The average induced electromotive force (emf) in the circular loop of wire is approximately 0.580 V. The principles of electromagnetic induction has significant applications in areas such as electric generators, transformers, and various electrical devices.
The average induced emf can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf in a loop of wire is proportional to the rate of change of magnetic flux through the loop.
The magnetic flux (Φ) through a circular loop of wire is given by the formula:
Φ = B * A
Where B is the magnetic field strength and A is the area of the loop.
Given:
Diameter of the circular loop (d) = 16.2 cm
= 0.162 m (radius = 0.081 m)
Magnetic field strength (B) = 1.44 T
Time taken to remove the loop from the field (Δt) = 0.140 s
The area of the circular loop can be calculated as:
A = π * r^2
The rate of change of magnetic flux can be obtained by dividing the change in magnetic flux by the time interval:
ΔΦ/Δt = (B * A) / Δt
Substituting the calculated values:
ΔΦ/Δt = (1.44 T) * (π * (0.081 m)^2) / (0.140 s)
Finally, the average induced emf can be calculated by multiplying the rate of change of magnetic flux by -1 (due to Lenz's law):
Average induced emf = - (ΔΦ/Δt)
The average induced electromotive force (emf) in the circular loop of wire, which is removed from a 1.44 T magnetic field in 0.140 s, is approximately 0.580 V. This calculation is based on Faraday's law of electromagnetic induction, which relates the induced emf to the rate of change of magnetic flux through the loop. By determining the change in magnetic flux and the time interval, the average induced emf can be evaluated. This concept is essential in understanding the principles of electromagnetic induction and has significant applications in areas such as electric generators, transformers, and various electrical devices.
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What is the instantaneous velocity of the sloth at time t = 8s?
The calculated value is AO=3.2sec.
The limit of average velocity as elapsed time approaches zero, or the derivative of x with respect to t: v (t) = d d t x (t), is the instantaneous velocity of an object. v (t) equals d d t x (t). Instantaneous velocity is a vector having a dimension of length per time, similar to average velocity.
In contrast to instantaneous velocity, which is determined by the slope of the tangent line, average velocity is defined as the change in position (or displacement) during the course of transit.
So, we obtain
cosθ= 5/4
= 4AO
So, we obtain AO=3.2sec.
The complete question is- Position-time graph is shown, which is a semicircle from t=2 to t=8 sec. Find time t at which the instantaneous velocity, is equal to average velocity over first t seconds.
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What is the net force on an object if two people are pushing on it to the right. Each applying a force of 50 N. There is a frictional force of 20 N to the left.1 80 N left2 120 N left3 100 N right 4 80 N right5 30 N right
Given,
The force applied by each person who is pushing the object, F=50 N
The frictional force, f=20 N
The frictional force is the force that opposes the motion of the object. As the people are pushing the object to the right, the frictional force will be directed to the left. And as the force with which people are pushing is greater than the frictional force, the net force will also be directed towards the right.
The net force is given by,
\(\begin{gathered} F_n=F+F-f \\ =2F-f \end{gathered}\)On substituting the known values,
\(\begin{gathered} F_n=2\times50-20 \\ =80\text{ N} \end{gathered}\)Therefore the net force on the object is 80 N to the right.
Thus the correct answer is option 4.
When a substance changes from one phase to another, which of the following occurs?
ОА.
The substance loses or gains heat.
OB.
The average kinetic energy of the substance changes.
OC.
The temperature of the substance changes.
OD
The molecular motion of the substance changes.
Answer:
D
Explanation:
Hope this helped
Is a kind of abstract expressionism that focuses on the emotional resonance of color?
A kind of abstract expressionism that focuses on the emotional resonance of color as exemplified by the work of Barnett Newman and Mark Rothko is known as chromatic abstraction.
Abstract Expressionism is a term applied to a movement in American painting that flourished in New York City after World War II, Also, Abstract Expressionism is a type of abstract art that originated in New York.
Abstract Expressionism is best known for large-scale paintings, Abstract Expressionism is often considered for its advancements in painting, its ideas had deep resonance in many mediums, including drawing and sculpture. This is all about it.
There were two types of Abstract Expressionism:
Action Painting and Color field Painting.
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Could you please help me with this question
what is meaning of saniya?
Answer:
“brilliant, radiant, splendid”.
Explanation:
Two angles are complementary. The first angle measures 39º and the second angle
is 3y. What is the value of y?
O 51°
O 470
O 11°
O 17°
Answer:
11
Explanation:
i've been in k12 and i got an a on this anser
Most often, a blackbody curve will have its maximum in what region of the electromagnetic spectrum?.
