Two long, parallel wires carry currents in the same direction. If I1 = 10 A, and I2 = 20 A, and they are d = 1.0 m apart, what is the magnetic field at a point P midway between them?

Answers

Answer 1

Answer:

The magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T

Explanation:

Given;

current in the first wire, I₁ = 10 A

current in the second wire, I₂ = 20 A

distance between the two wires, d = 1.0 m

Magnetic field at mid point between two parallel wires is calculated as;

[tex]B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r}(I_1 +I_2)[/tex]

where;

r is the midpoint between the wires, = 0.5 m

μ₀ is the permeability of free space, = 4π x 10⁻⁷

[tex]B = \frac{\mu_o }{2\pi r}(I_1 +I_2)\\\\B = \frac{4\pi*10^{-7} }{2\pi *0.5}(10 +20)\\\\B = \frac{4\pi*10^{-7} *30}{2\pi *0.5}\\\\B = 1.2 *10^{-5} \ T[/tex]

Therefore, the magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T


Related Questions

A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.

Answers

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

An 88.0 kg spacewalking astronaut pushes off a 645 kg satellite, exerting a 110 N force for the 0.450 s it takes him to straighten his arms. How far apart are the astronaut and the satellite after 1.40 min?

Answers

Answer:

The astronaut and the satellite are 53.718 m apart.

Explanation:

Given;

mass of spacewalking astronaut, = 88 kg

mass of satellite, = 645 kg

force exerts by the satellite, F = 110N

time for this action, t = 0.45 s

Determine the acceleration of the satellite after the push

F = ma

a = F / m

a = 110 / 645

a = 0.171 m/s²

Determine the final velocity of the satellite;

v = u + at

where;

u is the initial velocity of the satellite = 0

v = 0 + 0.171 x 0.45

v = 0.077 m/s

Determine the displacement of the satellite after 1.4 m

d₁ = vt

d₁ = 0.077 x (1.4 x 60)

d₁ = 6.468 m

According to Newton's third law of motion, action and reaction are equal and opposite;

Determine the backward acceleration of the astronaut after the push;

F = ma

a = F / m

a = 110 / 88

a = 1.25 m/s²

Determine the final velocity of the astronaut

v = u + at

The initial velocity of the astronaut = 0

v = 1.25 x 0.45

v = 0.5625 m/s

Determine the displacement of the astronaut after 1.4 min

d₂ = vt

d₂ = 0.5625 x (1.4 x 60)

d₂ = 47.25 m

Finally, determine the total separation between the astronaut and the satellite;

total separation = d₁ + d₂

total separation = 6.468 m + 47.25 m

total separation = 53.718 m

Therefore, the astronaut and the satellite are 53.718 m apart.

Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?

Answers

Answer:

Their frequency is 111.22 Hz

Explanation:

Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).

Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).

The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:

v = f * λ.

Then the frequency can be calculated as: f=v÷λ

In this case:

λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/s

Replacing:

[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]

Solving:

f=111.22 Hz

Their frequency is 111.22 Hz

A medieval city has the shape of a square and is protected by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by catapulting heated rocks over the wall (with an initial speed of 80 m/s). At what range of angles should you tell your men to set the catapult? (Assume the path of the rocks is perpendicular to the wall. Round your answers to one decimal place. Use g ≈ 9.8 m/s2. Enter your answer using interval notation. Enter your answer in terms of degrees without using a degree symbol.)

Answers

Answer:

  θ₁ = 85.5º       θ₂ = 12.98º

Explanation:

Let's analyze this projectile launch problem, the catapults are 100 m from the wall 15 m high, the objective is for the walls, let's look for the angles for which the rock stops touching the wall.

