Answer:
The energy stored in the solenoid is 7.078 x 10⁻⁵ J
Explanation:
Given;
diameter of the solenoid, d = 2.80 cm
radius of the solenoid, r = d/2 = 1.4 cm
length of the solenoid, L = 14 cm = 0.14 m
number of turns, N = 200 turns
current in the solenoid, I = 0.8 A
The cross sectional area of the solenoid is given as;
[tex]A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2[/tex]
The inductance of the solenoid is given by;
[tex]L = \frac{\mu_0 N^2A}{l} \\\\L = \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H[/tex]
The energy stored in the solenoid is given by;
E = ¹/₂LI²
E = ¹/₂(2.212 x 10⁻⁴)(0.8)²
E = 7.078 x 10⁻⁵ J
Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J
The energy stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800 A is 7.07 × 10⁻⁵ J.
A solenoid is a type of electromagnet, whose purpose is to generate a controlled magnetic field through a coil wound into a tightly packed helix.
First, given the diameter of the solenoid (d) is 2.80 cm (0.0280 m), we can calculate the cross-sectional area (A) using the following expression.
[tex]A =\pi \times (\frac{d}{2} )^{2} = \pi \times (\frac{0.0280}{2} )^{2} = 6.16 \times 10^{-4} m^{2}[/tex]
Next, we will calculate the inductance (L) of the solenoid using the following expression.
[tex]L = \frac{\mu _0 \times N^{2} \times A}{l} = \frac{(4\pi \times 10^{-7}H/m ) \times 200^{2} \times (6.16 \times 10^{-4}m^{2} )}{0.140m} = 2.21 \times 10^{-4} H[/tex]
where,
μ₀ is the vacuum permeabilityN is the number of turnsl is the length of the solenoidFinally, for a current (I) of 0.800 A, we can calculate the energy stored (E) using the following expression.
[tex]E = \frac{1}{2} \times L \times I^{2} = \frac{1}{2} \times (2.21 \times 10^{-4}H ) \times (0.800A)^{2} = 7.07 \times 10^{-5} J[/tex]
The energy stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800 A is 7.07 × 10⁻⁵ J.
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Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]|v| = 6.93[/tex]
Explanation:
From the question we are told that
The initial point is [tex](x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )[/tex]
The terminal point is [tex](x_2 , y_2 , z_2) = (-5 , 11, 8 )[/tex]
Generally the magnitude of the vector is mathematically represented as
[tex]|v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}[/tex]
=> [tex]|v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}[/tex]
=> [tex]|v| = 6.93[/tex]
which of the following require that you also state direction
A. speed
B. displacement
C. acceleration
D. instantaneous speed
SCIENCE QUESTION FOR 20 POINTS
Answer:
d
Explanation:
A 500 g ball and a 200 g ball are dropped from a tower. Both balls are the same size and there is no air resistance. When do they hit the ground? The 200 g mass hits the ground first. They both hit the ground at the same time. The 500 g mass hits the ground first.
Answer:
They both hit the ground at the same time.
Explanation:
if you drop a 100 g and 200 g mass simultaneously from the same height they’ll hit the ground at the same time.
Yes, there is a role of a very simple equation of kinematics i.e.
s=ut+1/2(at^2)
so, s means the height is the same for both masses(h), u(initial velocity) is obviously zero cause you are dropping it, a (acceleration) is due to the gravity here only ‘g’.
time is taken for a particular object to hit the ground,
t= (2h/g)^1/2 .
We can see that there is no term for the mass of objects, meanwhile, the mass of the object has nothing to do with the time taken in hitting the ground in this case.
What's the value of 96,745 joules in Btus?
Answer:
The value of 96,745 joules in Btus is 91.6965673 or 91.7 Btus.
Explanation:
Btus stands for British thermal unit. It is the unit of energy. 96,745 Joules is equal to 91.7 Btus of energy.
What is Energy?
