A 180-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,050 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable

Answers

Answer 1

Answer:

27yrs

Explanation:

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Potential Energy = 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2


Related Questions

Improved balance is a primary benefit of regular cardiovascular exercise .

Answers

Answer:

Cardiovascular exercise is the activity use that aerobic metabolism and cellular reaction.

Explanation:

Cardiovascular exercise is activity increase heart rate and raises oxygen large muscle group of the body.Cardiovascular exercise is that contain cardio improve to the health mental health, heart health.Cardiovascular exercise such as walking, swimming, running is that exercise is benefit to the health.Cardiovascular exercise to the internal body organs that the healthy heart for the function and performance of the heart.Cardiovascular exercise that having involve feet of the ground this type of activity is called high impact of cardio.Cardio is a good and maintaining exercise for the lungs and heart or healthy bones.Cardio exercise is performed that to a water in reduce to the gravity of that pull on the body weight.Cardiovascular daily to build the stronger muscle and that control the blood pressure.  

2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s

Answers

Answer:

9 Brainly hahaha ............huh

The kinetic energy of a particle of mass 500g is 4.8j. Determine the velocity of the particle

Answers

Answer:

4.38 m/s

Explanation:

The answer is 4.38 m/s

Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2. Compared to the incident ray, what happens to the refracted ray?

Answers

Answer:

It bends away from the normal

Explanation:

From Snell's law of Refraction, when a ray passes from a medium of lower Refractive index to a medium with higher Refractive index, the Refractive ray will bend towards the normal. However, when the ray passes from a medium of higher Refractive index to a medium of lower Refractive index, the Refractive ray will bend away from the normal.

Now, from the question we are told that Light passes from a material with index of refraction 1.3 into one with index of refraction 1.2.

This means from a higher Refractive index to a lower one and from Snell's law as earlier said, the refracted ray will bend away from the normal

The refracted ray is seen to bend away from the normal.

Let us recall that an optically denser medium will have a higher refractive index. This means that the medium with a refractive index of  1.3  is the denser medium and the medium with a refractive index of  1.2 is the less dense medium.

From the statement in the question, we can boldly say that light is travelling from a denser to less dense medium given the values of the refractive index given. When light is travelling from a denser to a less dense medium, the refracted ray bends away from the normal.

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Hi Friends!

please help me with these questions!

SUBJECT: Chemistry, Physics,Biology

Answers

Answer:

q.1 : Air near candle gets heated up and after this it rises by convection so the thermometer B will receive more heat than the thermometer A So, according to the given condition thermometer B will show a greater rise in temperature.

q.2 : x is the pure sample of compound . y is the pure sample of element . z is the mixture of different elements

q.3 : the saliva contains an enzyme salivary amylase (ptyalin) which converts starch in roti into maltose, isomaltose and small dextrins called a-dextrin.

If 1.7 kg of 238Pu is required to power the spacecraft's data transmitter, for how long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.

Answers

The question is incomplete. Here is the complete question.

The isotope of Plutonium 238Pu is used to make thermoeletric power sources for spacecraft. Suppose that a space probe was launched in 2012 with 3.5 kg of 238Pu.

(a) If the half-life of 238Pu is 87.7 yr, write a function of the form [tex]Q(t)=Q_{0}e^{-kt}[/tex] to model the quantity Q(t) of 238Pu  left after t years. Round ythe value of k to 3 decimal places. Do not round intermediate calculations.

(b) If 1.7kg of 238Pu is required to power the spacecraft's data transmitter, for low long after launch would scientists be able to receive data? Round to the nearest year. Do not round intermediate calculations.

Answer: (a) [tex]Q(t)=3.5e^{-0.0079t}[/tex]

              (b) 91 years.

Explanation:

(a) Half-life is time it takes a substance to decrease to half of itself, i.e.:

Q(t) = [tex]0.5Q_{0}[/tex]

[tex]0.5Q_{0}=Q_{0}e^{-87.7k}[/tex]

[tex]0.5=e^{-87.7k}[/tex]

[tex]ln(0.5)=ln(e^{-87.7k})[/tex]

[tex]ln(0.5)=-87.7k[/tex]

[tex]k = \frac{ln(0.5)}{-87.7}[/tex]

k = 0.0079

Knowing k and [tex]Q_{0}[/tex]=3.5kg, function is [tex]Q(t)=3.5e^{-0.0079t}[/tex]

(b) Using function:

[tex]Q(t)=3.5e^{-0.0079t}[/tex]

[tex]1.7=3.5e^{-0.0079t}[/tex]

[tex]e^{-0.0079t}=\frac{1.7}{3.5}[/tex]

[tex]e^{-0.0079t}=0.4857[/tex]

[tex]ln(e^{-0.0079t})=ln(0.4857)[/tex]

[tex]-0.0079t=-0.7221[/tex]

[tex]t = \frac{-0.7221}{-0.0079}[/tex]

t = 91.41

t ≈ 91 years

Scientists will be able to receive data for approximately 91 years.