Let's write the equations for motion for this point

X axis

          x = v₀ₓ t

          x = v₀ cos θ t

Y axis

         y = [tex]v_{oy}[/tex] t - ½ g t2

         y = v_{o} sin θ t - ½ g t²

let's substitute the values

         100 = 80 cos θ t

           15 = 80 sin θ t - ½ 9.8 t²

we have two equations with two unknowns, so the system can be solved

let's clear the time in the first equation

           t = 100/80 cos θ

         15 = 80 sin θ (10/8 cos θ) - 4.9 (10/8 cos θ)²

         15 = 100  tan θ - 7.656 sec² θ

we can use the trigonometric relationship

         sec² θ = 1- tan² θ

we substitute

       15 = 100 tan θ - 7,656 (1- tan² θ)

       15 = 100 tan θ - 7,656 + 7,656 tan² θ

        7,656 tan² θ + 100 tan θ -22,656=0

let's change variables

       tan θ = u

         

        u² + 13.06 u + 2,959 = 0

let's solve the quadratic equation

       u = [-13.06 ±√(13.06² - 4  2,959)] / 2

       u = [13.06 ± 12.599] / 2

        u₁ = 12.8295

        u₂ = 0.2305

now we can find the angles

         u = tan θ

         θ = tan⁻¹ u

        θ₁ = 85.5º

         θ₂ = 12.98º

When a particular wire is vibrating with a frequency of 6.3 Hz, a transverse wave of wavelength 53.3 cm is produced. Determine the speed of wave pulses along the wire.

Answers

Answer:

335.79cm/s

Explanation:

When a transverse wave of wavelength λ is produced during the vibration of a wire, the frequency(f), and the speed(v) of the wave pulses are related to the wavelength as follows;

v = fλ        ------------------(ii)

From the question;

f = 6.3Hz

λ = 53.3cm

Substitute these values into equation (i) as follows;

v = 6.3 x 53.3

v = 335.79cm/s

Therefore, the speed of the wave pulses along the wire is 335.79cm/s

Two carts are connected by a loaded spring on a horizontal, frictionless surface. The spring is released and the carts push away from each other. Cart 1 has mass M and Cart 2 has mass M/3.

a) Is the momentum of Cart 1 conserved?

Yes

No

It depends on M

b) Is the momentum of Cart 2 conserved?

Yes

No

It depends on M

c) Is the total momentum of Carts 1 and 2 conserved?

Yes

No

It depends on M

d) Which cart ends up moving faster?

Cart 1

Cart 2

They move at the same speed

e) If M = 6 kg and Cart 1 moves with a speed of 16 m/s, what is the speed of Cart 2?

0 m/s

4.0 m/s

5.3 m/s

16 m/s

48 m/s

64 m/s

Answers

Answer:

a) yes

b) no

c) yes

d)Cart 2 with mass [tex]\frac{M}{3}[/tex]   is expected to be more faster

e) u₂ = 48 m/s

Explanation:

a) the all out linear momentum of an arrangement of particles of Cart 1 not followed up on by external forces is constant.

b) the linear momentum of Cart 2 will be acted upon by external force by Cart 1 with mass M, thereby it's variable and the momentum is not conserved

c) yes, the momentum is conserved because no external force acted upon it and both Carts share the same velocity after the reaction

note: m₁u₁ + m₂u₂ = (m₁ + m₂)v

d) Cart 2 with mass [tex]\frac{M}{3}[/tex] will be faster than Cart 1 because Cart 2 is three times lighter than Cart 1.

e) Given

m₁=  M

u₁ = 16m/s

m₂ =[tex]\frac{M}{3}[/tex]

u₂ = ?

from law of conservation of momentum

m₁u₁= m₂u₂

M× 16 = [tex]\frac{M}{3}[/tex] × u₂(multiply both sides by 3)

therefore, u₂ = [tex]\frac{3(M .16)}{M}[/tex] ("." means multiplication)

∴u₂ = 3×16 = 48 m/s

Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the function V 230 sin (100t), where tis measured in seconds and Vin volts.What is the frequency of the voltage

Answers

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct

Answers

Complete question:

A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's interior, which statement is correct?