Energy is the quantitative property of an object which is transferred to an object or to a physical system, that is recognizable in the performance of the work done and in the form of heat and light generated by the object or system. Energy is a conserved quantity, according to the law of conservation of energy which states that energy can be transformed, but not created or destroyed.
A British thermal unit (Btu) is a measure of the heat content or energy of fuels or energy sources. 1 BTU is roughly equal to the amount of energy which is required to raise the temperature of one pound of water by 1 degree Fahrenheit.
1 Joule = 0.000947817 Btu
Therefore, 96,745 Joules = 91.696 or 91.7 Btus
The value of 96,745 joules in Btus is 91.6965673 or 91.7 Btus.
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What does BMI (Body Mass Index) attempt to measure?
O A. Target heart rate
O B. Energy levels
O C. Body fat percentage
O D. Weight
Answer: C. Body fat percentage
Explanation:
Answer:
your answer is C
Explanation:
A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bottom of the hill, it is noted that a person sitting on a scale in the car reads a value off the scale that is 80% more than normally when the scale is at rest in a bathroom. With what speed is the car traveling
Answer:
Explanation:
reading of scale = reaction force of surface R
centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .
R = mg + m v² / R
given ,
m v² / R = .80 mg
v² = .80 x g x R
= .8 x 9.8 x 9 = 70.56
v = 8.4 m /s
When a person wearing a helmet rides a bicycle, the helmet experiences which type of friction?
A. Fluid
B. Static
C. Sliding
D. Rolling
Answer:
B
Explanation:
Answer:
B.)Static
Hope I helped
A car is traveling at 15 m/sm/s . Part A How fast would the car need to go to double its kinetic energy
Answer:
21.21 m/s
Explanation:
Let KE₁ represent the initial kinetic energy.
Let v₁ represent the initial velocity.
Let KE₂ represent the final kinetic energy.
Let v₂ represent the final velocity.
Next, the data obtained from the question:
Initial velocity (v₁) = 15 m/s
Initial kinetic Energy (KE₁) = E
Final final energy (KE₂) = double the initial kinetic energy = 2E
Final velocity (v₂) =?
Thus, the velocity (v₂) with which the car we travel in order to double it's kinetic energy can be obtained as follow:
KE = ½mv²
NOTE: Mass (m) = constant (since we are considering the same car)
KE₁/v₁² = KE₂/v₂²
E /15² = 2E/v₂²
E/225 = 2E/v₂²
Cross multiply
E × v₂² = 225 × 2E
E × v₂² = 450E
Divide both side by E
v₂² = 450E /E
v₂² = 450
Take the square root of both side.
v₂ = √450
v₂ = 21.21 m/s
Therefore, the car will travel at 21.21 m/s in order to double it's kinetic energy.
For the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.
Given the data in the question;
Speed of the car; [tex]v_1 = 15m/s[/tex]
Speed for the car to double its kinetic energy; [tex]v_2 = \ ?[/tex]
Using the expression for kinetic energy:
[tex]K_E = \frac{1}{2}mv^2[/tex]
Where m is the mass and v is the velocity
Now, Initial Kinetic Energy will be;
[tex]K_E_1 = \frac{1}{2}mv_1^2 \\\\K_E_1 = \frac{1}{2}*m*(15)^2\\\\K_E_1 = \frac{1}{2}*m*225\\\\K_E_1 = 112.5*m[/tex]
For the kinetic energy to become double
[tex]K_E_2 = 2 * K_E_1\\\\\frac{1}{2}mv_2^2 = 2 * ( 112.5 * m)\\\\\frac{1}{2}mv_2^2 = 2m( 112.5)\\\\v_2^2 = 4( 112.5)\\\\v_2^2 = 450\\\\v_2 = \sqrt{450}\\\\v_2 = 21.21m/s[/tex]
Therefore, for the car to be able to double its kinetic energy, it would need to travel at a speed of approximately 21.21m/s.
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Why would a baseball traveling at 20 m/s have more kinetic energy than a baseball traveling at 5 m/s?
Answer:
because formula of kinetic energy is K=1/2mv^2 that means more the velocity then more kinetic energy
The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from equator, Longitude 121.
Answer:
The tangential speed at Livermore is approximately 284.001 meters per second.
Explanation:
Let suppose that the Earth rotates at constant speed, the tangential speed ([tex]v[/tex]), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:
[tex]v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi[/tex] (1)
Where:
[tex]\Delta t[/tex] - Rotation time, measured in seconds.
[tex]R[/tex] - Radius of the Earth, measured in meters.
[tex]\phi[/tex] - Latitude of the city above the Equator, measured in sexagesimal degrees.
If we know that [tex]\Delta t = 86160\,s[/tex], [tex]R = 6.371\times 10^{6}\,m[/tex] and [tex]\phi = 37.6819^{\circ}[/tex], then the tangential speed at Livermore is:
[tex]v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}[/tex]
[tex]v\approx 284.001\,\frac{m}{s}[/tex]
The tangential speed at Livermore is approximately 284.001 meters per second.
From the 1780s to the late 1800s, people thought the amount of land and resources in the West was limited
Answer:true
Explanation:
The figure below shows two forces F1 = 30.0 N and F2 = 20.0 N acting on an object that
can rotate around a center axle. Find the net torque in Nm. (Tolerance +/- 0.05)
Answer:
0.2Nm
Explanation:
F1 = 30.0N
F2 = 20.0N
R = 2.0cm = 0.02m
F1R1 - F2R2 = net Torque
30(0.02) - 20(0.02) = net Torque
0.6-0.4 = net Torque
Therefore net Torque = 0.2Nm
The figure below shows two forces [tex]F_1[/tex] = 30.0 N and [tex]F_2[/tex] = 20.0 N acting on an object that can rotate around a centre axle. The net torque in 0.2 N-m
What is torque?The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is a torque.
Torque is defined as Γ=r×F=r F sin(θ). In other words, torque is the cross product of the force vector, where 'a' is the angle between r and F, and the distance vector (the distance between the pivot point and the place where force is applied).
Torque due to [tex]F_1[/tex] = 30.0 N is 0.6 N-m outside the plane and due to [tex]F_2[/tex] = 20.0 N is 0.4 N-m inside the plane so net torque is 0.2 N-m outside the plane.
The figure below shows two forces [tex]F_1[/tex] = 30.0 N and [tex]F_2[/tex] = 20.0 N acting on an object that can rotate around a centre axle. The net torque in 0.2 N-m outside the plane.
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An astronaut with mass 84kg is at rest in space, before firing her water pack to move toward the shuttle. If the amount of water shot out is
2kg and it is fired at a speed of 10m/s, what will the speed of the astronaut be?
Answer:
The astronaut is moving at a speed of 0.238 m/s in a direction opposite the direction of the water shot out.
Explanation:
We are given;
Mass of astronaut; m1 = 84 kg
Mass of water shoot out; m2 = 2 kg
Initial speed of astronaut; u1 = 0 m/s
Initial speed of water shoot out; u2 = 0 m/s
Final speed of shoot out; v2 = 10 m/s
From law of conservation of momentum, we can say that;
Initial momentum = final momentum
Thus;
m1•u1 + m2•u2 = m1•v1 + m2•v2
Where v1 is the final speed of the astronaut
Plugging in the relevant values, we get;
(84 × 0) + (2 × 0) = (84 × v1) + (2 × 10)
0 = 84v1 + 20
-20 = 84v1
v1 = -20/84
v1 = -0.238 m/s
The negative sign indicates that the astronaut is moving 0.238 m/s in a direction opposite the direction of the water shot out.
Need help with this physics question!
A baseball is hit straight up at 60 m/s. At 9 seconds what is the displacement, the acceleration and velocity?