A blackbody curve will have its maximum in the visible region of the electromagnetic spectrum.
The blackbody graph illustrates how the intensity of light that is radiated changes as a blackbody is heated to a specific temperature.
According to the blackbody graph, a blackbody radiates energy over the whole electromagnetic spectrum once it reaches a temperature of 3,000 K. However, at a wavelength of about 1000 nm, the energy is released most strongly. This is a part of the electromagnetic spectrum's infrared range.
A few other essential points about the blackbody graph:
The total amount of energy radiated increases with temperature. No matter the temperature, the graph's red lines all follow the same pattern. The intensity of electromagnetic radiation peaks at a specific wavelength even if it is emitted across the entire spectrum.The wavelength with the highest radiation intensity varies with temperature. The radiation is most intense in the yellow-green region of the spectrum at 4,000 K. Due to the blackbody's tremendous radiation in the visible region of the electromagnetic spectrum, at 6,000 K, it would emit white hot. Keep in mind that white light is the simultaneous emission of all visible colors.Read more about Blackbody Graph:
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A stone is dropped from rest at a height of 2.0 m above the surface of a planet.
The planet has no atmosphere.
The speed of the stone just before reaching the surface of the planet is 3.8 m/s.
What is the acceleration of free fall on the planet?
A zero
B 1.9 m/s²
C 3.6 m/s²
D 7.2m/s²
The acceleration of free fall on the planet is 3.6 m/s^2. The correct answer is option C.
Let's assume that acceleration of free fall on the planet is a.
Potential energy of stone at height 2.0m is given be,
mah = 2ma
Kinetic energy of stone just before reaching the surface of the planet is,
Kinetic energy = mv^2/2
v= 3.8 m/s
Kinetic energy = 7.22m
By law of conservation of energy, Kinetic energy will be equal to Potential energy, and we get;
2ma = 7.22m
a = 3.6 m/s^2
Hence, the acceleration of free fall on the planet is is 3.6 m/s^2. Option C is the correct answer.
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It is better to use crow bar to displace a heavy load than with hands . Give reason .
Explanation:
Cause it multiples force which makes it easier to move.
The voltage in a circuit is 24 v and the current is 0.05 a. what's the power in this circuit? a circuit has a resistance of 100 ω and a current of 2 a. what's the power of this circuit?
Answer: 1.2 watt , 400 watt
Explanation:
1. The formula for calculating Power when voltage and current are given:
\(P = V^{2} /R\)
Now, to find R (resistance) we use the values that are given and get;
\(R = \frac{V}{I}\) = 24/0.05 = 480 ohms
So, the Power of this circuit is :
P = (24)^2 / 480 = 1.2 Watt
2. The formula for calculating Power when current and resistance are given:
\(P = I^{2} *R\)
Using the values given we get;
P = (2)^2*100 = 400 Watt
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7/21/22, 7:37 AMProblem Set ThreeNotes: Use 9.8 m/s 2 for the acceleration due to gravity. Formust be expressed in m/sLaw calculations, mass must be expressed in kg and velocity.An automobile weighs 2000lb. Calculate the weight of the car in N and it’s mass in kg.
Given:
The mass of the automobile is
\(m=2000\text{ lb}\)To find:
The weight of the car in N and its mass in kg
Explanation:
We know,
\(1\text{ lb=0.454 kg}\)So, the mass of the automobile is
\(\begin{gathered} m=2000\times0.454 \\ =908\text{ kg} \end{gathered}\)Derived quantities depend on...........physical quantity
Derived quantities depend on.( fundamental)..........physical quantity
Are you from Nepal?
Answer:
Derived quantities depend on two or more physical quantity.
A body moves in a straight line. At time, it's acceleration is given by a = 12t + 1. When t =0, the velocity of the body v is 2 metres per second and its displacement from the origin s is 0.
1. Express v and s in terms of t.
2. Determine velocity and displacement of the body after 3 seconds.
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
convert -13°F into (a) °C (b) kelvin
Answer:
-25ºC
Explanation:
If you walked at a speed of 3 m/s and 15 seconds of time has passed how far have you traveled
Answer: 45 meters
Explanation:
3 m/s ==> 3 meters/second
15 seconds * 3 meters/second=
15*3/1=
15*3=45 meters
Explain how humans produce speech sounds
The human vocal apparatus is an incredibly complex and efficient machine. The production of speech sounds begins with the diaphragm, a large, flat muscle located at the bottom of the lungs. When the diaphragm contracts, it increases the volume of the thoracic cavity and forces air into the lungs. The air then passes through the larynx, or voice box, which contains the vocal cords. The vocal cords are two thin bands of tissue that vibrate when air passes through them. The pitch of the resulting sound is determined by the tension of the vocal cords.
