Answer:
Emax = 13.209V
Explanation:
Given the following data;
Magnitude of the magnetic field, B = 0.35T
Number of turns, n = 200
Frequency, f = 60Hz
Area of the coil = 5cm² (to square meter = 5 *10^-4 = 0.0005m²)
[tex] Angular frequency, w = 2 \pi f[/tex]
Substituting into the above equation, we have;
[tex] Angular frequency, w = 2 *3.142 *60[/tex]
Angular frequency, w = 377.04
We know that as the coil rotates between 0 and 1; sin(wt) would vary sinusoidally. Therefore, the maximum EMF occurs when the value of sin(wt) is equal to 1.
Mathematically, the maximum EMF is given by;
[tex] Emax = nBAw[/tex]
Where;
Emax is the maximum EMF. n is the number of turns. B is the magnitude of the magnetic field. A is the area of the coil. w is the angular frequency.Substituting into the equation, we have;
[tex] Emax = 200 * 0.35 * 0.0005 * 377.4[/tex]
Emax = 13.209V
Therefore, the maximum EMF in the coil is 13.209 volts.
The maximum emf in the coil will be 13.209V.
What is induced emf?Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.
When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.
The given data in the problem is;
B is the magnitude of the magnetic field,= 0.35T
n is the number of turns, = 200
f is the Frequency = 60Hz
A is the area of the coil = 5cm² = 0.0005m²
The maximum emf is given as;
[tex]\rm E_{max}= nBA\omega \\\\ \rm E_{max}= 200 \times 0.35 \times 0.005 \times 2 \pi \times 60\\\\ \rm E_{max}= 13.209 \ V[/tex]
Hence the maximum emf in the coil will be 13.209V.
To learn more about the induced emf refer to the link;
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WILL MARK BRAINLIEST PLS HELPPP -- Which of Newton’s Laws explains why the satellite would collide with the moon if gravity is “turned off?”
picture attached
Answer:
b
Explanation:
If an object is not moving the forces acting on it are balanced.
This statement is true or false because?
Answer:
True
Explanation:
The answer is True
A Painting is on the wall using 2 strings, hanging perpendicular to the floor. 5 The strings have a tension force of 490N each, what is the mass of the hanged painting?
Answer:
100 kg
Explanation:
Two strings are used to hang the painting means the total weight of the painting is split into 2 the two string.
Each string has tension 490 N this means the total weight of the painting is
[tex]w=490\times 2=980\ \text{N}[/tex]
The tension of both the strings combimed will be the weight of the painting
Weight is given by
[tex]w=mg[/tex]
where,
m = Mass of the object
g = Acceleration due to gravity = [tex]9.8\ \text{m/s}^2[/tex]
Here
w = 980 N
[tex]980=m\times 9.8\\\Rightarrow m=\dfrac{980}{9.8}\\\Rightarrow m=100\ \text{kg}[/tex]
So, mass of the painting is 100 kg.
1. An object carries a charge of q = +4x10-8C.
How many electrons are needed to make it elec-
trically neutral?
Answer: i can help you
Explanation:
[tex]2.5[/tex]×[tex]10^{11}[/tex] electrons are needed to make an object neutral which is carrying a charge of [tex]q=+4[/tex]×[tex]10^{-8}[/tex]
What is a charge of an electron?The elementary charge (e) has a negative sign, and the electron's charge is equal to its magnitude. The charge of the electron is -1.602 x [tex]10^{-19}[/tex] coulombs (C), while the elementary charge is approximately 1.602 x [tex]10^{-19}[/tex] coulombs (C).Thus, -e can be used to represent the charge of an electron. The proton only possesses an e-charge despite being far more massive than the electron. The number of protons and electrons in neutral atoms is constant.The electron was unquestionably discovered by JJ Thomson. He only succeeded in obtaining the electron's charge-to-mass ratio, though, despite all the experiments he conducted on it. Robert Millikan holds the distinction of being the first to quantify the charge on an electron through his oil-drop experiment in 1909.To learn more about charge, refer to
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Please help this question
Answer:
Kantians would say that Sharkeisha acted immorally since she increased overall pain.
In a head-on collision, a ball of mass 0.3 kg travelling with velocity 2.8 m/s in the positive x-direction hits a stationary second ball of mass 0.4 kg. What is the velocity of the 0.3 kg ball after the collision? Assume collision is elastic.