Answer:
Acceleration: -9.8 m/s^2
Velocity: -28.2 m/s
Displacement: 143.1 m
Explanation:
The acceleration of gravity for any object close to earth is approximately -9.8 m/s^2.
Now, to find the velocity after 9 seconds, we can use a kinematics formula, where x is the final velocity:
Final Velocity = Initial Velocity + Acceleration * Time
x = 60 + -9.8*9
x = 60 - 88.2
x = -28.2
The velocity is -28.2 m/s.
Lastly, to find the displacement, we can use another kinematics formula, where y is the displacement:
Displacement = (Final Velocity + Initial Velocity)/2 * Time
y = (-28.2 + 60)/2 * 9
y = 143.1
The displacement is 143.1 meters.
it is rate for any motion to
a. stay the same for very long
b. change quickly
c. increase in velocity
d. decrease in speed
Answer:
a. stay the same for very long
Explanation:
It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.
For motion to remain constant, there must not be a net force acting on the body All the forces on the body must be balanced. This is very hard to come by. Motion changes very frequently.A 0.106-A current is charging a capacitor that has square plates 4.60 cm on each side. The plate separation is 4.00 mm.
(a) Find the time rate of change of electric flux between the plates in V·m/s
(b) Find the displacement current between the plates in A
Answer:
a
[tex]\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s[/tex]
b
[tex]I = 0.106 \ A[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 0.106 \ A[/tex]
The length of one side of the square [tex]a = 4.60 \ cm = 0.046 \ m[/tex]
The separation between the plate is [tex]d = 4.0 mm = 0.004 \ m[/tex]
Generally electric flux is mathematically represented as
[tex]\phi_E = \frac{Q}{\epsilon_o}[/tex]
differentiating both sides with respect to t is
[tex]\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}[/tex]
=> [tex]\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I[/tex]
Here [tex]\epsilon_o[/tex] is the permitivity of free space with value
[tex]\epsilon _o = 8.85*10^{-12} C/(V \cdot m)[/tex]
=> [tex]\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}[/tex]
=> [tex]\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s[/tex]
Generally the displacement current between the plates in A
[tex]I = 8.85*10^{-12} * 1.1977 *10^{10}[/tex]
=> [tex]I = 0.106 \ A[/tex]
a sample of steam at 15 bar pressure and 400°c temperature is first expanded at a constant enthalpy to 6 bar and then expanded isentropically where the dryness fraction of the steam is 0.90. find the change of enthalpy, the pressure of isentropically expanded steam and change of entropy.
Answer:
Answer.
Explanation:
Explanation.
A boy on a treasure hunt walks 12 paces West, then 21 paces North, then 25 paces East. What is the magnitude of boy’s resultant displacement in meters?
Answer: The displacement is 18.8 meters.
Explanation:
Let's define the North as the positive y-axis and the East as the positive x-axis.
We also can assume that if the point (x, y) represents the position of the boy, the initial position is (0 paces, 0 paces).
"A boy on a treasure hunt walks 12 paces West,"
West would be the negative x-axis in this case, then:
the position will be (-12 paces, 0 paces)
"then 21 paces North"
Now the position will be (-12 paces, 21 paces)
"then 25 paces East."
Now the position will be (-12 paces + 25 paces, 21 paces)
= (13 paces, 21 paces)
Now we have the initial and final positions of the boy, (0 paces, 0 paces) and (13 paces, 21 paces) respectively.
We want to find the displacement, which is defined as the distance between the final position and the initial position.
Remember that for two vectors (a, b) and (c, d), the distance is:
D = √( (a - c)^2 + (b - d)^2)
In this case, the displacement will be:
D = √( ( 13 paces - 0 paces)^2 + (21 paces - 0 paces)^2) = 24.7 paces.
But we want this quantity in meters, so we can use the relation:
1 pace = 0.762 meters.
