The light collecting area of the 100-m telescope will be 100 times the light collecting area of the 10-meter keck telescope.
Given the diameter of first telescope (d1) = 100m
The diameter of Keck telescope (d2) is = 10m
We know that the light collecting area of a telescope is represented as:
A = πD^2/4 where D is the diameter f telescope.
A1/A2 = πd1^2/πd2^2
A1/A2 = 100 x 100 / 10 x 10 such that A1/A2 = 100
A1 = 100A2
It represents that the light collecting area of the 100-m telescope will be 100 times the light collecting area of the 10-meter keck telescope.
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if the density (mass divided by volume) of the wall material is the same as that of pure water, what is the mass (in mgmg ) of the cell wall, assuming the cell to be spherical and the wall to be a very thin spherical shell?
The mass of cell wall is m = 6.082 x 10⁻¹⁶ kg = 6.082 x 10⁻¹⁰ mg , assuming the cell to be spherical and the wall to be a very thin spherical shell
What is semi permeable membrane ?Semi-permeable membrane - A membrane through which only smaller molecules like water can pass but not the bigger molecules like solutes is known as semi-permeable membranes (SPM)
--First, we find the the surface area of the cell wall. Since, the cell is spherical in shape. Therefore, surface area of cell wall will be:
A = 4πr²
where,
A = Surface Area = ?
r = Radius of Cell = Diameter/2
= 2.2 μm/2
= 1.1 μm
= 1.1 x 10⁻⁶ m
Therefore,
A = 4π(1.1 x 10⁻⁶ m)²
A = 15.2 x 10⁻¹² m²
Now, we find the volume of the cell wall. For that purpose, we use formula:
V = Volume of the Cell Wall = ?
t = Thickness of Wall = 40 nm
= 4 x 10⁻⁸ m
Therefore,
V = (15.2 x 10⁻¹² m²)(4 x 10⁻⁸ m)
V = 60.82 x 10⁻²⁰ m³
Now, to find mass of cell wall, we use formula:
ρ = m/V
m = ρV
where,
ρ = density of water = 1000 kg/m³
m = Mass of Wall = ?
Therefore,
m = (1000 kg/m³)(60.82 x 10⁻²⁰ m³)
m = 6.082 x 10⁻¹⁶ kg
= 6.082 x 10⁻¹⁰ mg
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when a planet, in its orbit, is closer to the sun, it: group of answer choices moves slower than average spins faster on its axis reflects less sunlight than average feels less gravitational pull than average moves faster than average
when a planet, in its orbit, is closer to the sun, it moves faster than average.
In general, the closer a planet is to the Sun, the shorter its orbital period (the time it takes to complete one orbit) will be. As a result, the closer a planet is to the Sun, the faster it will be moving in its orbit.
This is due to the fact that the force of gravity from the Sun on the planet is stronger the closer the planet is to the Sun, which causes the planet to move faster in its orbit.
Additionally, planets that are closer to the Sun will also have a higher surface temperature because they receive more sunlight, and they may have a stronger magnetic field due to the stronger solar wind near the Sun.
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Your question seems incomplete, but I assume the full question was:
"when a planet, in its orbit, is closer to the sun, it:
(group of answer choices)
moves slower than average
spins faster on its axis
reflects less sunlight than average
feels less gravitational pull than average
moves faster than average."
an object is thrown straight up with an initial velocity of 20.0 m/s, and there is an air resistance force which would cause an acceleration of 3.00 m/s2 opposite the direction of motion. with what speed does the object return to the ground?
As the object is thrown up with the initial velocity of 20.0 m/s, and there is an air resistance force that would cause an acceleration of 3.00 m/s2, the speed of the object as a return to the ground is 10.8 m/s.
Kinematic equations
The kinematic equations are a set of equations that describe the motion of an object with constant acceleration.
When we have an initial velocity value, it is written as Vo, while for the final velocity, we simply write V or Vt. As an object moves through the air, air resistance slows the object’s speed.
The formula of the kinematic equation used for solving this case is
Vt = V0 + at (the gravity is 10 m/s2)
[tex]Vt = Vo + (-g-a)t\\\0 = 20 + (-10-3)t\\0 = 20-13t\\\13t = 20\\\t = \frac{20}{13}[/tex]
After the time is known, now we can insert the value into the following formula :
[tex]Vt = Vo + (g-a)t\\Vt = 0 + 7.\frac{20}{13} \\Vt = 0 + \frac{140}{13} \\\Vt = 10.8 m/s[/tex]
Thus, the speed of the object returns to the ground after being thrown up with an initial velocity of 20.0 m/s and acceleration of 3.00 m/s2, which is 10.8 m/s.
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a basketball player grabbing a rebound jumps vertically upward. if the player leaves the ground with a speed of 3.53 m/s, how long, in seconds, will the player remain in air?
In order to calculate time, we need to know one more variable, the gravitational acceleration of earth which is 9.8 m/s^2 and the force exerted on the player's legs, which can be calculated by F=ma.
In order to determine how long the player will remain in the air, we would need to know the player's upward acceleration (which can be affected by factors such as the player's mass and the force exerted on the player's legs as they jump) as well as the player's initial upward velocity. Without this information, it is not possible to calculate how long the player will remain in the air.
In physics, an object's vertical motion can be described by the equation:
y = vi*t + (1/2)at^2
where y is the vertical displacement (or height) of the object, vi is the initial upward velocity, t is the time in the air, and a is the upward acceleration.
If we know the initial upward velocity (vi) and the acceleration (a), we can use this equation to solve for the time in the air (t). The acceleration of the object is determined by various factors such as the force applied on the object and the mass of the object.
In this case, we know the initial upward velocity of the basketball player is 3.53 m/s, but we don't have the acceleration value, which means we cannot solve for time. In order to calculate time, we need to know one more variable, the gravitational acceleration of earth which is 9.8 m/s^2 and the force exerted on the player's legs, which can be calculated by F=ma.
Therefore, without additional information, it is not possible to calculate the time the player will remain in the air.
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Two restaurant employees push a 730 kg wheeled dumpster along a horizontal surface. After they push the dumpster a distance of 5.5 m starting from rest, its speed is 0.75 m/s. What is the magnitude of the net force on the dumpster?
The net force that acts on the object is 36.5 N.
What is the net force?We have to note that the first thing that we would have to obtain is the acceleration of the object and we can be able to obtain this by the use of the formula;
v ^2 = u^2 + 2as
v = final velocity
u = initial velocity
a = acceleration
s = distance
(0.75)^2 = (0)^2 + 2 * a * 5.5
a = (0.75)^2 / 2 * 5.5
a = 0.5625/11
a = 0.05 m/s^2
The net force is ma
F = 730 kg * 0.05 m/s^2
F = 36.5 N
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an experimental rocket designed to land upright falls freely from a height of 2.98 102 m, starting at rest. at a height of 96.5 m, the rocket's engines start and provide constant upward acceleration until the rocket lands. what acceleration is required if the speed on touchdown is to be zero? (neglect air resistance.)
The required acceleration has a magnitude of 19.7 m/s2 directed upwards, correct to three significant figures.
The the purpose of analysis, the motion of rocket is divided into two phases.
Phase 1: the free fall motion of the rocket from the height 2.98*102m to a height 96.5m.
Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 96.5 m to the point of touchdown y = 0m.
The initial velocity of the rocket is 0m/s when it started to experience a fall from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.
The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.
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Define electroplating
Answer:
La galvanoplastia o electrodeposición es una aplicación práctica de la electroquímica. Se trata de una técnica basada en los principios electroquímicos, en donde se aplica una o varias capas de un metal seleccionado sobre un objeto receptor, por lo general, también metálico.
Explanation:
Espero te sirva de algo ;)
How can you differentiate the driver and driven gears in a gear train?
The driver gear is the gear that provides the input motion to the gear train, while the driven gear is the gear that receives motion from the other gears in the train and provides the output motion.
What is driver gear train?This refers to the main or primary gear that majorly drives the other gears found in the gear train.The driver gear is typically connected to a power source, such as an electric motor or a combustion engine, while the driven gear is connected to the load that the gear train is intended to drive. The driver gear is also the one that has the greater torque and the one that rotates faster.
Driven gears refer to the gear that receives motion from the other gears in the train and provides the output motion.
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what happens when you close the series circuit with a switch?
An object is experiencing an acceleration of 24m/s while traveling in a circle of radius 6. 0m. What is its velocity?
The velocity of an object experiencing an acceleration of 24m/s while traveling in a circle of radius 6.0m is 12m/s.
The required details for velocity in given paragraph
a = v²/r
Given a = 24m/s²; radius = 6m
24 = v²/6
Cross multiplying will give;
v² = 24×6
v² = 144
v =√144
v = 12m/s
The velocity of the object is 12m/s
How does the acceleration and radius of the circle relate to the object's velocity?
The acceleration of an object traveling in a circle is known as centripetal acceleration and is caused by the force of the object's velocity acting towards the center of the circle. The acceleration is directly proportional to the radius of the circle and the square of the velocity of the object. The equation for centripetal acceleration is a = v^2 / r, where a is the acceleration, v is the velocity, and r is the radius. So the bigger the radius of the circle, the lesser the object's velocity needs to be in order to maintain the same acceleration and vice versa.
The object's velocity and the radius of the circle are inversely proportional to each other.
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Mars rotates fast enough to make it an
electromagnet and have a magnetic
field, but it no longer has a liquid outer
core. Mars no longer has a magnetic
field to protect it.
What evidence do scientists have that
Mars used to have a magnetic field?
Mars has rocks on it that are magnetized.
The atmosphere of Mars contains evidence
of a magnetosphere.
There is no evidence. Scientists are just
speculating (guessing) based on videos
from the Mars roverto Settings to activate Windows.
The evidence that scientists have that Mars used to have a magnetic field is this:
B. The atmosphere of Mars contains evidence of a magnetosphere.
What evidence did scientists use to reach their conclusion?Scientists base whatever conclusions they reach on a subject matter on evidence. To reach their conclusion that mars had a magnetic field that protected it, they used a device known as an orbiter to measure the magnetic strength of mars.
Their discovery showed that the magnetic field on Mars was ten times as strong as the earth's field. This gives evidence of the fact that at a certain time, there was a strong magnetic field on Mars.
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State fourth law of Thermodynamics
Answer: The dissipative component of evolution is in a direction of steepest entropy ascent
Explanation:
According to our definition, every non-equilibrium state of a system or local subsystem for which entropy is well defined must be equipped with a metric in state space with respect to which the irreversible component of its time evolution is in the direction of the steepest entropy ascent permissible under the conservation constraints. We derive (nonlinear) expansions of Onsager reciprocity and fluctuation-dissipation relations to the far-non-equilibrium world inside the rate-controlled constrained-equilibrium approximation to demonstrate the force of the fourth law (also known as the quasi-equilibrium approximation).
Point A
Support
3.0m
2.5m
Figure 5
Weight of girl = 600 N
Weight of diving board = 800 N
Water
The support post pushes up on the diving board with a force of 3040 N. The force provides the anti-clockwise moment which keeps the board in equilibrium.
Calculate the distance that the support must be from point A.
The distance that the support must be from point A . 1.2010 3 N.
What is the distance that the support must be from point A?At x = 0, where the x axis is parallel to the diving board, we assume the force acting on the left pedestal to be F1. We define the right pedestal's force as F2 and its location as x=d. W is the diver's weight, which is found at x=L. Setting the total of forces (with upward positive) and the sum of torques (around x2) to zero yields the following two equations:
F 1 +F 2 −W=0
F 1 d+W(L−d)=0
(a) The result of the second equation is F 1 = d L d W = (2.5m 3.0m)
(600N) = 800N, which should be rounded to F 1 = 1.210 3 N. Consequently, F 1 = 1.2010 3 N
(b) F 1 is negative, suggesting a downward force.
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you are throwing paper airplanes in class, and you throw them hard enough they start out at a horizontal speed of 10fts . however, they have a horizontal deceleration of 4fts . what is the greatest distance away from the initial position that an object can be if it can be struck by the airplane?
The greatest distance away from the initial position that an object can be if it struck by the airplane is 12.5 ft, if the initial speed of the airplane is 10 ft/s.
Initial speed of the paper airplane, u = 10 ft/s
Final speed of the paper airplane, v = 0
deceleration of the airplane, a = -4 ft/s²
By the first equation of motion, v = u -at
Time of flight, t = (v-u)/a
t = (0 - 10)/-4 = 2.5 sec
Maximum distance travelled by the airplane, = S
S = ut + 0.5at²
S = 10 × 2.5 - 0.5 × 4 × 2.5²
S = 12.5 ft
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1. the magnitude (amplitude) of a vibration velocity signal is 0.23 ips at 1170 cpm. find the equivalent displacement amplitude in mils pk-pk and acceleration amplitude in g
In this case, the acceleration amplitude would be (0.23 ips x 1170 cpm)2 = 546.7 g.
What is acceleration?Acceleration is the rate of change of velocity over time. It is a vector quantity, meaning that it has both magnitude and direction. The magnitude of acceleration is the rate at which the velocity changes, and the direction of acceleration is the same as the direction of the velocity.
Displacement amplitude:
The formula for displacement amplitude is displacement amplitude = velocity amplitude / (2πf).
In this case, the displacement amplitude would be 0.23 ips / (2π x 1170 cpm) = 0.00196 inches pk-pk. Converting to mils pk-pk, this would be 1.96 mils pk-pk.
Acceleration amplitude:
The formula for acceleration amplitude is acceleration amplitude = (velocity amplitude x f)2.
In this case, the acceleration amplitude would be (0.23 ips x 1170 cpm)2 = 546.7 g.
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a uniform electric field of 1350 n/c pointing due north exists in a region of space. a point charge of 2.85 nc is then placed in that original electric field. what is the magnitude of the net electric field now at a point 15.5 cm due east of the charge?
The magnitude of the net electric field now at a point 15.5 cm due east of the charge is 1721.15 N/C.
The physical field that surrounds electrically charged particles and pulls or attracts all other charged particles in the vicinity is known as an electric field. The physical field for a system of charged particles is another usage of the term.
Conductive materials are affected by electric fields, which change how electric charges are distributed at their surface. As a result, current travels through the body and to the ground. Circulating currents are induced within the human body by low-frequency magnetic fields.
A uniform electric field of 1350 n/c is already given,
[tex]E_1 = 1350[/tex]
Now, for the point charge, By Coulomb's law,
We get,
[tex]E_2=\frac{Kq}{r^{2} }[/tex]
here, [tex]K = 9*10^{9}[/tex]
According to the question,
[tex]q=2.85*10^{-9} C[/tex]
[tex]r = 0.155 m[/tex]
Putting these values in the formula,
We get,
⇒ [tex]E_2=\frac{(9*10^{9})*(2.85*10^{-9} ) }{(0.155)^{2}}[/tex]
⇒ [tex]E_2= 1067.6 N/C[/tex]
Now,
the net electric field,
⇒ [tex]E = \sqrt{(E_1^{2})+(E_2)^{2} }[/tex]
⇒ [tex]E=\sqrt{(1067.6)^{2} +(1350)^{2} }[/tex]
⇒ [tex]E= 1721.15 N/C[/tex]
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You are to drive 280km on an expreway. The interview i at 11:15 a.m. You plan to drive at 100km/h, o you leave at 8:00 a.m. to allow ome extra time. You drive at
The least speed is required to cover the rest of the trip to arrive in time for the interview is [tex]114.05km/hr[/tex]
Distance and displacement are two quantities that seem to mean the same but are distinctly different with different meanings and definitions. Distance is the measure of “how much ground an object has covered during its motion” while displacement refers to the measure of “how far out of place is an object.”
Here distance formula is,
Δ[tex]d=d_{1} +d_{2}[/tex]
Here we need to find the time left and the distance covered:
T[tex]1[/tex] [tex]=\frac{100}{100}=1hr=60 min[/tex]
T[tex]2[/tex] [tex]=\frac{43}{41} =1.04hr=62.92 min[/tex]
Total time was [tex]3.25hr[/tex][tex]=195min[/tex]
So, Time left is: [tex]1.21hr =72.08min[/tex]
Total distance left to cover is:
D[tex]=280-100-43=137[/tex][tex]km[/tex]
Now minimum speed required is:
[tex]s=\frac{distance}{time} =\frac{137}{1.21} =114.05km/hr[/tex]
Therefore, the least speed is required to cover the rest of the trip to arrive in time for the interview is [tex]114.05 km/hr[/tex].
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Complete question: you are to drive to an interview in another town, at a distance of 280km on a expressway. The interview is at 11:15 a.m. you plan to drive at 100km/h, but then construction work forces you to slow to 41.0km/h for 43.0km what would be the least speed needed for the rest of the trip to arrive in time for the interview?
you rub a balloon on your head, and the balloon gains a charge of 30 nc . how many electrons were transferred during this process? express your answer to two significant figures.
The electrons transferred during the process of rubbing a balloon on your head are -1.88×10¹¹.
The electrons, which are negatively charged particles, whirl about the nucleus' periphery. Because of how swiftly they rotate, it might be challenging for scientists to keep an eye on them. The tiny atoms in an atom are drawn to the positive ions of the protons; you can fit 2000 of them in a proton.
Charge on the balloon q = 30 nc
Charge on the electron = -1.6 × 10⁻¹⁹ C
The number of electrons is determined by adding the charge of one particle's entire value.
Suppose there are n electrons,
Number of electrons = (30× 10⁻⁹)/-1.6 × 10⁻¹⁹ = -(3× 10⁻⁸)/(1.6 × 10⁻¹⁹) = -1.88×10¹¹
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which method of naming/locating stars will be the best for giving someone the location of the star?
Celestial navigation is the best method of naming or locating stars, if we have to give someone the location of a star.
With the help of "sights" or time angular measurements taken typically between a celestial body (e.g. the Sun, a planet, the moon or a star) and the visible horizon, Celestial navigation locate a star or any celestial body. However it can also take advantage of measurements between celestial bodies without taking the Earth horizon as reference, such as when the Moon is used in the practice, then it is called "lunars" or lunar distance method, which is used for determining precise time when time is unknown.
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if a capacitor has opposite 5.2 microcoulomb charges on the plates and an electric field of 2.0 kv/mm is desired between the p lates, what must each plate's area be?
The area of each plate must be 2.6 x 10^-9 m^2 in order to achieve an electric field of 2.0 kV/mm between the plates.
What is electric field?Electric field is a physical quantity that is used to describe the force that an electric charge exerts on other charges in its vicinity. It is a vector field, meaning it has both magnitude and direction. Electric fields are created by electric charges, or by time-varying magnetic fields.
The electric field (E) between the plates of a capacitor is equal to the charge (Q) divided by the area (A) of the plates:
E = Q/A
We can rearrange this equation to solve for A:
A = Q/E
Since the charge on the plates is 5.2 microcoulombs (5.2 x 10^-6 C) and the electric field desired is 2.0 kV/mm (2.0 x 10^3 V/mm), we can plug in these values to find A:
A = 5.2 x 10^-6 C/2.0 x 10^3 V/mm = 2.6 x 10^-9 m^2
Therefore, the area of each plate must be 2.6 x 10^-9 m^2 in order to achieve an electric field of 2.0 kV/mm between the plates.
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look at attached photo pls. physics, momentum, acceleration. 30 points:)
1. The deceleration of the aeroplane in the 35s is -1.6m/s²
2. The force acting on the aeroplane is 4.0× 10⁵N
3. The momentum of the aeroplane when its speed is 6.0m/s is 15 ×10⁶kgm/s
What are the equation of motion?The equation of motion are used in solving problems related to motion. The equations of motion are
1. v = u+ at
2 S = ut + 1/2at²
3. v² = u²+2as
where v is the final velocity
u is the initial velocity
S is the distance
t is time
a is the acceleration
The deceleration of the plane after 35s can be calculated as;
v = u+at
v= 6m/s, u= 62m/s , t = 35s
6 = 62+35a
35a = 6-62
35a =- 56
a = - 56/35
a = -1.6m/s²
the negative sign shows that the plane decelerates.
The force acting on the plane is calculated as;
F = ma
F = 2.5×10⁵× 1.6
F = 4× 10⁵N
The momentum of the plane at 6m/s is ;
p = mv
p = 2.5×10⁵ × 6
p = 1.5× 10⁶ kgm/s
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as a heart chamber contracts, what happens to the pressure of the fluid within it?
The pressure of the fluid inside the heart chamber gets increased when the heart chamber contracts.
A typical heart has 4 chambers, two upper and two lower chambers. The upper two chambers, the right and left, receive incoming blood. The lower two chambers, the more muscular right and left, pump the blood out of the heart. The heart valves, which maintain blood flowing in the right direction, are gates at the chamber openings. We know when a closed chamber which is filled with a fluid is compressed, the fluid inside the chamber tries to come out of the chamber. Which increases the blood pressure in case of a heart chamber.
