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
Reika [66]
2 years ago
9

What is the definition of mutual flux?​

Physics
1 answer:
olchik [2.2K]2 years ago
6 0

Answer:

Is where two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another

You might be interested in
Which of the following is most likely to happen when energy is transferred to
ankoles [38]

Explanation:

the object will begin to move

4 0
3 years ago
Read 2 more answers
A net force of 125 N accelerates a 25.0 kg mass. What is the resulting acceleration?
Neko [114]

Answer: a=5 m/s^2

Explanation:

The acceleration of an object can be calculated by using Newton's second law:

F=ma

where

F is the net force applied on the object

m is the mass of the object

a is its acceleration

In this problem, we have F=125 N and m=25.0 kg, so we can rearrange the equation to calculate the acceleration:

a=\frac{F}{m}=\frac{125 N}{25.0 kg}=5 m/s^2

5 0
3 years ago
Read 2 more answers
Which of the following describes resistance force?
Verdich [7]
Force applied by the machine to over come resistance
5 0
3 years ago
Read 2 more answers
A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force.  We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as  588 newtons  or as 
132.3 pounds.  That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is 

                                     y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is     y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is          y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

                                                                = - (4 π² M / 15²)  sin(2π t/15)

                                                                = - 0.1755 M sin(2π t/15) .

There's the important number ... the  0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of  0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of  65kg, when in reality it's only  60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

                        0.08333 G  =  0.1755 M

The 'M' is what we need to find.

Divide each side by  0.1755 :          M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
                                       M = (0.08333 / 0.1755) (9.8) =  4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0
2 years ago
What is a cost of urbanization
maw [93]
Urbanization often requires extensive ecological change and many times destruction and displacement to occur. This can have a negative impact on the environment, and the animals within ecosystems undergoing urbanization.

Hope this helps!
4 0
3 years ago
Other questions:
  • What best describes desert pavements
    7·1 answer
  • A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com
    10·1 answer
  • 2 differences between calorimeter and thermometer ?
    9·1 answer
  • What is the average acceleration during the time interval 0 seconds to 10 seconds?
    12·1 answer
  • This problem uses the same concepts as Multiple-Concept Example 17, except that kinetic, rather than static, friction is involve
    7·1 answer
  • If a circling object is released, centrifugal force will make it travel away from the center of its original path.
    13·1 answer
  • camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin
    5·1 answer
  • Liquids, when heated,
    14·1 answer
  • Suppose we observe a cepheid variable in a distant galaxy. The cepheid brightens and dims with a regular period of about 10 days
    10·1 answer
  • If the ball reaches the ground in 8 seconds and the velocity of the ball before it hits the ground is 78.4. What is its accelera
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!