The sound of the vibrating vocal cords is then amplified by the resonating cavities of the head and mouth. The shape of the mouth cavity plays a significant role in the production of speech sounds. For example, the placement of the tongue, teeth, and lips all contribute to the creation of different vowel sounds. The hard and soft palate, along with the tongue, also help to create different consonant sounds.
If a car started to move from rest position ,after 6 second its acceleration became 5/m2.What will be final velocity of car
Answer:
the final velocity of the car after 6 seconds of acceleration is 30 m/s.
Explanation:
The final velocity of the car can be found using the kinematic equation:
v = u + at
where u is the initial velocity (which is zero since the car starts from rest), a is the acceleration (which is 5 m/s^2), t is the time elapsed (which is 6 seconds).
Substituting the given values, we get:
v = 0 + 5 x 6
v = 30 m/sA 2. 0 kg box is at rest on a table. The static friction coefficient His between the box and table is 0. 40, and the kinetic friction coefficient He is 0. 20. Then, a
25 N horizontal force is applied to the box.
What is the best estimate of the magnitude of the box's acceleration?
Round the answer (if non-zero) to two significant digits.
O 8. 6 m/s
O 17 m/s
O 11 m/s
0 m/s
The best estimate of the magnitude of the box's acceleration rounded to two significant digits is 11 m/s².
Mass of the box = m = 2 Kg
Coefficient of static friction between the box and table = μ = 0.40
Coefficient of kinetic friction between the box and table = μ' = 0.20
Horizontal force = f = 25 N
The kinetic frictional force = f' =
= f' = μ' X f
= f' = μ' X m X g
= f' = 0.20 X 2 X 9.81
= f' = 3.924 N
By applying the Newton's second law of motion, the total or net force is given as =
= ∑Fx = Applied horizontal force - kinetic frictional force = m X a
= Applied horizontal force - kinetic frictional force = m X a
= ( f - f' ) = ma
Acceleration of the box = a =
= a = (25 - 3,924) / 2
= a = 10.538 m/s²
= a ≈ 11 m/s²
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what is the electric field strength at a distance of .50 m from a 1.00x10^-6 c charge?
Answer:
e = k q / r² then,
e = 9×10^9 * 1×10^-6 / 50² = 3.6 N/C
hope this helps ❤.
he first right-hand rule relates which two quantities?
A.
current (thumb) to magnetic field (fingers)
B.
magnetic field (thumb) to current (fingers)
C.
current (thumb) to force (fingers)
D.
force (thumb) to current (fingers)
Reset Next
Answer: A.
current (thumb) to magnetic field (fingers)
Answer:
A. current (thumb) to magnetic field (fingers)
Explanation:
According to the Right-Hand Thumb rule, if we are holding a current- carrying straight conductor in our right hand such that the thumb points toward the direction of current, then the fingers will wrap around the conductor in the direction of the field lines of the magnetic field.
Hope it helps.
If you have any query, feel free to ask.
The lowest frequency of a guitar string with a length 0. 65 m is 248 hz. what is the speed of the wave on this string?
The speed of the wavelength on this string 322.4 m/s.
Given,
Half wavelength = 0.65m,
So, Wavelength = 2*.65=1.3m, frequency=248Hz
Speed=wavelength*frequency=1.3*248=322.4 m/s.
WavelengthThe wavelength, or spatial period, of a periodic wave in physics refers to the length over which the wave's structure repeats. It is the separation between neighbouring wave points that correspond to the same phase, such as two adjacent crests, troughs, or zero crossings. It is a property of both travelling waves and standing waves as well as other spatial wave patterns. The spatial frequency is the wavelength's reciprocal. The Greek letter lambda (), which represents wavelength, is frequently used. When describing modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids, the term wavelength is also occasionally used.
The lowest frequency of a guitar string with a length 0. 65 m is 248 hz. what is the speed of the wave on this string?
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A group of household electrical appliances has been connected as in the circuit shown below, which of the following is correct for the location of the fuse and the value of the current intensity that the fuse can bear to operate all devices safely?
The correct position that the fuse should be to ensure safety from the image is b.
What does the fuse do?A fuse is a type of electrical safety device used to guard against excessive current flow in electrical circuits. When there is a surge or overload of power, its main purpose is to interrupt the electrical current and break the circuit.
The fuse wire or element inside the fuse warms up due to resistance when the current flowing through a circuit exceeds the rated limit of the fuse. This opens the circuit and destroys the electrical connection by causing the fuse wire to melt or burst. The fuse does this to stop an excessive amount of current from passing through the circuit, which may cause overheating, harm to electrical equipment, or even provide a fire risk.
