Answer:
Option (c)
Explanation:
1.The transformer core is assembled from steel sheets to:
(a) Reduced power loss due to eddy current
(b) Reduced power loss due to hysteresis
(c) Reduced power loss due to current flowing through the winding
(d) Reduce all types of power loss.
A transformer is a device which converts the low voltage into high and vice versa.
There are two types of a transformer.
Step up: It is used to convert low voltage into high.
Step down It is used to convert high voltage into high.
It depends on the number of turns in primary and the secondary coil.
The core of the transformer is laminated and it is in the form of sheets.
By using such type of core, the power loss due to the windings is reduced.
option (c) .
Find the starting pressure of CCl4 at this temperature that produces a total pressure of 1.1 atm at equilibrium. Express the pressure in atmospheres to three significant figures.
The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.
Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.
Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
Explanation:
The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.
[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]
Initial: a 0 0
Equilibrium: (a - x) 0 2x
Total pressure = (a - x) + 2x = a + x
As it is given that the total pressure is 1.1 atm.
So, a + x = 1.1
a = 1.1 - x
Now, expression for equilibrium constant for this equation is as follows.
[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]
Hence, the value of 'a' is calculated as follows.
a + x = 1.1 atm
a = 1.1 atm - x
= 1.1 atm - 0.31 atm
= 0.79 atm
Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.
What is the work done if a Boulder of mass 100 kilogram is rolled 40 meter up slope an angle of 20 degrees assuming the force of friction is negligible
Answer:
The work done is 13680.8 J.
Explanation:
The work done can be calculated as follows:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 40 m
The force acting on the boulder is given by:
[tex] F = mgsin(\theta) [/tex]
Where:
m: is the mass = 100 kg
g: is the acceleration due to gravity = 10 m/s²
θ: is the angle = 20°
Then, the work is:
[tex] W = mgsin(\theta)d = 100 kg*10 m/s^{2}*sin(20)*40 m = 13680.8 J [/tex]
Therefore, the work done is 13680.8 J.
I hope it helps you!
The drawings show (in cross section) two solid spheres and two spherical shells. Each object is made from copper and has a net charge, as the plus and minus signs indicate. Which drawing correctly shows where the charges reside when they are in equilibrium?
a) shows a lot of negative signs in the interior of circle
b) shows a lot of positive signs in the interior of circle
c) shows a hollowed out "hole" in the interior of the circle, with negative signs surrounding the opening.
d) shows a hollowed out "hole" in the interior of the circle, with positive signs surrounding the exterior edge
Answer:
d
Explanation:
The minimum energy configuration in electrostatics states that Charges always reside on the surface of a conductor. If anyhow they were inside, an electric field would exist inside and would act to move them to the surface,
Therefore, the drawing that shows where the charges reside when they are in equilibrium is a hollowed-out "hole" in the interior of the circle, with positive signs surrounding the exterior edge. This means that the d part is the correct answer.
A merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg-m2
and is rotating at 10 rev/min. A 25-kilogram child at rest jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round?
Answer:
dont be lose because the person who lose will win the match
A 0.20 mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to be 0.20 . If, instead, a 0.40 mass were used in this same experiment, choose the correct value for the maximum speed.
a. 0.40 m/s.
b. 0.20 m/s.
c. 0.28 m/s.
d. 0.14 m/s.
e. 0.10 m/s.
Answer:
d
Explanation:
Ya gon find the Kenitic Energy first
K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004
and now the replacement:
0.004=½×0.4V²====> v²=0.02===> V=0.14m/s
2. How do the phytochemicals present in various foods help us?
Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.
a concrete has a height of 5m and has unit area 3m² supports a mass of 30000kg.
Determine the stress, strain and change in height
Answer:
stress = 98000 N/m^2
strain = 3.92 x 10^-6
change in height = 0.0196 mm
Explanation:
Height, h = 5 m
Area, A = 3 m²
mass, m = 30000 kg
Stress is defined as the force per unit area.
[tex]stress = \frac{mg}{A}\\\\stress = \frac{30000\times 9.8}{3}\\\\stress = 98000 N/m^2[/tex]
Young's modulus of concrete is Y = 2.5 x 10^10 N/m^2
Young's modulus is defined as the ratio of stress to the strain.