Then 24.7 paces, are 24.7 times 0.762 meters.
D = 24.7*0.762 m = 18.8 meters.
A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At what angle from the beam axis will the first order peak occur if the tube emits light with wavelength of 405.3 nm? Tries 0/20 At what angle will the second order peak occur?
Answer:
θ₁ = 5.4°
θ₂ = 10.86°
Explanation:
The angle ca be found by using grating equation:
mλ = d Sinθ
where,
m = order of diffraction
λ = wavelength = 405.3 nm = 4.053 x 10⁻⁷ m
d = grating element = 1/230 lines/mm = 0.0043 mm/line = 4.3 x 10⁻⁶ m/line
θ = angle = ?
FOR m = 1:
(1)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁
Sin θ₁ = 0.09425
θ₁ = Sin⁻¹(0.09425)
θ₁ = 5.4°
FOR m = 2:
(2)(4.053 x 10⁻⁷ m) = (4.3 x 10⁻⁶ m/line) Sin θ₁
Sin θ₂ = 0.1885
θ₂ = Sin⁻¹(0.1885)
θ₂ = 10.86°
In a Young's double slit experiment, the separation between the slits is 1.20 × 10–4 m; and the screen is located 3.50 m from the slits. The distance between the central bright fringe and the second-order bright fringe is 0.0375 m. What is the wavelength of the light used in this experiment?
Answer:
Explanation:
separation between slit d = 1.2 x 10⁻⁴ m
distance of screed D = 3.5 m
distance of second order right fringe
= 2λD / d where λ is wavelength of light .
2λD / d = .0375
2 x λ x 3.5 / 1.2 x 10⁻⁴ = .0375
λ = 6428 x 10⁻¹⁰ m
= 643 nm .
A bag of groceries has a weight of 44 Newton’s. Find it’s mass in kilograms
Answer:
It's mass would be 4.486768kg..
What is the approximate efficiency of the engine?
A heat engine has a temperature of 1800 K. Some of the
heat from the engine flows to the surroundings, which
are at a temperature of 160 K.
0 16%
O 91%
O 103%
O 109%
Answer:91% I think
Explanation:
The typical engine efficiency for a modern petrol combustion engine is from 20 to 30 percent. The remaining 70 to 80 percent of the heat energy of the petrol is released from the engine as friction loss, mechanical sound energy, or exhaust heat. Thus, option B is correct.
What approximate efficiency of the engine?An engine's efficiency is measured by the heat produced divided by the amount of useful work done. Is the task complete. Is the task finished. Please be aware that the phrase “work done” refers to the force applied to the clutch or driveshaft.
Utilizing the following formula, efficiency may be represented as a ratio: output input. Output, often called work output, is the entire quantity of productive work completed, excluding waste and spoilage.
You may determine the efficiency by figuring out what proportion of the energy input results in the desired output: Efficiency is defined as Energy Out/Energy In 100%.
Therefore, This implies that the work produced by thermodynamic expansion is reduced by the friction and other losses.
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Selection
boy throws a 1.2 kg ball straight up at 12 m/s. How high does the ball go? 1 point
(Answer to two digits)
Answer:
69
Explanation:
A film of soapy water ( n= 1.33 ) on top of a plastic cutting board has a thickness of 233 nm. What color is most strongly reflected if it is illuminated perpendicular to its surface?
Answer:
620nm
Explanation:
According to the question,
Refractive index, n = 1.33Thickness, t = 233 nmAs we know,
→ [tex]t = \frac{\lambda}{2n}[/tex]
or,
→ [tex]\lambda = 2tn[/tex]
By putting the values, we get
[tex]= 2\times 233\times 10^{-9}\times 1.33[/tex]
[tex]= 620 \ nm[/tex]
Thus the above response is correct.