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.


Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?

Answers

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the smallest observable detail (in m) utilizing 495 MHz radar?

Answers

Answer:

0.61 m

Explanation:

The smallest observable length by the radar must be at least equal to or greater than the wavelength of the radar.

using the relationship

c = fλ

where

c is the speed of light in vacuum = 3 x 10^8 m/s

f is the frequency of the wave = 495 MHz = 4.95 x 10^8 Hz

λ is the wavelength = ?

λ = c/f = (3 x 10^8)/(4.95 x 10^8) = 0.61 m

answer to your question is 0.6m

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Answers

Answer:

A boat having stones floats in water. If stones are unloaded in water, what will happen to the level of water?​

Explanation:

An airplane propeller is rotating at 2200 rpm . You may want to review (Pages 255 - 259) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Rotation of a compact disc.
A. How many seconds does it take for the propeller to turn through 49.0?
t = 4.41x10^-3 S
B. Compute the propeller's angular velocity in rad/s
w = 194 rad/s

Answers

Answer:

a) 3.7 x 10^-3 s

b) 230.41 rad/s

Explanation:

The angular speed N = 2200 rpm (revolution per minute)

==> 2200/60 revolutions per sec = 36.67 rps

The total angle turned in one second = 36.67 x 360° = 13201.2°

if it takes 1 sec to revolve 13201.2°

then it will take t sec to rotate 49.0°

time t = 49/13201.2 = 3.7 x 10^-3 s

conversion to rad/s = 2πN/60 = (2 x 3.142 x 2200)/60 = 230.41 rad/s

Which statement about friction is true? (1 point)
o
Static friction and kinetic friction in a system always act in opposite directions of each other and in the same direction as the
applied force
Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the
applied force
Static friction and kinetic friction in a system always act in opposite directions of each other and in the opposite direction of the
applied force
O
Static friction and kinetic friction in a system always act in the same direction as each other and in the same direction as the
applied force.

Answers

Answer:static friction and kinetic friction in a system always act in the same direction as each other and n the opposite direction of the applie force . Is the correct answer

Explanation:

Static friction and kinetic friction in a system always act in the same direction as each other and in the opposite direction of the applied force. The correct option is B.

What is friction?

Friction is the force that prevents one hard material from scooting or rolling over the other.

Frictional forces, such as the locomotion required to walk without dropping, are advantageous, but they also create a significant amount of resistance to motion.

We can control cars because of friction between the tires and the road: more precisely, because there are three types of friction: rolling friction, starting friction, and sliding friction.

Friction reduces the speed of moving objects and can even stop them from moving. The friction between the objects generates heat. As a result, energy is wasted in the machines. Friction will cause wear and tear on the machine parts.

In a system, static and kinetic friction always act in the same direction and in the opposite direction of the applied force.

Thus, the correct option is B.

For more details regarding friction, visit:

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Which of the following statements is true vibrations ?

Answers

Answer:

C. Neither ultrasonic nor infrasonic vibrations can be heard by humans.

Explanation:

The complete question is

Which of the following statements is true of vibrations? A. The frequency of infrasonic vibrations is much too high to be heard by humans. B. Ultrasonic vibrations have a frequency lower than the range for normal hearing. C. Neither ultrasonic nor infrasonic vibrations can be heard by humans. D. Infrasonic vibrations are used in sonar equipment and to detect flaws in steel castings.

Ultrasonic vibrations have frequencies higher than our range of hearing, while infrasonic vibrations have frequencies lower than our range of hearing. Ultrasonic vibrations or sound is used in sonar equipment, and is used for detecting hidden flaws in steel castings and structures. Both infrasonic and ultrasonic fall below and above our normal hearing range respectively, and are only audible to dogs, cats, and some other mammals.

Answer:

Answer is " Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate."

