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2 years ago

what device is made of coils of wire with an electrical current passing through it in order to generate a magnetic field

Physics

1 answer:

Umnica [9.8K]2 years ago

8 0

Transformer

<u>Explanation:</u>

A transformer is a device with two or more magnetically coupled windings. A time varying current in one coil (primary winding) generates a magnetic field which induces a voltage in the other coil (secondary winding). Transformers are capable of either increasing or decreasing the voltage and current levels of their supply, without modifying its frequency, or the amount of electrical power being transferred from one winding to another via the magnetic circuit. There are two types of transformer:

1. Step up transformer - increases voltage

2. Step down transformer - decreases voltage

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What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t

yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity = ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

Required Information:

time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R = ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R = (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

6 0

2 years ago

What is the hang time when the person moves 6 m horizontally during a 1.25 m high jump?

AlekseyPX

Answer:

1 sec

Explanation:

Horizontal distance (x) = 6m

Vertical distance (y) = 1.25m

Hang time is the duration the object is in the air before it reaches maximum height.

The time of free fall is given by

t = √2y/g

g = acceleration due to gravity

t = √(2*1.25)/9.8

t = √2.5/9.8

t = 0.5secs

Hang time = 2*0.5

= 1 sec

3 0

3 years ago

A snowball accelerates at

ANTONII [103]

Answer:

0. 1226495726kg

Explanation:

Force is the product of mass and acceleration.

Mathematically,

Force(F) = mass (m)×acceleration(a)

Substituting the values into the equation

2. 87=m×23. 4

2. 87=m (23. 4)

2. 87/23. 4=m (23. 4)/23. 4

2. 87/23. 4=m

0. 1226495726=m

8 0

2 years ago

The time, t, required to drive a fixed distance varies inversely as the speed, r. It takes 3 hr at a speed of 14 km/h to driv

kenny6666 [7]

Answer:

time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes

Explanation:

Given:

Time is inversely proportional to the speed

mathematically,

t ∝ (1/r)

let the proportionality constant be 'k'

thus,

t = k/r

therefore, for case 1

time = 3 hr

speed = 14 km/hr

3 = k/14

also,

for case 2

let the time be = t

r = 23 km/h

thus,

we have

t = k/23

on dividing equation 2 by 1

we get

$\frac{t}{3}=\frac{k/23}{k/14}$

$t=\frac{14\times3}{23}$

t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)

4 0

2 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago