Answer:
"13.48 Kwhr" is the right solution.
Explanation:
The given values are:
Average rate of heat energy,
= 0.063 W/m²
Diameter,
= 8m
Efficiency of conversion,
= 50%
Now,
The area of hotspot will be:
⇒ [tex]A=\frac{\pi}{4} d^2[/tex]
On substituting the values, we get
⇒ [tex]=\frac{3.14}{4} (8)^2[/tex]
⇒ [tex]=0.785\times 64[/tex]
⇒ [tex]=50.24 \ m^2[/tex]
Total heat generation rate will be:
⇒ [tex]Q=q\times A[/tex]
[tex]=0.063\times 50.24[/tex]
[tex]=3.16 \ W[/tex]
hence,
The electricity generation capacity will be:
⇒ [tex]P=\eta Q t[/tex]
On substituting the values, we get
⇒ [tex]=0.5\times 3.16\times 8760[/tex]
⇒ [tex]=13840.8 \ Whr[/tex]
On converting into Kwhr, we get
⇒ [tex]=13.84 \ Kwhr[/tex]
A piece of irregularly shaped metal weighs 300N in air. When the metal is completely submerged in water, it weighs 232.5N. Find the volume and specific gravity of the metal.
Answer:
Volume of metal piece = 0.0069 m³ (Approx.)
Explanation:
Given:
Weight of metal in air = 300 N
Weight of metal in water = 232.5 N
Find:
Volume of metal piece
Specific gravity of metal
Computation:
We know that;
Density of water = 1,000 kg/m³
Buoyant force applied on metal piece = Weight of metal in air - Weight of metal in water
Buoyant force applied on metal piece = 300 N - 232.5 N
Buoyant force applied on metal piece = 67.5 N
Buoyant force = Volume of metal x Density of water x Gravitational force
67.5 = Volume of metal x 1,000 x 9.8
Volume of metal piece = 0.0069 m³ (Approx.)
A 1.4-cm-tall object is 22 cm to the left of a lens with a focal length of 11 cm . A second lens with a focal length of -5 cm is 37 cm to the right of the first lens.
Calculate the distance between the final image and the second lens.
Calculate the image height.
Answer:
can you explain yourself more please
Explanation:
large roll of fabric
Costume Rendering
Scissors
Pattern
Bolt
Answer:
Do we have to choose out of those three?
Explanation:
what causes the moon to change its appearance over a month?
Answer:
The reason we see different phases of the Moon here on Earth is that we only see the parts of the Moon that are being lit up by the Sun. When the Moon is between Earth and the Sun, the lit side is hidden from us. As it moves around Earth, more and more of the lit side comes into view. Then it begins to disappear again, creating the different phases we see.
Explanation:
balance Al +N₂ --> AIN
Answer:
[tex]2Al + N_{2} ==> 2AlN[/tex]
Explanation:
3. At an early age, what happened to Cesar’s home?
Answer: He died later in 78 BC and was accorded a state funeral. Hearing of Sulla's death, Caesar felt safe enough to return to Rome. Lacking means since his inheritance was confiscated, he acquired a modest house in the Subura, a lower-class neighbourhood of Rome. hope this helps. Can u give me brainliest
Explanation:
Calculate the equivalent resistance for each of the following circuits.
Answer:
9. 4.8 Ω
10. 0.92 Ω
Explanation:
9. Determination of the equivalent resistance.
Resistor 1 (R₁) = 10 Ω
Resistor 2 (R₂) = 20 Ω
Resistor 3 (R₃) = 30 Ω
Resistor 4 (R₄) = 40 Ω
Equivalent Resistance (R) =?
The equivalent resistance can be obtained as follow:
1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/R = 1/10 + 1/20 + 1/30 + 1/40
Find the least common multiple (lcm) of 10, 20, 30 and 40. The result is 120. Divide 120 by each of the denominator and multiply the result obtained by the numerator as shown below:
1/R = (12 + 6 + 4 + 3) / 120
1/R = 25 / 120
Invert
R = 120 / 25
R = 4.8 Ω
Thus, the equivalent resistance is 4.8 Ω
10. Determination of the equivalent resistance.
Resistor 1 (R₁) = 2 Ω
Resistor 2 (R₂) = 3 Ω
Resistor 3 (R₃) = 4 Ω
Equivalent Resistance (R) =?