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Calculate the distance traveled by Max. He left his house and drove west 3 miles, then north 2 miles, and then east 3 miles.
The term "distance" refers to how far you move. The rate is a measurement of your trip speed. Time is measured by how far you travel.
What is Distance?Time must be measured in hours if the rate in the problem is given in miles per hour (mph). To find the number of hours before calculating the problem, divide the time, if it is given in minutes, by sixty.
Algebra word problems frequently take the form of distance word problems. They involve a situation in which you must determine the speed, distance, or duration of one or more objects' travels.
Because one of the most well-known distance problems involves determining the precise moment when two trains traveling in opposite directions cross paths, these are frequently referred to as "train problems."
Therefore, The term "distance" refers to how far you move. The rate is a measurement of your trip speed. Time is measured by how far you travel.
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what would be the magnitude of this gravitaional force if earth and the mon were seprated by adistance of 1.02 times 10^8 meters
Two objects attract each other with a gravitational force of magnitude 1.00 × 1 0 − 8 1.00 \times 10 ^ { - 8 } 1.00×10−8 N when separated by 20.0 cm.
What is meant by magnitude?
Magnitude is simply "distance or quantity," according to the definition given in physics. It shows how an object moves when it is in motion, whether that movement is absolute, relative, or of a certain size. It serves as a way to describe something's size or scope. Magnitude is a broad term used in physics to describe size or distance.Size can be described as magnitude. A automobile is travelling quicker than a bike, for instance, in terms of speed. The car is currently moving faster than the bike in this situation by a significant margin. It provides information about the motion of an item in terms of size, direction, and relative or absolute dimensions.To learn more about magnitude refer to
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How far could a force of 3N pull a wooden block before transferring 12J of energy?
The distance which a force of 3N will pull a wooden block before transferring 12J of energy is 4m.
What is Force?This is referred to as an influence that can change the motion of an object and the unit is Newton.
It has a relationship with distance and work which can be seen below:
Work = Force × distance
12J = 3N × d
d = 12/3 = 4m.
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The products have the same ___ as the reaction
Answer:
they have the same atoms
Help! I have no idea to go about this... Please answer both questions with clear working so that I understand!
(i) The power wasted in heating the wire if the current in part (a) flows in the wire is 53.8 MW.
(ii) The power wasted in heating the wire if the current in part (b) flows in the wire is 17,800 MW.
What is the value of the power wasted in the wire?The value of the power wasted in heating the wire is calculated by applying the following formula.
P = I²R
where;
I is the value of the current flowing in the wireR is the resistance of the wireFor the first case when the resistance, R = 9 Ω and current, I = I,
The power wasted is calculated as;
P = I² x 9
P = 9I²
For 22 kV, the current, I = (22,000 V) / 9 Ω
I = 2,444.44 A
P = 9 x (2,444.44)²
P = 5.38 x 10⁷ W
P = 53.8 x 10⁶ W
P = 53.8 MW
For the second case when the resistance, R = 9 Ω and current, I = I,
The power wasted is calculated as;
P = I² x 9
P = 9I²
For 400 kV, the current, I = (400,000 V) / 9 Ω
I = 44,444.44 A
P = 9 x (44,444.44)²
P = 1.78 x 10¹⁰ W
P = 17,800 x 10⁶ W
P = 17,800 MW
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The half life of a material is 100 years. If you have 1000g, how much will remain after 500 years?
define the two physical characteristics of sound, and identify how they determine our awareness of loudness and pitch.
what was the definition of a planet according to the ancient greeks? group of answer choices an object that orbits the sun. an object that orbits a star, is massive enough to be round, and dominates its orbital region. an object that wanders across the zodiac.
The definition of a planet according to the ancient Greeks is an object that orbits the sun.
Massive, circular celestial objects that are neither stars nor their remnants are known as planets. The planetary nebulae hypothesis, that states that an interstellar cloud bursts out of a nebulae to produce a young protostar encircled by a protoplanetary disc, is currently the most promising theory for planet formation. The steady accretion of matter accelerated by gravity, or accretion, is how planets expand within this disc.
Sun's core undergoes nuclear fusion events, transforming it into a nearly perfect ball of hot plasma that is incandescent. The Sun is the most significant source of energy for life on Earth, radiating this energy mostly as light, uv, and infrared rays.
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