Inductance is usually denoted by L and is measured in SI units of henries (also written henrys, and abbreviated H), named after Joseph Henry, a contemporary of Michael Faraday. The EMF E produced in a coil with inductance L is, according to Faraday's law, given byE=−LΔIΔt.
Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
What EMF is produced if a waffle iron that draws 2.5 amperes and has an inductance of 560 millihenries is suddenly unplugged, so the current drops to essentially zero in 0.015 seconds?

Answers

Answer 1
Answer:

Answer:

Explanation:

E= −L ΔI / Δt.

L = E Δt / ΔI

Hence the unit of inductance may be V s A⁻¹

or volt s per ampere .

In the given case

change in current ΔI = - 2.5 A

change in time = .015 s

L = .56 H

E = − L ΔI / Δt.

= .56 x 2.5 / .015

= 93.33 V .


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A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?

Answers

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 516 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.6 ft deep, estimate the magnitude of his average acceleration as he slowed to a stop (that is while he was impacting the snow).

Answers

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 32.174* 516+0^2)\n\Rightarrow v=182.218\ ft/s

v^2-u^2=2as\n\Rightarrow a=(v^2-u^2)/(2s)\n\Rightarrow a=(0^2-182.218^2)/(2* 3.6)\n\Rightarrow a=-4611.58\ ft/s^2

Magnitude of acceleration while stopping is 4611.58 ft/s²

A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes

Answers

Answer:

T1 = 499.9N, T2 = 865.8N, T3 = 1000N

Explanation:

To find the tensions we need to find the vertical and horizontal components of T1 and T2

T1x = T1 cos60⁰, T1y = T1 sin60⁰

Also, T2x = T2 cos30⁰, T2y = T2 sin30⁰

For the forces to be in equilibrium,

the sum of vertical forces must be zero and the sum of horizontal forces must also be zero

Sum of Fx = 0

That is, T1x - T2x=0

NB: T2x is being subtracted because T1x and T2x are in opposite directions

T1 cos60⁰ - T2 cos30⁰ = 0

0.866T1 - 0.5T2 = 0 ............ (1)

Sum of Fy = 0

T1y + T2y - 1000 = 0

T1 sin60⁰ + T2 sin30⁰ - 1000 = 0

NB: The weight of the bag of cement is also being subtracted because it's in an opposite direction.

0.5T1 - 0.866T2 - 1000 = 0 ........(2)

From (1)

make T1 the subject

T1 = 0.5T2/0.866

Substitute T1 into (2)

0.5 (0.5T2/0.866) - 0.866T2 = 1000

(0.25/0.866)T2 - 0.866T2 = 1000

0.289T2 - 0.866T2 = 1000

1.155T2 = 1000

T2 = 865.8N

Then T1 = 0.5 x 865.8 / 0.866

T1 = 499.9N

T3 = 1000N

NB: The weight of the bag is the Tension above the rope, which is T3

Coach ulcer paces the sidelines. Sarting at the 30 yd. line (A), he moves to the 10 yd. line (B), back to the 50 yd. line (C) and finally to the 20yd. Line (D) in 200 seconds. Determine his average speed and velocity.

Answers

Answer:

See the answer below

Explanation:

Average speed = total distance traveled/total time taken

In order to determine the total distance traveled by the coach, consider the attached image.

Distance covered:

30 yd. line to 10 yd. line (A to B)= 20 yds

10 yd. line to 50 yd. line (B to C) = 40 yds

50 yd. line to 20 yd. line (C to D) = 30 yds

Total distance covered = 20 + 40 + 30 = 90 yds

Time taken = 200 seconds

Average Speed = 90/200 = 0.45 yd/s

Velocity = speed with direction

Hence,

His Velocity = 0.45 yd/s to the left of his starting point.

5. (Serway 9th ed., 7-3) In 1990, Walter Arfeuille of Belgium lifted a 281.5-kg object through a distance of 17.1 cm using only his teeth. (a) How much work was done on the object by Arfeuille in this lift, assuming the object was lifted at constant speed? (b) What total force was exerted on Arfeuille’s teeth during the lift? (Ans. (a) 472 J; (b) 2.76 kN)

Answers

Para resolver este problema es necesario aplicar los conceptos de Fuerza, dados en la segunda Ley de Newton y el concepto de Trabajo, como expresión de la fuerza necesaria para realizar una actividad en una distancia determinada.

El trabajo se define como

W = F*d

Where,

F = Force

d = Distance

At the same time we have that the Force by second's Newton law is equal to

F = mg

Where,

m = mass

g = Gravitational acceleration

PART A) Using our values and replacing we have that

W = F*d\nW = mg*d\nW=281.5*9.8(17.1*10^(-2)\nW = 471.738 J\approx 472J

PART B) Using Newton's Second law we have that,

F = mg \nF= 281.5*9.8\nF= 2758.7 N \approx 2.76kN

A substance that does NOT conduct an electric current when it forms a solution is a(n) ____. a electrolyte

b nonelectrolyte

c liquid

d solid

Answers

Answer:

B. Nonelectrolyte.

Explanation:

Nonelectrolytes do not dissociate into ions in solution, hence, nonelectrolyte solutions don't conduct electricity.

A non-electrolyte doesn’t conduct electric current even when it forms a solution.

Answer: Option B

Explanation:

Where electrolytes are defined as the compounds that can conduct electric current with mobile ions existing in its solution, non-electrolytes are the compounds that don’t behave the same either in the aqueous solution or in the molten state.  

This is all because these compounds don’t produce mobile ions to flow from one electrode to the other and hence conduct electric flow in the solution. Sugar and ethanol are the best examples of non-electrolytes that don’t induce electric current even after getting dissolved in water.