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
galina1969 [7]
3 years ago
15

a) When we were examining the Electromagnetic Tab, we saw that a flow of electrons or a current as we say it, creates a magnetic

field. What about the converse, can a magnetic field be involved in the creation of a flow of electrons/current? Therefore is it reasonable to suggest that we can create a magnetic field by having a flow of current and this can be used to make more current? Explain how this can occur
Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

Magnetic field can be used to produce current, infact a changing magnetic field can produce current.

A changing magnetic field in a loop causes the flux linked with the loop to change in turn generating a emf in the loop and therefore a current.

For a loop of area A and resistance R.

I =dPhi/dt/R

В. А

I = AcosФ/R .dB /dt

But it isn't reasonable to say that we can create a magnetic field by having a flow of current and this can be used to make more current because the current generated due to change in magnetic field created by increase/decrease in flow of current will be in a direction such that it will counter act the change in magnetic field caused by increase/decrease in current flow.(lenz's law).

We were unable to transcribe this image

Ф= В. А

I = Acos dB Rd

You might be interested in
PLS PLS PLS PLS PLS PLS PLS PLS PLS HELP FREE POINTS
Ne4ueva [31]

Answer:

see below

Explanation:

First: Leave a couple inches of wire loose at one end and wrap most of the rest of the wire around iron  u-shaped bar and make sure not to overlap the wires.

Second:Cut the wire (if needed) so that there is about a couple inches loose at the other end too.

Third: Now remove about an inch of the plastic coating from both ends of the wire and connect the one wire to one end of a battery and the other wire to the other end of the battery.

3 0
4 years ago
8. A car starts from rest and accelerates at 5.5ms in an easterly direction.
snow_lady [41]
  • Initial velocity=0m/s=u
  • Acceleration=a=5.5m/s^2

#8.1

  • t=12s

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+5.5(12)

\\ \sf\longmapsto v=66m/s

#8.2

  • u=0m/s
  • v=18m/s
  • a=5.5m/s^2

Use third equation of kinematics

\\ \sf\longmapsto v^2-u^2=2as

\\ \sf\longmapsto s=\dfrac{v^2-u^2}{2a}

\\ \sf\longmapsto s=\dfrac{18^2-0^2}{2(5.5)}

\\ \sf\longmapsto s=\dfrac{324}{11}

\\ \sf\longmapsto s=29.4m

8 0
3 years ago
Read 2 more answers
If the space station is 190 m in diameter, what angular velocity would produce an "artificial gravity" of 9.80 m/s2 at the rim?
ozzi

Answer:

The angular velocity produced is 0.321 rad/s.

Explanation:

Given :

Diameter of space station , D = 190 m.

Therefore, radius , R=\dfrac{D}{2}=\dfrac{190}{2}=95\ m.

Also, acceleration , a=9.8\ m/s^2.

We know, angular velocity , \omega=\sqrt \dfrac{a}{R}.

Putting value of g and R in above equation.

We get ,

\omega=\sqrt \dfrac{9.8\ m/s^2}{95\ m}

\omega=0.321\ rad/s.

Hence, this is the required solution.

6 0
3 years ago
. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin
Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

T-(m*g)=m*a

Where:

T= thrust =3 *10^{4} N

m= module's mass =1 *10^{4} N

g= moon's gravity acceleration =1.623 \frac{m}{s^2}

a= module's acceleration during takeoff

Then, we can find the acceleration like this:

a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}

a=1.377 \frac{m}{s^2}

The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N

Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

8 0
4 years ago
14
romanna [79]

Answer: 0.00068 N

Explanation: Universal gravitational constant=6.674 *10^(-11)

Force=Gm1m2/(r^2)

Force= 6.67*25000*40000*10^(-11)/(10^2)

Force=0.00068 N

6 0
3 years ago
Other questions:
  • Ultrasound with a frequency of 4.257 MHz can be used to produce images of the human body. If the speed of sound in the body is t
    12·1 answer
  • If the coefficient of static friction is 0.40, and the same ladder makes a 51° angle with respect to the horizontal, how far alo
    7·1 answer
  • Why doesn't an object thrown in an upward direction fall the same distance in each time interval as it descends toward Earth?
    6·1 answer
  • What is the acceleration of a car that maintains a constant velocity of 100 km/h for 10 sec?
    15·1 answer
  • You have four fixed-volume containers at STP . Container A has 0.5 mol of gas in 11.2 L. Container B has 2 mol of gas in 22.4 L.
    9·1 answer
  • 2. A 3.5 m long string is fixed at both ends and vibrates in its 7th harmonic with an amplitude (at an antinode) of 2.4 cm. If t
    7·1 answer
  • Answer meeeeeeeeeeeeeee
    12·2 answers
  • How would the seasons be different if Earth were not tilted on its Axis?
    12·2 answers
  • A 9 V battery produces a current of 18 amps. What is the resistance?
    7·1 answer
  • Being willing to sincerely apologize for mistakes errors and blunders goes a long way toward
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!