A. the interior field points in a direction parallel to the exterior field

B. There is no electric field on the interior of the conducting sphere.

C. The interior field points in a direction perpendicular to the exterior field.

D. the interior field points in a direction opposite to the exterior field.

Answer:

B. There is no electric field on the interior of the conducting sphere.

Explanation:

Conductors are said to have free charges that move around easily. When the conductor is now placed in a static electric field, the free charges react to attain electrostatic equilibrium (steady state).

Here, a solid conducting sphere is placed in an external uniform electric field. Until the lines of the electric field are perpendicular to the surface, the free charges will move around the spherical conductor, causing polarization. There would be no electric field in the interior of the spherical conductor because there would be movement of  free charges in the spherical conductor in response to any field until its neutralization.

Option B is correct.

There is no electric field on the interior of the conducting sphere.

can I get help please?​

Answers

Answer shown on photo

An electron, moving west, enters a magnetic field of a certain strength. Because of this field the electron curves upward. What is the direction of the magnetic field?

Answers

Answer:

Towards the west.

Explanation:

The direction of a magnetic field lines is the direction north end of a compass needle points. The magnetic field exert force on positive charge.

Using the magnetic rule,which indicate that in order to find the direction of magnetic force on a moving charge, the thumb of the right hand point in the direction of force, the index finger in the direction of velocity charge and the middle finger in the direction of magnetic field.

According to the right hand rule, the electron moving moving west which is the thumb, the direction of the electron is west which is the middle finger and it is upward

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energy that is equal to the proton rest energy, the speed of the kaon is most nearly:___________.
A. 0.25c
B. 0.40c
C. 0.55c
D. 0.70c
E. 0.85c

Answers

Answer:

0.85c

Explanation:

Rest mass of Kaon [tex]M_{0K}[/tex] = 494 MeV/c²

Rest mass of proton [tex]M_{0P}[/tex]  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy [tex]E_{0K}[/tex] = 494c² MeV

for the proton, rest energy [tex]E_{0P}[/tex] = 938c² MeV

Recall that the rest energy, and the total energy are related by..

[tex]E[/tex] = γ[tex]E_{0}[/tex]

which can be written in this case as

[tex]E_{K}[/tex] = γ[tex]E_{0K}[/tex] ...... equ 1

where [tex]E[/tex] = total energy of the kaon, and

[tex]E_{0}[/tex] = rest energy of the kaon

γ = relativistic factor = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex]

where [tex]\beta = \frac{v}{c}[/tex]

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

[tex]E_{K}[/tex] = [tex]E_{0P}[/tex] ......equ 2

where [tex]E_{K}[/tex] is the total energy of the kaon, and

[tex]E_{0P}[/tex] is the rest energy of the proton.

From [tex]E_{K}[/tex] = [tex]E_{0P}[/tex] = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = [tex]\frac{1}{\sqrt{1 - \beta ^{2} } }[/tex] = 1.89

1.89[tex]\sqrt{1 - \beta ^{2} }[/tex] = 1

squaring both sides, we get

3.57( 1 - [tex]\beta^{2}[/tex]) = 1

3.57 - 3.57[tex]\beta^{2}[/tex] = 1

2.57 = 3.57[tex]\beta^{2}[/tex]

[tex]\beta^{2}[/tex] = 2.57/3.57 = 0.72

[tex]\beta = \sqrt{0.72}[/tex] = 0.85

but, [tex]\beta = \frac{v}{c}[/tex]

v/c = 0.85

v = 0.85c

The compressor of an air conditioner draws an electric current of 16.2 A when it starts up. If the start-up time is 1.45 s long, then how much electric charge passes through the circuit during this period

Answers

Answer:

Q = 23.49 C

Explanation:

We have,

Electric current drawn by the air conditioner is 16.2 A

Time, t = 1.45 s

It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,

[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]

So, the charge of 23.49 C is passing through the circuit during this period.

Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?

Answers

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

An aging coyote cannot run fast enough to catch a roadrunner. He purchases on eBay a set of jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 m/s2. The coyote starts at rest 70.0 m from the edge of a cliff at the instant the roadrunner zips past in the direction of the cliff.

