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A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?

Answers

Answer 1

Answer:

Answer:

The current flows in the second wire is $1.3*10^(10)\ A$

Explanation:

Given that,

Upward current = 24 A

Force per unit length $(F)/(l) =88*10^(4)\ N/m$

Distance = 7.0 cm

We need to calculate the current in second wire

Using formula of magnetic force

$F=ILB$

$(F)/(l)=(\mu I_(1)I_(2))/(2\pi r)$

Where,

$(F)/(l)$ =force per unit length

I₁= current in first wire

I₂=current in second wire

r = distance between the wires

Put the value into the formula

$88*10^(4)=(4\pi*10^(-7)*24* I_(2))/(2\pi *7*10^(-2))$

$I_(2)=(88*10^(4)*7*10^(-2))/(2**10^(-7)*24)$

$I_(2)=1.3*10^(10)\ A$

Hence, The current flows in the second wire is $1.3*10^(10)\ A$

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Which of the following is correct? *PLEASE HELP MEEEE
1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

Answers

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the projectile a freely falling body?Yes or No(b) What is its acceleration in the vertical direction? (Let up be the positive direction.)____? m/s2(c) What is its acceleration in the horizontal direction?

Answers

Answer:

A) No

B)-9,81 m/s^2

C)0 m/s^2

Explanation:

A free fallin object has only velocity on the vertical axis so any object that is moving in the Y and X axis has projetile motion not free falling, and when dealing with projectile motion the object is experiencing acceleration towards the ground of -9,81m/s^2 and in the Y axis, in the X axis there´s is only acceleration if the air is providing resistance, since it states that it isnot, then the accleration is 0.

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?

Answers

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

Final answer:

Positive, negative, and zero terms in the energy equation. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.

Explanation:

To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:

Positive terms:

ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
ΔUs - the change in elastic potential energy is positive if there is any stretch or compression in the system.

Negative terms:

ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.

Zero terms:

ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.

To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in gravitational potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.

The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the metabolic power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.

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The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode along, what would the car's acceleration be?

Answers

Final answer:

The problem discusses the change in acceleration when a passenger is added to a car. It requires understanding of Newton's second law of motion, force equals mass times acceleration, and then recalculating the acceleration with the passenger added to the total mass.

Explanation:

This problem pertains to Newton's second law of motion, stating that the force applied on an object equals its mass times its acceleration (F = ma). Given that the initial acceleration of the Lamborghini Huracan with a driver is 0.80g or 0.80*9.80 m/s², we can calculate the force applied by the car. By multiplying the car's mass (1510 kg) with its acceleration, we will find the force.

Οnce we have the force, we can calculate the new acceleration if the 80 kg passenger rode along. Given that the force is constant, we determine the car's new acceleration by dividing this force with the new total mass (car mass + passenger's mass). So the question ultimately requires an application of the concepts of force, mass, and acceleration.

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Final answer:

The new acceleration of the Lamborghini Huracan with an added passenger can be calculated by finding the initial force using the car's mass and acceleration, and then using this force with the increased mass (original mass + passenger's mass) to find the new acceleration. The new acceleration will be less than the initial acceleration due to the increased mass.

Explanation:

To determine the new acceleration of the Lamborghini Huracan with an added passenger, we first calculate the initial force acting on the car. This can be done by using Newton's second law which states that Force = mass * acceleration. Initially, the acceleration is 0.80g (where g is acceleration due to gravity = 9.81 m/s²), and the mass is 1510 kg (including the driver). Therefore, the initial force = 1510 kg * 0.8 * 9.81 m/s².

However, when an 80-kg passenger rides along, the total mass becomes 1510 kg + 80 kg = 1590 kg. To find the new acceleration, we keep the force constant (as it is not affected by the introduction of the passenger) and rearrange the formula F = m*a as a = F/m. Use the increased mass to find the new acceleration. Please note that the new acceleration will be less than the initial acceleration due to increased mass.

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Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.1 m/s2 while Kathy maintains an acceleration of 4.99 m/s. 2 (a) Find the time at which Kathy overtakes Stan. s from the time Kathy started driving (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him. Kathy m/s Stan m/s

Answers

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement

t₁ : Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁: Stan acceleration

d₂: car displacement

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂: Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ = 1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1=equation 2

1.55*t₁² = 2.495* t₂² , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² =2.495* t₂²

1.55* (t₂² +2t₂+1) =2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0 Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁ t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂

vf₂ =0+4.99* 3.87

vf₂ = 19.31m/s : Kathy's final speed

A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The painter is standing 2 m from the first rope and 8 m from the second rope. What is the tension in the first rope (in Newtons)