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In a two-dimensional Cartesian system, the x-component of a vector is known, and the angle between vector and x-axis is known. Which operation is used to calculate the magnitude of the vector?
a. dividing by cosine
b. dividing by sine
c. multiplying by cosine
d. multiplying by sine
100 POINTS + BRAINLIEST, HEART, & FIVE STAR.
In the one pully system when you move the mass from the 20 cm mark to the 15 cm mark, it moves 5 cm. How far did you pull the string.
Question 1 options:
5 cm
10 cm
15 cm
20 cm
In the two pully system if you lifted the weight from the 20 cm mark to the 15 cm mark moving 5 cm. How far did you pull the rope down?
Question 2 options:
5 cm
10 cm
15 cm
20 cm
Answer:
Question 1 the answer is 5 cm for the one pulley system, I just took the quiz
Question 2 is 10 cm
Explanation:
The rope only needs to move 5 cm for the mass to move from 15 to 20.
For the 2 pulley system the string has to move twice as far making it 10 cm. I took the quiz and got 100%
What is the ratio of the radii of curvature for a proton and an an electron traveling through this apparatus
Answer:
hello your question is incomplete below is the missing part of the question
Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric field and then injected into a region of constant magnetic field with a field strength of 0.55 T.
answer : ≈ 42
Explanation:
Ratio of the Radii of curvature for a proton and an electron travelling through this apparatus
= [tex]\frac{Rp}{Re}[/tex] = 0.025 m / 5.94 * 10^-4 m = 42.08
Rp = 0.025 m ( calculated )
Re = 5.94 * 10^-4 ( calculated )
A person swings a 0.546-kg tether ball tied to a 4.56-m rope in an approximately horizontal circle. If the maximum tension the rope can withstand before breaking is 11.0 N, what is the maximum angular speed of the ball
Answer:
2.1 rad/s
Explanation:
Given that,
Mass of a tether ball, m = 0.546 kg
Length of a rope, l = 4.56 m
The maximum tension the rope can withstand before breaking is 11.0 N
We need to find the maximum angular speed of the ball. Let v is the linear velocity. The maximum tension is balanced by the centripetal force acting on it. It can be given by :
[tex]F=\dfrac{mv^2}{r}\\\\v=\sqrt{\dfrac{Fr}{m}} \\\\v=\sqrt{\dfrac{11\times 4.56}{0.546}} \\\\=9.584\ m/s[/tex]
Let [tex]\omega[/tex] is the angular speed of the ball. The relation between the angular speed and angular velocity is given by :
[tex]v=r\omega\\\\\omega=\dfrac{v}{r}\\\\=\dfrac{9.584}{4.56}\\\\=2.1\ rad/s[/tex]
So, the maximum angular speed of the ball is 2.1 rad/s.
By what factor must you increase the intensity of a sound in order to hear a 1.0-dB rise in the sound level?
Answer:
The right approach will be "1.3". A further explanation is given below.
Explanation:
As we know,
In Decibels, the change in sound volume will be:
= [tex]10log\frac{I_1}{I_2}[/tex]
Now,
According to the question,
⇒ [tex]1=10 log \frac{I_1}{I_2}[/tex]
By applying cross multiplication and putting the value of log, we get
⇒ [tex]10^{\frac{1}{10} }=\frac{I_1}{I_2}[/tex]
⇒ [tex]1.26=\frac{I_1}{I_2}[/tex]
⇒ [tex]\frac{I_1}{I_2}=1.3[/tex]
A local AM radio station broadcasts at a frequency of 627 kHz. Calculate the wavelength at which it is broadcasting. Wavelength
Answer:
Wavelength = 478.46 m
Explanation:
It is given that,
A local AM radio station broadcasts at a frequency of 627 kHz, f = 627000 Hz
We need to find the wavelength at which it is broadcasting. The wavelength is given by :
[tex]\lambda=\dfrac{c}{f}\\\\=\dfrac{3\times 10^8}{627\times 10^3}\\\\=478.46\ m[/tex]
So, the wavelength is 478.46 m.