Required:
a. Determine the minimum constant speed the roadrunner must have to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. The coyote’s skates remain horizontal and continue to operate while he is in flight.
b. The cliff is 100 m above the flat floor of the desert. Determine how far from the base of the cliff the coyote lands.
c. Determine the components of the coyote’s impact velocity

Answers

Answer:

a)  v_correcaminos = 22.95 m / s ,  b)  x = 512.4 m ,

c) v = (45.83 i ^ -109.56 j ^) m / s

Explanation:

We can solve this exercise using the kinematics equations

a) Let's find the time or the coyote takes to reach the cliff, let's start by finding the speed on the cliff

         v² = v₀² + 2 a x

they tell us that the coyote starts from rest v₀ = 0 and its acceleration is a=15 m / s²

         v = √ (2 15 70)

         v = 45.83 m / s

with this value calculate the time it takes to arrive

        v = v₀ + a t

        t = v / a

        t = 45.83 / 15

        t = 3.05 s

having the distance to the cliff and the time, we can find the constant speed of the roadrunner

         v_ roadrunner = x / t

         v_correcaminos = 70 / 3,05

         v_correcaminos = 22.95 m / s

b) if the coyote leaves the cliff with the horizontal velocity v₀ₓ = 45.83 m / s, they ask how far it reaches.

Let's start by looking for the time to reach the cliff floor

            y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

             

in this case y = 0 and the height of the cliff is y₀ = 100 m

          0 = 100 + 45.83 t - ½ 9.8 t²

          t² - 9,353 t - 20,408 = 0

we solve the quadratic equation

         t = [9,353 ±√ (9,353² + 4 20,408)] / 2

         t = [9,353 ± 13] / 2

         t₁ = 11.18 s

        t₂ = -1.8 s

Since time must be a positive quantity, the answer is t = 11.18 s

we calculate the horizontal distance traveled

        x = v₀ₓ t

        x = 45.83 11.18

        x = 512.4 m

c) speed when it hits the ground

         vₓ = v₀ₓ = 45.83 m / s

we look for vertical speed

         v_{y} = [tex]v_{oy}[/tex] - gt

         v_{y} = 0 - 9.8 11.18

         v_{y} = - 109.56 m / s

   

         v = (45.83 i ^ -109.56 j ^) m / s

A "laser cannon" of a spacecraft has a beam of cross-sectional area A. The maximum electric field in the beam is 2E. The beam is aimed at an asteroid that is initially moving in the direction of the spacecraft. What is the acceleration of the asteroid relative to the spacecraft if the laser beam strikes the asteroid perpendicularly to its surface, and the surface is not reflecting

Answers

Answer:

Acceleration of the asteroid relative to the spacecraft = 2ε[tex]E^{2}[/tex]A/m

Explanation:

The maximum electric field in the beam = 2E

cross-sectional area of beam = A

The intensity of an electromagnetic wave with electric field is

I = cε[tex]E_{0} ^{2}[/tex]/2

for [tex]E_{0}[/tex] = 2E

I = 2cε[tex]E^{2}[/tex]    ....equ 1

where

I is the intensity

c is the speed of light

ε is the permeability of free space

[tex]E_{0}[/tex]  is electric field

Radiation pressure of an electromagnetic wave on an absorbing surface is given as

P = I/c

substituting for I from above equ 1. we have

P = 2cε[tex]E^{2}[/tex]/c = 2ε[tex]E^{2}[/tex]    ....equ 2

Also, pressure P = F/A

therefore,

F = PA    ....equ 3

where

F is the force

P is pressure

A is cross-sectional area

substitute equ 2 into equ 3, we have

F = 2ε[tex]E^{2}[/tex]A

force on a body = mass x acceleration.

that is

F = ma

therefore,

a = F/m

acceleration of the asteroid will then be

a = 2ε[tex]E^{2}[/tex]A/m

where m is the mass of the asteroid.

