Answer:
a. The amount of heat energy produced on completely burning of 1 kg of fuel is called the calorific value
b. The combustion in which no external heat is given is known as spontaneous combustion.
c. Hydrogen being a highly inflammable gas and oxygen being a supporter of combustion.
d. Fuel must be heated to its ignition temperature before it starts burning.
e. Fire produced by oil cannot be controlled by water
Hope this helps you. Do mark me as brainliest.
In which type of collision is no kinetic energy converted to heat or sound
energy?
Answer:
elastic collision
An elastic collision occurs when the two objects "bounce" apart when they collide. Two rubber balls are a good example. In an elastic collision, both momentum and kinetic energy are conserved. Almost no energy is lost to sound, heat, or deformation.
I hope it's helpful!
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate 1.1 m/s2 of until it reaches 95 km/h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at-2.0 m/s2 Assume it stops at each intermediate station for 22 s.
Answer:
a) t = 746 s
b) t = 666 s
Explanation:
a)
Total time will be the sum of the partial times between stations plus the time stopped at the stations.Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:[tex]v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)[/tex]
Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:[tex]t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)[/tex]
Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:[tex]x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)[/tex]
In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:[tex]t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)[/tex]
We can find the distance traveled while the train was decelerating as follows:[tex]x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)[/tex]
Finally, we need to know the time traveled at constant speed.So, we need to find first the distance traveled at the constant speed of 26.4m/s.This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:[tex]t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)[/tex]
Total time between two stations is simply the sum of the three times we have just found:t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:tm = t*5 = 131.6 * 5 = 658.2 s (9)Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)b)
Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.We can find the distance traveled at constant speed, rewriting (6) as follows:x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:[tex]t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)[/tex]
Total time between two stations is simply the sum of the three times we have just found:t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:tm = t*3 = 207.4 * 3 = 622.2 s (14)Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)can you please tell me what this is
Answer:
200000 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of roller coaster = 1000 Kg
Velocity (v) of roller coaster = 20 m/s
Kinetic energy (KE) =?
Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:
KE = ½mv²
Where
KE => is the kinetic energy.
m =>is the mass of the object
V => it the velocity of the object.
With the above formula, we can obtain the kinetic energy of the roller coaster as follow:
Mass (m) of roller coaster = 1000 Kg
Velocity (v) of roller coaster = 20 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 1000 × 20²
KE = 500 × 400
KE = 200000 J
Therefore, the kinetic energy of the roller coaster is 200000 J.
She left the cubes in the water for three hours which of the following describes a heat flow that took place during those three hours?
Answer:
veffvevfevve
Explanation:
Explanation:
wht are the choices
attach the choices and illbe happy to help you
in fact i most definitely will
The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.
Answer:
[tex]26852726.19\ \text{N}[/tex]
[tex]57.52^{\circ}[/tex]
Explanation:
r = Radius of circle = 7 m
w = Width of dam = 60 m
h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]
A = Area = [tex]rw[/tex]
V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]
Horizontal force is given by
[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]
Vertical force is given by
[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]
Resultant force is
[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]
The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].
The direction is given by
[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]
The line of action is [tex]57.52^{\circ}[/tex].
Learning Task 5: Read each situation and encircle the letter of the best
answer. Do this activity on your notebook,
1. Our team is the champion because of our captain. No matter how fast his
Opponent shoot the ball he can do the same, he never misses it and hits it
right back. What emotion is shown in the situation?
A. Happiness
B. Anger
C. Sadness
D. Afraid
Answer:
I think the tone is happiness using the clues of "champion". But depending on who is saying it and background information it could be something different but based on the text here and the answer choices personally the best answer in my opinion is happiness.
If there was jealousy or optimism it would be a harder decision.