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Suppose you discover a Type Ia supernova in a distant galaxy. At maximum brilliance, the supernova reaches an apparent magnitude of 10. How far away in the galaxy
Type Ia supernova is approximately 72.38 million parsecs away in the distant galaxy.
To determine the distance to a Type Ia supernova in a distant galaxy with an apparent magnitude of 10, we will use the distance modulus formula. The distance modulus formula relates the apparent magnitude (m), absolute magnitude (M), and distance (d) in parsecs. The formula is:
m - M = 5 * log10(d) - 5
For a Type Ia supernova, the absolute magnitude is roughly -19.3. Now, we have the values for m and M, and we can solve for d:
10 - (-19.3) = 5 * log10(d) - 5
Next, isolate log10(d):
29.3 = 5 * log10(d) - 5
34.3 = 5 * log10(d)
Now, divide by 5:
6.86 = log10(d)
To find d, raise 10 to the power of 6.86:
d = 10^6.86
Finally, calculate the distance:
d ≈ 72.38 million parsecs
So, the Type Ia supernova is approximately 72.38 million parsecs away in the distant galaxy.
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Four landing sites are proposed for a lander. Data about the sites are listed in the table. A 3 column table with 4 rows. The first column is labeled landing site with entries W, X, Y, Z. The second column is labeled height above surface in meters with entries 32, 16, 35, 12. The last column is labeled acceleration due to gravity in meters per second squared with entries 1. 6, 3. 7, 1. 6, 3. 7. At which landing site would the lander have the greatest amount of gravitational potential energy? W X Y Z.
Answer:
its B just finished
Explanation:
Answer:
The answer above is correct! The correct option is B.
Explanation:
Just took the review on edge :)
Hope this helped - brainliest would be appreciated!
Have a great day everyone :D
which kind of spectrum is produced by a white household incandescent lightbulb?
Can someone explain to me why Ohms law stops working when the temperature of the component increases?
Answer:
Ohm's law verifies that current directly varies with voltage, when resistance increases, current current decreases.
When temperature increases, the electrons gain more average kinetic energy increases and current increases as well.
In the same situation, resistance also increases because atoms of a component vibrate at higher amplitudes about their mean positions hence reducing the free path of electrons. This increases resistance, hence ohms law violated or disobeyed.
This generates the temperature coefficient of resistance
in rutherford's famous set of experiments the fact that some alpha particles were deflected at large angles indicated that
In Rutherford's famous set of experiments, the fact that some alpha particles were deflected at large angles indicated that the atom contains a dense, positively charged nucleus.
Rutherford conducted experiments where he bombarded a thin gold foil with alpha particles. According to the prevailing model at the time, the Thomson model, it was believed that the positive charge in an atom was spread uniformly throughout the atom, much like plum pudding.
However, Rutherford's observations revealed that some alpha particles experienced significant deflections and even bounced back at large angles. This unexpected result could not be explained by the Thomson model.
Rutherford proposed a new atomic model known as the nuclear model, suggesting that the atom consists of a tiny, dense, positively charged nucleus at the center and the majority of the atom is empty space. This explained the deflection of alpha particles, as they were repelled or deflected by the positive charge concentrated in the nucleus.
The deflection of alpha particles at large angles indicated the presence of a compact and positively charged nucleus within the atom, leading to a fundamental revision of the understanding of atomic structure.
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What role does gravity play in the orbit of planets?
Answer: Gravity is what holds the planets in orbit around the sun and what keeps the moon in orbit around Earth
Explanation:
what techniques can the sonographer utilize to demonstrate acoustic shadowing with small gallstones
It's important for the sonographer to employ a combination of these techniques while considering patient factors, gallstone characteristics, and equipment capabilities to effectively demonstrate acoustic shadowing with small gallstones during the ultrasound examination.
The sonographer can utilize several techniques to demonstrate acoustic shadowing with small gallstones during an ultrasound examination. Acoustic shadowing occurs when the sound waves encounter a highly reflective or attenuating structure, such as a gallstone, causing a shadow to appear behind it. Here are some techniques commonly used:
Adjusting imaging angle: Changing the angle of the ultrasound beam relative to the gallstone can help accentuate the shadowing effect. By angling the transducer appropriately, the sonographer can optimize the visualization of the gallstone and the resulting shadow.
Utilizing higher-frequency transducers: Higher-frequency transducers provide better resolution and are more sensitive to small structures like gallstones. Using a high-frequency transducer can enhance the ability to visualize and demonstrate acoustic shadowing from small gallstones.
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