Answer:
The final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
Explanation:
Given;
mass of the first object, m₁ = 0.3 kg
initial velocity of the first ball, u₁ = 2.8 m/s
mass of the second ball, m₂ = 0.4 kg
initial velocity of the second ball, u₂ = 0
let the final velocity of the first ball, = v₁
let the final velocity of the second ball, = v₂
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.3 x 2.8) + (0.4 x 0) = 0.3v₁ + 0.4v₂
0.84 = 0.3v₁ + 0.4v₂
2.8 = v₁ + 1.333v₂ -------equation (1)
Apply one-direction velocity;
u₁ + v₁ = u₂ + v₂
2.8 + v₁ = 0 + v₂
v₂ = 2.8 + v₁
substitute the value of v₂ into equation (1)
2.8 = v₁ + 1.333v₂
2.8 = v₁ + 1.333(2.8 + v₁)
2.8 = v₁ + 3.732 + 1.333v₁
2.8 - 3.732 = v₁ + 1.333v₁
-0.932 = 2.333v₁
v₁ = -0.932 / 2.333
v₁ = -0.4 m/s
Therefore, the final velocity of the first ball (0.3 kg) is 0.4 m/s in negative x direction.
A 50-kg cart is on an incline of 30 degrees above the horizontal. The cart is at rest. The static coefficient of the slope is 0.85 and the kinetic coefficient is 0.50. What is the force due to friction on the cart?
The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude w), the normal force (mag. n), and the friction force (maximum mag. f ).
In the horizontal direction, we have
n cos(120º) + f cos(30º) = 0
-1/2 n + √3/2 f = 0
n = √3 f
and in the vertical,
n sin(120º) + f sin(30º) + (-w) = 0
n sin(120º) + f sin(30º) = (50 kg) (9.80 m/s²)
√3/2 n + 1/2 f = 490 N
Substitute n = √3 f and solve for f :
√3/2 (√3 f ) + 1/2 f = 490 N
2 f = 490 N
f = 245 N
(pointed up the incline)
what is used to insulate or laminate the core of transformer?
Answer:
Explanation:
Laminations used in most power transformers (electrical and electronic) have a layer of insulating varnish to keep individual lamination sheets separate from each other.
hope this helps ;)
50
1. Force = ? N mass = 65 kg acceleration = 25 m/s2 (Record your answer
by number only and don t.include units...)
Answer: el pepe
Explanation:
Answer:
8125 N.
Explanation:
F = M A
M is mass
A is acceleration
F = 65 X 25
F = 8125 N.
Aquatic organisms are going extinct five times faster than land organisms due to water pollution.
True
False
Answer:
true 知恵を yfyfhdftufvdughf
Which statement best describe one of George Berkeley's arguments against materialism?
O A. There cannot be two different substances; my mind exists; therefore, matter does not exist.
OB. To be is to be perceived; we don't perceive matter; therefore, matter does not exist.
OC. I can only perceive my own mind; therefore, matter does not exist.
OD.
I can only perceive other minds; therefore, matter does not exist.
Reset
Next
Answer:
the answer is B
Explanation:
materialism is just wanting to always buy new things B does the opposite
A penny is dropped from the top of the
Chrysler Building (320 m high). How fast is
it moving when it hits the ground?
Answer:
As the penny falls 320 meters, its vertical velocity increases at the rate of 9.8 m/s
Explanation:
A girl pulls on a 10 kg wagon with a constant force of 20 N. What is the wagon’s acceleration
Answer:
2
Explanation:
İf system ideal ( no frictional force )
Fnet=m.a
20=10.a
a=2m/s2
The acceleration of the wagon pulled by the girl is 2m/s².
Given the data in the question;
Mass of wagon; [tex]m = 10kg[/tex]Force applied; [tex]F = 20N[/tex]Acceleration of wagon; [tex]a =\ ?[/tex]
To determine the acceleration of the wagon, we use the expression from the Newton's Second law of motion:
[tex]F = m * a[/tex]
Where m is mass and a is acceleration
We substitute our given values into the equation and solve for "a"
[tex]20N = 10kg * a\\\\20kg.m/s^2 = 10kg * a\\\\a = \frac{20kg.m/s^2}{10kg}\\\\a = 2m/s^2[/tex]
Therefore, the acceleration of the wagon pulled by the girl is 2m/s².