The equivalent resistance can be obtained as follow:
1/R = 1/R₁ + 1/R₂ + 1/R₃
1/R = 1/2 + 1/3 + 1/4
Find the least common multiple (lcm) of 2, 3 and 4. The result is 12. Divide 12 by each of the denominator and multiply the result obtained by the numerator as shown below:
1/R = (6 + 4 + 3) / 12
1/R = 13 / 12
Invert
R = 12 / 13
R = 0.92 Ω
Thus, the equivalent resistance is 0.92 Ω
Where does lymph come from?
What would be a good reason to increase friction between surfaces? to allow objects to slide past each other easily to reduce wear and tear to provide traction so that slipping does not occur to avoid contact that results in increased temperature and overheating
Answer:
B to provide traction so that slipping does not occur
Explanation:
To increase friction between two surfaces, allow the objects to slide past each other thereby force that is called friction.
What is friction?Friction is a type of force acting on an object to hinder its motion . It is negative by direction, since this force, hinder the motion of the object. Friction is also known as an opposing force because it always acts in the opposite direction of a moving or attempting to move body.
The virtue of friction causes a moving body to slow down. Friction is useful at times because it prevents car tires from skidding on the road and also allows us to walk on the pavement without slipping.
To increase friction , make an uneven, rugged, or sticky point of contact. When two or more bodies slide or rub against each other, three things can happen: small irregularities, nooks and crannies on the surfaces can catch on each other.
To find more on friction, refer here:
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You set a tuning fork into vibration at a frequency of 683 Hz and then drop it off the roof of the Physics building where the acceleration due to gravity is 9.80 m/s2. Determine how far the tuning fork has fallen when waves of frequency 657 Hz reach the release point
Answer:
The distance traveled by the tuning fork is 9.37 m
Explanation:
Given;
source frequency, [tex]f_s[/tex] = 683 Hz
observed frequency, [tex]f_o[/tex] = 657 Hz
The speed at which the tuning fork fell is calculated by applying Doppler effect formula;
[tex]f_o = f_s [\frac{v}{v + v_s} ][/tex]
where;
[tex]v[/tex] is speed of sound in air = 343 m/s
[tex]v_s[/tex] is the speed of the falling tuning fork
[tex]657 = 683[\frac{343}{343 + v_s} ]\\\\\frac{657}{683} = \frac{343}{343 + v_s}\\\\0.962 = \frac{343}{343 + v_s}\\\\0.962(343 + v_s) = 343\\\\343 + v_s = \frac{343}{0.962} \\\\343 + v_s = 356.55\\\\v_s = 356.55 - 343\\\\v_s = 13.55 \ m/s[/tex]
The distance traveled by the tuning fork is calculated by applying kinematic equation as follows;
[tex]v_s^2 = v_o^2 + 2gh[/tex]
where;
[tex]v_o[/tex] is the initial speed of the tuning fork = 0
g is acceleration due to gravity = 9.80 m/s²
[tex]v_s^2 = 0 + 2gh\\\\h = \frac{v_s^2}{2g} \\\\h = \frac{13.55^2 }{2\times 9.8} \\\\h = 9.37 \ m[/tex]
Therefore, the distance traveled by the tuning fork is 9.37 m
Helppp answer question pic in photo
Which of the following is NOT a geological event caused by the movement of tectonic plates?
A) Sand Dunes
B) Mountain forming
C) Earthquakes
D) Volcanoes
a 0.1 kg object oscillates as a simple harmonic motion along x axis with a frequency f=3.185 hz. At a position x1 , the object has a kinetic energy of o.7 j and a potential energy 0.3 J.The amplitude of oscillation A is:
Answer:
The total energy must be .7 J + .3 J = 1 J for a particle at the endpoint or midpoint of motion.
Also, omega = (k / m)^1/2
f = omega / (2 * pi)
omega^2 = 4 pi^2 * f^2 = k / m
k = 4 * pi^2 * f^2 * m = 40.05
Max KE or PE = 1/2 k A^2
A^2 = 2 * E / k = 2 * 1 / k = .0499
A = .223 meters
Research the main categories of mental disorders and their evaluation and assessments. For example, you can use your favorite Internet search engine to look up the following sites:
National Institute of Mental Health
Mental Help.net
American Psychological Association Help Center
Step 2 Report on methods of diagnosis and assessments of the major mental disorders.
Write a one and one-half-page report on mental disorders, their diagnosis, and treatment. Identify the following elements in your paper:
The major categories of mental disorders. ( Hint: Anxiety disorders are one category.)
The commonly used methods of diagnosis.
The treatment of choice (meaning the favorite current treatment) for this disorder.
Answer:
D
Explanation:
It’s D
According to Kepler's second law at the two shaded areas above are equal then...
A. it takes the same amount of time for planet A to travel between T1 and T2 as it does to travel between T3 and T4
B. Planet A will cause an eclipse on Planet B during those time intervals
C. it takes Planet A more time to travel between T1 and T2 as it does to travel between T3 and T4
D. it takes Planet A less time to travel between T1 and T2 as it does to travel between T3 and T4
Answer:
A. Kepler's second law says that the radius vector sweeps equal areas in equal times.
The volumes of two bodies are measured to be
V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³. Calculate sum and difference in
volumes with error limits.
Answer:
sum of volumes = (16.6 ± 0.03) cm³
and difference of volumes = (3.8 ± 0.03) cm³
Explanation:
Here,
V₁ = (10.2 ± 0.02) cm³ and V₂ = (6.4 ± 0.01) cm³.
Now,
∆V = ± (∆V₁ + ∆V₂)
= ± (0.02 + 0.01) cm³
= ± 0.03 cm³
V₁ + V₂ = (10.2 + 6.4) cm³ = 16.6 cm³ and
V₁ - V₂ = (10.2 - 6.4) cm³ = 3.8 cm³
Thus, sum of volumes = (16.6 ± 0.03) cm³
and difference of volumes = (3.8 ± 0.03) cm³
-TheUnknownScientist
Answer:
I hope it's helpful .............
Which feature of electromagnets makes them more useful than permanent
magnets in many modern technologies?
A. Electromagnets are not dependent on a power supply.
B. The strength of the magnetic field is more consistent in
electromagnets.
C. Electromagnets cannot be turned on and off.
D. The current can be adjusted to control the strength of the
magnetic field.
Answer:
The current can be adjusted to control the strength of the
magnetic field.
3. A ball thrown vertically upward returns to its starting point in 4s. Find its initial speed. [4]
Answer:
9.8 ×4 equal 39.2 m/s This is v intial
A lens with f= 20.0 cm creates a
virtual image at -37.5 cm (in front
of the lens). What is the object
distance?
Answer:
13.04
Explanation:
credit to the comment above
A lens with f= 20.0 cm creates a virtual image at -37.5 cm, then the object distance is approximately 0.0767 cm.
To determine the object distance in this scenario, we can use the lens formula:
[tex]\rm \frac{1}{f} =\frac{1}{v} -\frac{1}{u}[/tex]
Here, we have:
f = 20.0 cm (positive for a converging lens)
v = -37.5 cm (negative because it is a virtual image formed in front of the lens)
Substituting the values into the lens formula:
1÷20.0 = 1 ÷ (-37.5) - 1 ÷ u
Simplifying the equation:
1 ÷ 20.0 + 1÷ 37.5 = 1 ÷ u
Multiplying through by the common denominator:
(37.5 + 20.0) ÷ (20.0 * 37.5) = 1 ÷ u
57.5 ÷ 750 = 1 ÷ u
Dividing both sides by 57.5:
1 ÷ u = 750 ÷ 57.5
u = 57.5 ÷ 750
u ≈ 0.0767 cm
Thus, the object distance is approximately 0.0767 cm.
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Two identical objects, A and B, move along straight, parallel, horizontal tracks. The graph above represents the position as a function of time for the two objects.
(a) At a time of 2 seconds, where the lines intersect, do the displacements of the two objects from their initial positions have the same magnitude? Briefly explain your answer.
(b) At a time of 2 seconds, where the lines intersect, do the velocities of the two objects have the same magnitude? Briefly explain your answer.
(c) At a time of 2 seconds, where the lines intersect, which object, if either, has a net force with a greater magnitude exerted on it? If the net force has the same magnitude for both objects, indicate this explicitly
(d) In a clear, coherent paragraph-length response, explain your response to part (c). Be sure to reference and compare the graphed information for both objects A and B.
Answer:
After a little online search, I've found the graph of this question, the graph can be seen below.
a) The displacement is defined as the distance between the final position and the initial position.