Explanation:

My options were:

A)  Forced vibrations, such as those between a tuning fork and a large cabinet surface, result in a much lower sound than was produced by the original vibrating body.

B)  Resonance occurs as a result of sympathetic vibrations.

C)  A non-vibrating object can begin to vibrate as a result of forced vibrations.

D)  Two bodies with the same vibration frequency that are placed next to each other will exhibit sympathetic vibrations as one body causes the other to vibrate.

A is correct

.

what is the average flow rate in of gasoline to the engine of a plane flying at 700 km/h if it averages 100.0 km/l

Answers

Answer:

1.94cm³/s

Explanation:

1L = 1000cm³

Ihr = 3600s

So

Using

Average flow rate

Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L

= 1.94cm³/s

At the B end of the recessed horizontal bar, forces F1 and F2 of magnitudes 3 KN and 2KN respectively are applied and oriented as shown in the figure. Determine the magnitude of the resulting force and its orientation with respect to the horizontal.

Answers

Answer:

2.98 kN, 69.1° below the horizontal

Explanation:

Sum of forces in the x direction:

Fₓ = 3 kN cos 30° − 2 kN cos 40°

Fₓ ≈ 1.07 kN

Sum of forces in the y direction:

Fᵧ = -3 kN sin 30° − 2 kN sin 40°

Fᵧ = -2.79 kN

Magnitude of the resultant force is:

F² = Fₓ² + Fᵧ²

F = 2.98 kN

The direction of the resultant force is:

tan θ = Fᵧ / Fₓ

θ = -69.1°

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed?

Answers

Answer:

a

   [tex]m = 0.169 \ kg[/tex]

b

  [tex]|v_{max} |= 0.5653 \ m/s[/tex]

Explanation:

From the question we are told that

    The  spring constant is  [tex]k = 14 \ N/m[/tex]

     The  maximum extension of the spring is  [tex]A = 6.0 \ cm = 0.06 \ m[/tex]

     The number of oscillation is  [tex]n = 30[/tex]

      The  time taken is  [tex]t = 20 \ s[/tex]

Generally the the angular speed of this oscillations is mathematically represented as

           [tex]w = \frac{2 \pi}{T}[/tex]

where T is the period which is mathematically represented as

     [tex]T = \frac{t}{n}[/tex]

substituting values

     [tex]T = \frac{20}{30 }[/tex]

     [tex]T = 0.667 \ s[/tex]

Thus  

       [tex]w = \frac{2 * 3.142 }{ 0.667}[/tex]

       [tex]w = 9.421 \ rad/s[/tex]

this angular speed can also be represented mathematically as

       [tex]w = \sqrt{\frac{k}{m} }[/tex]

=>   [tex]m =\frac{k }{w^2}[/tex]

substituting values

      [tex]m =\frac{ 15 }{(9.421)^2}[/tex]

      [tex]m = 0.169 \ kg[/tex]

In SHM (simple harmonic motion )the equation for velocity is  mathematically represented as

        [tex]v = - Awsin (wt)[/tex]

The  velocity is maximum when  [tex]wt = \(90^o) \ or \ 1.5708\ rad[/tex]

     [tex]v_{max} = - A* w[/tex]

=>   [tex]|v_{max} |= A* w[/tex]

=>    [tex]|v_{max} |= 0.06 * 9.421[/tex]

=>   [tex]|v_{max} |= 0.5653 \ m/s[/tex]

If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y direction, the wave is traveling in what direction? Explain your answer.

Answers

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field  = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E  x B is along k direction or z  - axis so wave is propagating along z - axis .

The direction of motion of electromagnetic wave will be +z-direction.

Electromagnetic waves are waves that consist of the electric field and magnetic field.

The electric and magnetic fields are perpendicular to each other and the wave propagates in the direction perpendicular to both the fields.

Now, the direction of wave motion can be estimated by taking the cross-product of directional unit vectors of the electric and magnetic fields.

The electric field is in the +y direction and the magnetic field is in the +x-direction.

So, the direction of the wave will be,

[tex]i\times j=k[/tex]

Therefore, the direction of motion of electromagnetic wave will be +z-direction.

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A circular loop of wire has radius of 9.50 cmcm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0215 W/m2W/m2, and the wavelength of the wave is 6.90 mm.What is the maximum emf induced in the loop?
Express your answer with the appropriate units.