[tex]Y = \frac{stress}{strain}\\\\2.5\times 10^{10}= \frac{98000}{strain}\\\\strain = 3.92\times 10^{-6}[/tex]
let the change in height is h'.
Strain is defined as the ratio of change in height to the original height.
[tex]3.92\times 10^{-6} = \frac{h'}{5}\\\\h' = 1.96\times 10^{-5}m = 0.0196 mm[/tex]
At what distance x from the center of the coil, on the axis of the coil, is the magnetic field half its value at the center?
Answer:
The value of x is 2.1 cm from the center of the coil.
Explanation:
Radius, R = 2.7 cm
Number of turns, N = 800
The magnetic field at the axis is half of the magnetic field at the center.
[tex]B_{axis}=\frac{B_{center}}{2}\\\\\frac{\mu o}{4\pi}\times \frac{2 \pi I N R^2}{\left (R^2 + x^2 \right )^{\frac{3}{2}}} = 0.5\frac{\mu o}{4\pi}\times\frac{2\pi N I}{R}\\\\\frac{R^2}{(R^2 + x^2)^\frac{3}{2}} = \frac{1}{2R}\\\\4R^6 = (R^2+x^2)^3\\\\1.6 R^2 = R^2 + x^2\\\\x^2 = 0.6 \times 2.7\times 2.7 \\\\x = 2.1 cm[/tex]
Find the refractive index of a medium
having a velocity of 1.5 x 10^8*
Explanation:
refractive index ,is the ratio of velocity of light in vacuum to the velocity of light a medium
suppose you have a block resting on a horizontal smooth surface. th block with a mass m is attached to a horizontal spring which is fixed at one end. the spring can be compressed and stretched. the mass is pulled to one side then released what is the formula required
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
What is meant by spring constant ?The spring constant of a spring is defined as the measurement of ratio of the force that is exerted on the spring to the displacement caused by it.
Here,
The mass of the block = m
Let F be the applied force on the spring and k be the spring constant.
When the mass attached to the spring is pulled to one side then released, it executes SHM.
Therefore we can write that, the applied force,
F = kx
Restoring force = -kx
According to Newton's law, we know that,
F = ma
So,
ma = -kx
Therefore, the acceleration,
a = (-k/m) x
For an SHM, the acceleration is given as,
a = -ω²x
Therefore, we can write that,
-ω²x = (-k/m) x
ω² = k/m
So, the time period of the spring,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex][√(m/k)]
Hence,
The time period of the spring is 2[tex]\pi[/tex][√(m/k)].
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You are holding one end of a horizontal stretched string. Flicking your wrist will send a pulse down the string. Which actions will make the pulse travel faster
Answer:
Use a lighter string of the same length, under the same tension.
Stretch the string tighter to increase the tension
Explanation:
The wave speed depends on propertices of the medium, not on how you generate the wave. For a string
Increasing the tension or decreasing the linear density (lighter string) will increase the wave speed.
A pulse will be sent down a horizontal extended thread if a person flicks the wrist. To raise the tension, pull the string tighter.
Define pulse.Pulse is the same thing as monitoring heartbeat. The pulse rate could also be measured via auscultation, which includes hearing to the heart rhythm using a stethoscope then counting it for one minute.
A pulse will be sent down a horizontal extended thread if a person holds one end of the rope and flicks their wrist.
To raise the tension, pull the string tighter.
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An object is suspended by a string from the ceiling of an elevator. If the tension in the string is equal to 25 N at an instant when the elevator is accelerating downward at a rate of 2.0 , what is the mass of the suspended object
By Newton's second law, the net force on the object is
∑ F = T - mg = - ma
where
• T = 25 N, the tension in the string
• m is the mass of the object
• g = 9.8 m/s², the acceleration due to gravity
• a = 2.0 m/s², the acceleration of the elevator-object system
Solve for m :
25 N - m (9.8 m/s²) = - m (2.0 m/s²)
==> m = (25 N) / (9.8 m/s² - 2.0 m/s²) ≈ 3.2 kg
A 300 kg block of dimensions 1.5 m × 1.0 m × 0.5 m lays on the table with its largest face.
Calculate:
Area of the largest face
Answer:
1.5
x 1.0
1.50
x 0.5
075.00
answer: 75.00m
Explanation:
I hope this help