In the graph, the vertical axis represents the distance. We also can see that both of the lines start in position 0, so at any given time, the displacement of the objects is given by the vertical position in the graph.
Thus, at t = 2 seconds, both lines have the same y-value, this means that the displacements have the same magnitude.
b) The velocity is related to the slope of the curve,
We can clearly see that the slope of graph A and the slope of gaph B are different at t = 2 seconds (graph A is steeper) then we can conclude that the velocities do not have the same magnitude.
c) By Newton's second law, we know that F = m*a
Force equals mass times acceleration.
Acceleration is the rate of change of velocity. When the position graph is a straight line, in any point of the line the slope will be the same, thus the object has always the same velocity, thus the object is not accelerated.
If we do not have a straight line (like in graph A) then the velocity is changing, then we have acceleration, then we have a force.
Then object A has a greater net force (because object B has a net force equal to 0)
d) It is already explained in point c.
A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the bike parallel to the road with a force of 25N.
A. How much work does the rider do on the bike?
B. How much work is done by the force of gravity on the bike?
Answer:
A. W = 6875.0 J.
B. W = -14264.6 J.
Explanation:
A. The work done by the rider can be calculated by using the following equation:
[tex] W_{r} = |F_{r}|*|d|*cos(\theta_{1}) [/tex]
Where:
[tex]F_{r}[/tex]: is the force done by the rider = 25 N
d: is the distance = 275 m
θ: is the angle between the applied force and the distance
Since the applied force is in the same direction of the motion, the angle is zero.
[tex] W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J [/tex]
Hence, the rider does a work of 6875.0 J on the bike.
B. The work done by the force of gravity on the bike is the following:
[tex] W_{g} = |F_{g}|*|d|*cos(\theta_{2}) [/tex]
The force of gravity is given by the weight of the bike.
[tex] F_{g} = -mgsin(24) [/tex]
And the angle between the force of gravity and the direction of motion is 180°.
[tex] W_{g} = |mgsin(24)|*|d|*cos(\theta_{2}) [/tex]
[tex] W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J [/tex]
The minus sign is because the force of gravity is in the opposite direction to the motion direction.
Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.
I hope it helps you!
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process. Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m3 to 0.026 m3 while doing work on a piston.
This question is incomplete, the complete question is;
The entropy of an ?-ideal gas changes in the following way as a function of temperature and volume:
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
For an adiabatic process, the change in T is determined by the change in V. In this problem you will compute the contributions to S from the V and T terms separately, then add them up to find the total entropy change for an adiabatic process.
Argon gas, initially at pressure 100 kPa and temperature 300 K, is allowed to expand adiabatically from 0.01 m³ to 0.026 m³ while doing work on a piston.
1) What is the change in entropy due to the volume change alone, ignoring any effects of changing internal energy? ΔS = ? J/K
2) For this adiabatic expansion, what is the final temperature? T[tex]_f[/tex] = ? K
Answer:
1) the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) the final temperature is 158.66 K
Explanation:
Given the data in the question;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
P[tex]_i[/tex] = 100 kPa = 100000 Pa
V[tex]_i[/tex] = 0.01 m³
V[tex]_f[/tex] = 0.026 m³
T[tex]_i[/tex] = 300 K
1) the change in entropy due to the volume change alone
from the question; ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex]) + ∝nRln(T[tex]_f[/tex]/T[tex]_i[/tex])
so change in entropy due to the volume change alone is;
ΔS = nRln(V[tex]_f[/tex]/V[tex]_i[/tex])
we know that, from ideal gas law; PV = nRT
so, nR = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] ---- let this be equation 1
∴ ΔS = P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] × ln(V[tex]_f[/tex]/V[tex]_i[/tex])
we substitute
ΔS = [( 100000 Pa × 0.01 m³) / 300 K ] × ln(0.026m³ / 0.01m³ )
ΔS = 3.185 J/K
Therefore, the change in entropy due to the volume change alone, ignoring any effects of changing internal energy is 3.185 J/K.