What is the change in internal energy of an engine if you put 15 gallon of gasoline into its tank? The energy content of gasoline is 1.5 x 106 J/gallon. All other factors, such as the engine’s temperature, are constant. How many hours the engine can work if the power of the engine’s motor is 600 W? (8 marks)

Answers

Answer:

ΔU = 2.25 x 10⁸ J

t = 104.17 s

Explanation:

The change in internal energy of the engine can be given by the following formula:

ΔU = (Mass of Gasoline)(Energy Content of Gasoline)

ΔU = (1.5 x 10⁶ J/gallon)(15 gallon)

ΔU = 2.25 x 10⁸ J

Now, for the time of operation, we use the following formula of power.

P = W/t = ΔU/t

t = ΔU/P

where,

t = time of operation = ?

ΔU = Change in internal energy = 2.25 x 10⁸ J

P = Power of motor = 600 W

Therefore,

t = (2.25 x 10⁸ J)/(600 W)

t = (375000 s)(1 h/3600 s)

t = 104.17 s

Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?

Answers

Answer:

94.248 g/sec

Explanation:

For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:

[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]

And, the velocity of blood pumping is 30 cm^2

Now apply the following formula to solve the total current

[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]

Q =  94.248 g/sec

Basically we applied the above formula So, that the total current could come

Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.
Assuming that the skateboarder's acceleration is gsin 18.0 ∘, find his speed when he reaches the bottom of the ramp in 3.50 s .

Answers

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

For a skateboarder who starts from the rest, the speed when he reaches the bottom of the ramp will be 10.6 m/s.

What are Velocity and Acceleration?

The term "velocity" refers to a vector measurement of the rate and direction of motion. Velocity is the rate of movement in a single direction, to put it simply. Velocity can be used to determine how fast a rocket is heading into space and how fast a car is moving north on a congested motorway.

There are several types of velocity :

Instantaneous velocityAverage VelocityUniform VelocityNon-Uniform Velocity

The pace at which a person's velocity changes is known as acceleration. This implies that an object is accelerating if its velocity is rising or falling. An object that is accelerating won't have a steady change in location every second like an item moving at a constant speed does.

According to the question, the given values are :

Time, t = 3.50 sec

Initial Velocity, u = 0 m/s

Use equation of motion :

v = u+at

v = 0+ g sin 18 × 3.5

v = 10.6 m/s.

So, the final velocity will be 10.6 m/s.

To get more information about Velocity and Acceleration :

https://brainly.com/question/14683118

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At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.

Answers

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

     

The Law of Biot-Savart shows that the magnetic field of an infinitesimal current element decreases as 1/r2. Is there anyway you could put together a complete circuit (any closed path of current-carrying wire) whose field exhibits this same 1/r^2 decrease in magnetic field strength? Explain your reasoning.

Answers

Answer and Explanation:

There is no probability of obtaining such a circuit of closed track current carrying wire whose field of magnitude displays i.e.  [tex]B \alpha \frac{1}{r^2}[/tex]

The magnetic field is a volume of vectors

And [tex]\phi\ bds = 0[/tex]. This ensures isolated magnetic poles or magnetic charges would not exit

Therefore for a closed path,  we never received magnetic field that followed the [tex]B \alpha \frac{1}{r^2}[/tex] it is only for the simple current-carrying wire for both finite or infinite length.

A skater on ice with arms extended and one leg out spins at 3 rev/s. After he draws his arms and the leg in, his moment of inertia is reduced to 1/2. What is his new angular speed

Answers

Answer:

The new angular speed is [tex]w = 6 \ rev/s[/tex]

Explanation:

From the  question we are told that

      The angular velocity of the spin is  [tex]w_o = 3 \ rev/s[/tex]

       The  original moment of inertia is  [tex]I_o[/tex]

        The new moment of inertia is  [tex]I =\frac{I_o}{2}[/tex]    

Generally angular momentum is mathematically represented as

      [tex]L = I * w[/tex]

Now according to the law of conservation of momentum, the initial momentum is equal to the final momentum hence the angular momentum is constant so

         [tex]I * w = constant[/tex]

=>       [tex]I_o * w _o = I * w[/tex]

where w is the new angular speed  

  So  

          [tex]I_o * 3 = \frac{I_o}{2} * w[/tex]

=>        [tex]w = \frac{3 * I_o}{\frac{I_o}{2} }[/tex]

=>         [tex]w = 6 \ rev/s[/tex]

Use Coulomb’s law to derive the dimension for the permittivity of free space.



