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Morgarella [4.7K]
2 years ago
15

There is a square loop of wire with side length L moving to the right at a speed of v. The resistance in the wire is R. To the r

ight of it is a region with a magnetic field with field strength B into the page. Find the current in the wire loop (a) before the loop reaches the magnetic field, (b) while the loop is entering the field, (c) while the loop is entirely within the field, (d) while the loop is exiting the field, and (e) after the loop leaves the field. Be sure to include the direction of the current (i. s. clockwise or counter-clockwise). All your answers will be in terms of L, v, B, and R.
Physics
1 answer:
topjm [15]2 years ago
8 0

There is a square loop of wire with side length L moving to the right at a speed of v. The resistance in the wire is R. To the right of it is a region with a magnetic field with field strength B into the page.  of anxiety (low vs. high) and stress (low vs. moderate vs. high) on test.

Everyone experiences anxiety occasionally, but persistent anxiety can reduce your quality of life. Though likely best known for altering behavior, worry can have negative effects on our physical health. Anxiety speeds up our heartbeat and breathing, resistance blood flow to the parts of our brains that need it. You are getting ready for a challenging situation by having this extremely bodily reaction. Test performance may be impacted by anxiety. According to studies, pupils with low levels of test anxiety perform better on multiple-choice resistance (MCQ) exams than pupils with high levels of anxiety. Studies have indicated that female students have greater levels of test anxiety than male students.

Learn more about anxiety here:

brainly.com/question/4913240

#SPJ4

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t
Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The  Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6×10^{24} be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67×10^{-11} N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v=\sqrt{\frac{Gmm}{R+h} }         [(R+h) is the distance of the satellite   from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = h_{i} = 98 km = 98000 m

∴Initial Energy (E_{i})  = \frac{1}{2}mv^{2} + \frac{-GMm}{(R+h_{i} )}

Substituting v=\sqrt{\frac{GMm}{R+h_{i} } } in the above equation and simplifying we get,

E_{i} = \frac{-GMm}{2(R+h_{i}) }

Similarly for final condition,

h=h_{f} = 198km = 198000 m

∴Final Energy(E_{f}) = \frac{-GMm}{2(R+h_{f}) }

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = E_{f} - E_{i}

        = \frac{GMm}{2}(\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} })

Substituting ,

M = 6 × 10^{24} kg

m = 1036 kg

G = 6.67 × 10^{-11}

R = 6400000 m

h_{i} = 98000 m

h_{f} = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = \frac{1}{2}m[v_{f} ^{2} - v_{i} ^{2}]

                                                          = \frac{GMm}{2}[\frac{1} {R+h_{f} } - \frac{1} {R+h_{i} }]

                                                          = -ΔE                                                            

                                                          = - 484.438 MJ

c)  Change in Potential Energy (ΔPE) = GMm[\frac{1}{R+h_{i} } - \frac{1}{R+h_{f} }]

                                                             = 2ΔE

                                                             = 968.907 MJ

3 0
4 years ago
A drawback to using nuclear fission is that it...
Westkost [7]

Answer:

D. creates radioactive waste.

Explanation:

Nuclear energy can create nuclear radioactive waste

8 0
1 year ago
The blades in a blender rotate at a rate of 6800 rpm . When the motor is turned off during operation, the blades slow to rest in
tangare [24]
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
5 0
3 years ago
Infer the direction of the net force acting on a car as it slows down and turns right.
Kobotan [32]
Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.
6 0
3 years ago
Read 2 more answers
Some students set up a circuit and decided to measure the voltage at different points around
Fiesta28 [93]

Answer:

 V₁ = 6 V ,  V₂ = V₃ = 3 V

Explanation:

To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.

* a serial circuit there is only one path for current

in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term

* Parallel circuit in this there are two or more paths for the current

in this circuit the voltage is constant and the east is divided between each branch

with these principles let's analyze the proposed circuit

The DC battery is in parallel with resistor R1 and the equivalent of the other branch,

as in a parallel circuit the voltage is constant

               V₁ = 6 V

in the other branch (23) it forms a series construction, where the current is constant

               6 = iR₂ + iR₃

as they indicate that each resistance has the same value

              6 = 2 iR

              V = V₂ = V₃ = 3 V

3 0
3 years ago
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