Learn more: https://brainly.com/question/13617200
don't you ever feel like thursday is like the number 7?
and friday like 8? or 6?
and monday maybe 3?
hi people see bye 18 45 36
Answer:
bye
Explanation:
a 90 gram object moves around a circle whose diameter is 2.56 m at a constant speed off 16 m/s. what is the object acceleration? how much time will take to complete the around?
Answer:
[tex]a=200\ m/s^2[/tex] and t = 0.5 s
Explanation:
Given that,
Mass of the object, m = 90 gram = 0.09 kg
Diameter of the circle, d = 2.56 m
Radius, r = 1.28 m
Speed, v = 16 m/s
We need to find the object acceleration. On the circular path, acceleration is given by :
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(16)^2}{1.28}\\\\a=200\ m/s^2[/tex]
So, the object's acceleration is [tex]200\ m/s^2[/tex].
Let t is time it will take to complete the around. Speed is given by :
[tex]v=\dfrac{2\pi r}{t}\\\\t=\dfrac{2\pi r}{v}\\\\t=\dfrac{2\pi \times 1.28}{16}\\\\t=0.5\ s[/tex]
So, time to cover one complete round is 0.5 s.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is [tex]\tau = 34.3 \ N\cdot m[/tex]
Explanation:
From the question we are told that
The mass of the steel ball is [tex]m = 3.0 \ kg[/tex]
The length of arm is [tex]l = 70 \ cm = 0.7 \ m[/tex]
The mass of the arm is [tex]m_a = 4.0 \ kg[/tex]
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as
[tex]r = \frac{l}{2}[/tex]
=> [tex]r = \frac{ 0.7}{2}[/tex]
=> [tex]r = 0.35 \ m[/tex]
Generally the magnitude of torque about the athlete shoulder is mathematically represented as
[tex]\tau = m_a * g * r + m * g * L[/tex]
=> [tex]\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70[/tex]
=> [tex]\tau = 34.3 \ N\cdot m[/tex]
*Earth Science Question*
Uses an array of mirrors to allow the collection of more light*
A. Reflecting
B. Refracting
C. Multiple Mirror
D. CCD
Some help me on this please
What can be said for certain about the affect of an external net force on
an object? *
A.It changes the object's mass.
B.It changes the object's kinetic energy.
C.It changes the object's temperature.
D.It changes the object's momentum.
Answer:
Changes in the object's momentum (answer D)
Explanation:
A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.
So, select the last option (D) in the given list.
What will be the effect on the acceleration due to
gravity of the earth if it is compressed to a the
size the moon?
Answer:
If the present earth is squeezed to the size of moon, the acceleration due to gravity will be increased by about 14 times since, acceleration due to gravity is inversely proportional to the square of radius if mass is constant.
A bicycle starts at 2.5m/s and accelerates along a straight path to a speed of 12.5m/s in a time of 4.5 seconds. What is the bicyclist’s acceleration to the nearest tenth of a m/s^2 ?
Answer:
The bicyclist's acceleration is 2.2m/s^2
Explanation:
Given
[tex]u = 2.5m/s[/tex] ---- Initial Velocity
[tex]v = 12.5m/s[/tex] ---- Final Velocity
[tex]t = 4.5s[/tex] ---- Time
Required
Determine the acceleration
This will be solved using the first equation of motion
[tex]v = u + at[/tex]
Substitute values for v, u and a
[tex]12.5 = 2.5 + a * 4.5[/tex]
[tex]12.5 = 2.5 + 4.5a[/tex]
Collect Like Terms
[tex]4.5a = 12.5 - 2.5[/tex]
[tex]4.5a = 10.0[/tex]
Solve for a
[tex]a = 10.0/4.5[/tex]
[tex]a = 2.2m/s^2[/tex] ---- (approximated)
Hence, the bicyclist's acceleration is 2.2m/s^2
What is the ratio of a 0.7 in circle in centimeters.
Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the UK where it is often very cloudy.
Answer:
In the clarification portion elsewhere here, the definition of the concern is mentioned.
Explanation:
So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.
Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:
Mostly at night would they have been seen. Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.