Answers

Answer:

Coulomb's law is:

[tex]F = \frac{1}{4*pi*e0} *(q1*q2)/r^2[/tex]

First, force has units of Newtons, the charges have units of Coulombs, and r, the distance, has units of meters, then, working only with the units we have:

N = (1/{e0})*C^2/m^2

then we have:

{e0} = C^2/(m^2*N)

And we know that N = kg*m/s^2

then the dimensions of e0 are:

{e0} = C^2*s^2/(m^3)

(current square per time square over cubed distance)

And knowing that a Faraday is:

F = C^2*S^2/m^2

The units of e0 are:

{e0} = F/m.

find the value of k for which the given pair of vectors are not equal
2ki +3j​ and 8i + 4kj

Answers

Answer:

5

Explanation:


When looking at the chemical symbol, the charge of the ion is displayed as the
-superscript
-subscript
-coefficient
-product

Answers

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

Answer:

superscript

Explanation:

An ice skater spinning with outstretched arms has an angular speed of 5.0 rad/s . She tucks in her arms, decreasing her moment of inertia by 11 % . By what factor does the skater's kinetic energy change? (Neglect any frictional effects.)

Answers

Answer:

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

         [tex]L_{f}[/tex] = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values ​​we can calculate the change in kinetic energy

         [tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

An asteroid that has an orbit with a semi-major axis of 4 AU will have an orbital period of about ______ years.

Answers

Answer:

16 years.

Explanation:

Using Kepler's third Law.

P2=D^3

P=√d^3

Where P is the orbital period and d is the distance from the sun.

From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.

P=?

P= √4^3

P= √256

P= 16 years.

Orbital period is 16 years.

A wave with a frequency of 1200 Hz propagates along a wire that is under a tension of 800 N. Its wavelength is 39.1 cm. What will be the wavelength if the tension is decreased to 600 N and the frequency is kept constant

Answers

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]

when the tension is decreased to 600 N, that is T₂ = 600 N

[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]

Therefore, the wavelength will be 33.9 cm

An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then released, it vibrates vertically. The equation of motion is s = 9 cos(t) + 9 sin(t), t ≥ 0, where s is measured in centimeters and t in seconds. (Take the positive direction to be downward.) (a) Find the velocity and acceleration at time t.

Answers

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

2. A pair of narrow, parallel slits sep by 0.25 mm is illuminated by 546 nm green light. The interference pattern is observed on a screen situated at 1.3 m away from the slits. Calculate the distance from the central maximum to the

Answers

Answer:

for the first interference m = 1   y = 2,839 10-3 m

for the second interference m = 2   y = 5,678 10-3 m

Explanation:

The double slit interference phenomenon, for constructive interference is described by the expression

                d sin θ = m λ

where d is the separation between the slits, λ the wavelength and m an integer that corresponds to the interference we see.

In these experiments in general the observation screen is L >> d, let's use trigonometry to find the angles

           tan θ = y / L

with the angle it is small,

          tan θ = sin θ / cos θ = sin θ

   

we substitute

         sin θ = y / L

         d y / L = m λ

the distance between the central maximum and an interference line is

        y = m λ L / d

let's reduce the magnitudes to the SI system

     λ = 546 nm = 546 10⁻⁹ m

     d = 0.25 mm = 0.25 10⁻³ m

let's substitute the values

      y = m 546 10⁻⁹ 1.3 / 0.25 10⁻³

      y =  m 2,839 10⁻³

the explicit value for a line depends on the value of the integer m, for example

for the first interference m = 1

the distance from the central maximum to the first line is y = 2,839 10-3 m

for the second interference m = 2

the distance from the central maximum to the second line is y = 5,678 10-3 m

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