17.28=(8.314)n*ln(3 )\\17.28=(8.314)n*1.0986\\n=\frac{17.28}{8.314*1.0989}\\n=1.89 mole\\

6 0
3 years ago
A student stretches an elastic band by 0.8 m in 0.5 seconds. The spring constant of the elastic band is 40 N/m. What was the pow
miv72 [106K]

Answer:

The power exerted by the student is 51.2 W

Explanation:

Given;

extension of the elastic band, x = 0.8 m

time taken to stretch this distance, t = 0.5 seconds

the spring constant, k = 40 N/m

Apply Hook's law;

F = kx

where;

F is the force applied to the elastic band

k is the spring constant

x is the extension of the elastic band

F = 40 x 0.8

F = 32 N

The power exerted by the student is calculated as;

P = Fv

where;

F is the applied force

v is velocity = d/t

P = F x (d/t)

P = 32 x (0.8 /0.5)

P = 32 x 1.6

P = 51.2 W

Therefore, the power exerted by the student is 51.2 W

3 0
3 years ago
Other questions:
  • What two factors determine how fast weathering occurs
    9·1 answer
  • PLEASE ANSWER ASAP!!
    14·2 answers
  • The worked examples of charged-particle motion are relevant to:______. a. a transistor. b. a cathode-ray tube. c. magnetic reson
    11·1 answer
  • What is the maximum eccentricity an ellipse can have
    8·1 answer
  • The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e
    10·1 answer
  • Please helppppppppppp​
    5·1 answer
  • The chimney of an oil lamp cracks when touched with rhe brade of cold knife?why<br>​
    10·2 answers
  • An 8 kg toddler is running at a speed of 10 m/s.<br> how much energy does he have?
    14·1 answer
  • Give an example for each of the following, where the force:
    13·2 answers
  • An object, which is at the origin at time t=0
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!