Answers

Answer:

The induced emf  is  [tex]\epsilon = 0.1041 \ V[/tex]  

Explanation:

From the question we are told that

   The  radius of the circular loop is  [tex]r = 9.50 \ cm = 0.095 \ m[/tex]

     The  intensity of the wave is  [tex]I = 0.0215 \ W/m^2[/tex]

      The wavelength is  [tex]\lambda = 6.90\ m[/tex]

Generally the intensity is mathematically represented as

         [tex]I = \frac{ c * B^2 }{ 2 * \mu_o }[/tex]

Here  [tex]\mu_o[/tex] is the permeability of free space with value  

         [tex]\mu_o = 4 \pi *10^{-7} N/A^2[/tex]

B is the magnetic field which can be mathematically represented from the equation as

          [tex]B = \sqrt{ \frac{ 2 * \mu_o * I }{ c} }[/tex]

substituting values

          [tex]B = \sqrt{ \frac{ 2 * 4\pi *10^{-7} * 0.0215 }{ 3.0*10^{8}} }[/tex]

          [tex]B = 1.342 *10^{-8} \ T[/tex]

The  area is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

       [tex]A = 3.142 * (0.095)^2[/tex]

       [tex]A = 0.0284[/tex]

The angular velocity is mathematically represented as

        [tex]w = 2 * \pi * \frac{c}{\lambda }[/tex]

substituting values          

       [tex]w = 2 * 3.142 * \frac{3.0*10^{8}}{ 6.90 }[/tex]  

        [tex]w = 2.732 *10^{8} rad \ s^{-1}[/tex]  

Generally the induced emf is mathematically represented as

        [tex]\epsilon = N * B * A * w * sin (wt )[/tex]

At maximum induced emf  [tex]sin (wt) = 1[/tex]

    So

         [tex]\epsilon = N * B * A * w[/tex]

substituting values

         [tex]\epsilon = 1 * 1.342 *10^{-8} * 0.0284 *2.732 *10^{8}[/tex]  

         [tex]\epsilon = 0.1041 \ V[/tex]  

         

A dentist uses a concave mirror (focal length 2 cm) to examine some teeth. If the distance from the object to the mirror is 1 cm, what is the magnification of the tooth

Answers

Answer:   2

Explanation:

1/2=1/1 +1/x

x=-2

magnification= 2/1

magnification=2

Can abnormality exist outside of a cultural context

Answers

you should search this up and put your own thoughts into it, it’s always good to learn something new!!

Which nucleus complete the following equation

Answers

(C) [tex]^{208}_{84}\text{Po}[/tex]

Explanation:

[tex]^{212}_{86}\text{Rn} \rightarrow \:^4_2\text{He} + \:^{208}_{84}\text{Po}[/tex]

an electron starts from rest from a fixed point charge with q. what total potential difference accelerates the electron to being very far away from Q

Answers

Answer:

V = (k*Q)/R

Explanation:

Total potential difference accelerates the electron to being very far away from Q is;

V = (k*Q)/R

Where,

V is the Potential Difference in Joules per Coulomb

k is the constant

Q is the charge in Coulomb

R is Electron distance in cm or m

Example

An electron starts from rest 66.1 cm from a fixed point charge with Q = -0.120 μC. What total potential difference accelerates the electron from being very far away from Q

For k = 9.0*10^9 N*m^2/C^2

V = (k*Q)/R

V = (9.0*10^9 * -0.120*10^-6)/0.661

V = -1633.9 Volt.

The answer will change to positive because V = (k*Q)/R is negative at the outset and Zero far away.

The electron (with a negative charge) has a positive energy in the beginning and that gets converted into a positive kinetic energy "far away".

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387 N. Between the box of mass m2 and the box of mass m3, the force meter reads F23=2304 N. Assume that the ropes and force meters are massless.

Answers

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

[tex]F_{pull}=m_{T}.a[/tex]

[tex]m_{T}=\frac{F_{pull}}{a}[/tex]

[tex]m_{T}=\frac{3615}{1.516}[/tex]

[tex]m_{T}=2384.5[/tex]

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For [tex]m_{1}[/tex]:

The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

[tex]m_{1} = \frac{F_{12}}{a}[/tex]

[tex]m_{1} = \frac{1387}{1.516}[/tex]

[tex]m_{1}[/tex] = 915kg

For [tex]m_{2}[/tex]:

There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]

[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]

[tex]m_{2}[/tex] = 605kg

For [tex]m_{3}[/tex]:

[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]

