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ahrayia [7]
2 years ago
10

Any help????????????

Physics
2 answers:
kolezko [41]2 years ago
5 0
(-2,2) is the answer
pashok25 [27]2 years ago
3 0
The answer should be D
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An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The curren
Maurinko [17]

i

CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

3 0
3 years ago
In a heat engine, if 500 J of heat enters the system, and the piston does 300 J of work.
Alex Ar [27]

Answer: The final answer is 1,700 J

Explanation: According to first law of thermodynamics:

=Final energy-initial energy=Change in internal energy

q = heat absorbed or released

w = work done by or on the system

w = work done by the system=  {Work done by the system is negative as the final volume is greater than initial volume}

q = +500J   {Heat absorbed by the system is positive}

w = work done by the system = -300J

U2- 1500J= + 500J --300J

U2= 1700J

7 0
3 years ago
Read 2 more answers
Which of these objects are malleable? Check all that apply.
g100num [7]

Answer:

The tin fork and knife, the copper coin, and the steel fence pole.

Explanation:

Those are both what people would call soft metals so they are malleable to the extent of probably not needing heavy duty equipment. It depends on you description of malleable because the steel fence pole could be malleable with the correct equipment and not snap in half if bent slowly enough.

The definition of malleable: (of a metal or other material) able to be hammered or pressed permanently out of shape without breaking or cracking.

But the glass table, marble sculpture and antique ceramic vase are nowhere near malleable because if you tried bending them they wouldn't bend but would shatter and break into pieces.

6 0
3 years ago
Read 2 more answers
A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, w
jek_recluse [69]

Answer:

XY=636N

Explanation:

From the question we are told that:

Distance d=0.35m

Angle \theta=32\textdegree

Force F=750N

Generally the equation for magnitude of the stabilizing component of the brachialis force is mathematically given by

 XY=Fcos\theta

 XY=750cos 32\textdegree

 XY=636N

4 0
3 years ago
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