2) Final temperature
we know that, in an adiabatic expansion;
[tex]PV^Y[/tex] = K
where Y = 5/3
so
[tex]P_i[/tex][tex]V_i^{(5/3)[/tex] = [tex]P_f[/tex][tex]V_f^{(5/3)[/tex]
[tex]P_f[/tex] = [tex]P_i[/tex][tex]( \frac{V_i}{V_f})^{(5/3)[/tex]
we substitute
[tex]P_f[/tex] = ( 100000 Pa) [tex]( \frac{0.01 m^3}{0.026 m^3})^{(5/3)[/tex]
[tex]P_f[/tex] = 20341.255 Pa
Also from ideal gas law;
PV = nRT
T = PV / nR
so
T[tex]_f[/tex] = P[tex]_f[/tex]V[tex]_f[/tex] / nR
but from equation 1; nR = PV/T
so
T[tex]_f[/tex] = (P[tex]_f[/tex]V[tex]_f[/tex]) / (P[tex]_i[/tex]V[tex]_i[/tex]/T[tex]_i[/tex] )
T[tex]_f[/tex] = ( P[tex]_f[/tex]V[tex]_f[/tex]T[tex]_i[/tex] / P[tex]_i[/tex]V[tex]_i[/tex] )
we substitute
T[tex]_f[/tex] = ( 20341.255 Pa × 0.026 m³ × 300 K) / 100000 Pa × 0.01 m³ )
T[tex]_f[/tex] = 158.66 K
Therefore, the final temperature is 158.66 K
Electron spin: Radio astronomers can detect clouds of hydrogen too cool to radiate optical wavelengths of light by means of the 21 cm spectral line corresponding with the flipping of the electron in a hydrogen atom from having its spin parallel to the proton spin to having it antiparallel. From this wavelength, and thus E between states, find the magnetic field experienced by the electron in a hydrogen atom
Answer:
the magnetic field experienced by the electron is 0.0511 T
Explanation:
Given the data in the question;
Wavelength λ = 21 cm = 0.21 m
we know that Bohr magneton μ[tex]_B[/tex] is 9.27 × 10⁻²⁴ J/T
Plank's constant h is 6.626 × 10⁻³⁴ J.s
speed of light c = 3 × 10⁸ m/s
protein spin causes magnetic field in hydrogen atom.
so
Initial potential energy = -μ[tex]_B[/tex]B × cos0°
= -μ[tex]_B[/tex]B × 1
= -μ[tex]_B[/tex]B
Final potential energy = -μ[tex]_B[/tex]B × cos180°
= -μ[tex]_B[/tex]B × -1
= μ[tex]_B[/tex]B
so change in energy will be;
ΔE = μ[tex]_B[/tex]B - ( -μ[tex]_B[/tex]B )
ΔE = 2μ[tex]_B[/tex]B
now, difference in energy levels will be;
ΔE = hc/λ
2μ[tex]_B[/tex]B = hc/λ
2μ[tex]_B[/tex]Bλ = hc
B = hc / 2μ[tex]_B[/tex]λ
so we substitute
B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]
B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]
B = 510556326.09
B = 0.0511 T
Therefore, the magnetic field experienced by the electron is 0.0511 T
1. If a spring has a spring constant of 2 N/m and it is stretched 5 cm, what is the force of
the spring?
4
What happens because the Earth's orbit
is NOT a perfect circle?
Help Resources
A. We are always the exact same distance away from the
Sun throughout the entire year.
B. At certain timesh during the year, we are closer to the Sun
than at other times.
C. The shape of Earth's orbit does not affect the distance
from the Sun.
PLEASE HELP ME PLEASE! BRAINLEST
Answer:
10kg
Explanation:
Weight is "how much does gravity drag this down".
Mass is "how much matter is there here".
The relation is:
[tex]F_g = mg[/tex]
where [tex]F_g[/tex] is the weight, [tex]m[/tex] is the mass and [tex]g[/tex] is the gravitational acceleration (roughly equal to 10N/kg on Earth).
From the task we know that:
[tex]F_g = 100N\\g = 10\frac{N}{kg}[/tex]
So let's input it into the relation:
[tex]100N = m\cdot 10\frac{N}{kg}\\10N = m \cdot 1\frac{N}{kg}\\10N \cdot \frac{kg}{N} = m\\~\\m = 10kg[/tex]
Please please help me please
d. What is the net force on the bowling ball rolling lane
Answer:
Friction.
Explanation:
Help answer question in licture
Answer:
D
Explanation:
D is high pitch c is low pitch
Answer:
D creo que esa es la nota mas alta, la segumda
HELP ME PLEASE <333, NUMBER 3 ONLY!
Answer:
b
Explanation:
is correct .....................
i think