[tex]m_{3} = 2384.5-1520.0[/tex]

[tex]m_{3}[/tex] = 864.5kg

If one could transport a simple pendulum of constant length from the Earth's surface to the Moon's, where acceleration due to gravity is one-sixth (1/6) that on the Earth, by what factor would be the pendulum frequency be changed

Answers

Answer:

The frequency will change by a factor of 0.4

Explanation:

T = 2(pi)*sqrt(L/g)

Since g(moon) = (1/6)g(earth), the period would change by sqrt[1/(1/6)] = sqrt(6) ~ 2.5 times longer on the moon. Since the period & frequency are inverses, the frequency would be 1/2.5 or 0.4 times shorter on the moon.

You're conducting an experiment on another planet. You drop a rock from a height of 1 m and it hits the ground 0.4 seconds later. What is acceleration due to gravity on the planet ?

Answers

Answer:

Here,

v (final velocity) = 0

u (initial velocity) = u

a = ?

s = 1m

t = 0.4s

using the first equation of motion,

0 = u + 0.4a

= -0.4a = u

using the second equation of motion:

1 = 0.4u + 0.08a

from the bold equation

1 = 0.4(-0.4a) + 0.08a

1 = -0.16a + 0.08a

1 = -0.08a

a = -1/0.08

a = -100/8

a = -12.5 m/s/s

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A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?

Answers

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{1}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{1}=0.01021\ Wb[/tex]

We need to calculate the magnetic flux

Using formula of flux

[tex]\phi_{2}=BA\cos\theta[/tex]

Put the value into the formula

[tex]\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0[/tex]

[tex]\phi_{2}=0.00127\ Wb[/tex]

We need to calculate the magnitude of the average induced emf

Using formula of emf

[tex]\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})[/tex]

Put the value into t5he formula

[tex]\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})[/tex]

[tex]\epsilon=0.016867\ V[/tex]

[tex]\epsilon=16.87\ mV[/tex]

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

[tex]I=\dfrac{\epsilon}{R}[/tex]

Put the value into the formula

[tex]I=\dfrac{0.016867}{2.5}[/tex]

[tex]I=0.00675\ A[/tex]

[tex]I=6.75\ mA[/tex]

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

A wave travelling along the positive x-axis side with a
frequency of 8 Hz. Find its period, velocity and the distance covered
along this axis when its wavelength and amplitude are 40 and 15 cm
respectively.​

Answers

Explanation:

The frequency is given to be f = 8 Hz.

Period is the inverse of frequency.

T = 1/f = 0.125 s

Velocity is wavelength times frequency.

v = λf = (0.40 m) (8 Hz) = 3.2 m/s

The wave travels 3.2 meters every second.

What is an understood decimal

Answers

The decimal place accuracy of a number is the number of digits to the right of the decimal point. ... If there is no decimal point, it is understood to be after the last digit on the right and there is no place (zero place) accuracy. The significant digits of a number are those digits that are most accurate.

Heat and thermodynamics Numerical ​

Answers

Answer:

K = 227.04 W/m.°C

Explanation:

First we need to find the heat required to melt the ice:

q = m H

where,

q = heat required = ?

m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg

H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg

Therefore,

q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)

q = 2839 J

Now, we find the heat transfer rate through rod:

Q = q/t

where,

t = time = (10 min)(60 s/1 min) = 600 s

Q = Heat Transfer Rate = ?

Therefore,

Q = 2839 J/600 s

Q = 4.73 W

From Fourier's Law of Heat Conduction:

Q = KA ΔT/L

where,

K = Thermal Conductivity = ?

A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²

L = Length of rod = 60 cm = 0.6 m

ΔT = Difference in temperature = 100°C - 0°C = 100°C

Therefore,

4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m

K = (4.73 W)/(0.0208 m.°C)

K = 227.04 W/m.°C

What is tensor quantity?
Is Inertia a tensor? give reason​

Answers

Answer:

A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts. Stress and strain are both tensor quantities. ... A tensor is a quantity, for example a stress or a strain, which has magnitude, direction, and a plane in which it acts.

Inertia Tensor. where I = the inertia tensor. The angular momentum of a rigid body rotating about an axis passing through the origin of the local reference frame is in fact the product of the inertia tensor of the object and the angular velocity. ... As shown in [7], the inertia tensor is symmetric.

Explanation:

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Answers

Answer:

Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.

Explanation:

As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.

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