Explanation:
Hope this helped :)
A certain car can accelerate from 0 to 100 km/hr in 6.0 seconds what is the average acceleration of the car in m/s^2
Answer:
GIVEN :-Initial velocity = 0 × 5/18 = 0 m/s
Final Velocity = 100 × 5/18 = 27.77 m/s
Time = 6 sec
TO FIND :-Acceleration
SOLUTION :We know that
a = v - u/t
a = 27.77 - 0/6
a = 27.77/6
a = 4.6 m/s²
[tex] \\ [/tex]
The average acceleration of the car is equal to 4.63 m/s².
What is the average acceleration?The average acceleration is the change in velocity of an object divided by the elapsed time. Acceleration for an object can be described as the rate of change of velocity of the body.
The average acceleration for an interval is the change in velocity for that interval per time. So, the mathematical expression for the average acceleration can be written as:
Average acceleration (a) = △v/△t
If 'u' is the initial velocity and 'v' is the final velocity, and t is the time for the change of the velocity from u to v then a = v - u.
Given, the initial velocity of the car, u = 0
The final velocity of the car, v = 100 Km/hr = 27.77 m/s
The time is taken to change the velocity from 0 to 100 km/hr, t = 6 s
The average acceleration = (27.77 - 0)/6 = 4.63 m/s²
Therefore, the average acceleration of the car is equal to 4.63 m/s².
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In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass m
Answer:
[tex]\mathbf{v_a = 4.06 \times 10^7 \ m/s}[/tex]
Explanation:
[tex]\text{The missing part of the question is attached below.} \\ \\ \text{ mass m = 4u and charge q = 2e. Suppose a uranium nucleus with 92 protons decays into }[/tex][tex]\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }[/tex][tex]\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}[/tex][tex]\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }[/tex]
[tex]\text{So, from the above infromation;}[/tex]
[tex]radius (r) = \dfrac{15 \ fm }{2}[/tex]
[tex]radius (r) = 7.5 \times 10^{-15} \ m[/tex]
[tex]\text{Using the formula for potential energy}[/tex]
[tex]U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}[/tex]
[tex]U = 5.529 \times 10^{-12} \ J[/tex]
[tex]\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}[/tex][tex]\dfrac{1}{2}mv_a^2= 5.529 \times 10^{-12} J[/tex]
[tex]v_a = \sqrt{\dfrac{2(5.529 \times 10^{-12} \ J)}{4(1.67 \times 106{-37} \ kg)}}[/tex]
[tex]\mathbf{v_a = 4.06 \times 10^7 \ m/s}[/tex]
NEED HELP When a guitar string is plucked, what part of the standing wave is found at the fixed ends of the string?(1 point)
the nodes
the resonance
the antinodes
the harmonics
nWhat are four key features of our solar system that any theory of solar system formation must be able to explain
Answer: There are vast numbers of asteroids and comets.
Explanation:
There are clear patterns to the rotation and orbits of large bodies in the solar system.
There are a few notable "exceptions to the rules."
Planets fall into two major categories.
Calculate the acceleration due to gravity at Earth due to the Moon. (b) Calculate the acceleration due to gravity at Earth due to the Sun.
Answer:
a) g’= 3.44 10⁻⁵ m / s²
b) g ‘’ = 5.934 10⁻³ m / s²
Explanation:
For this exercise let's use the law of universal gravitation
F = [tex]G \frac{m M}{r^2}[/tex]
where m is the mass of the body under study, M the mass of the body that creates the force and r the distance between the bodies
F = [tex]m \ ( G \frac{M}{r^2} )[/tex]
the attractive force is called weight W = m g,
Thus
g = [tex]G \frac{M}{r^2}[/tex]
is called the acceleration of gravity
a) the acceleration created by the moon
g' = G \frac{M}{r^2}
the mass of the moon is M = 7.36 10²² kg
the distance from the moon to the Earth's surface is
r = D -R_e
r = 3.84 10⁸ -6.37 10⁶
r = 3.7763 10⁸ m
we calculate
g’= [tex]6.67 \ 10^{-11} \ \frac{7.36 \ 10^{22}}{ (3.7763 \ 10^8)^2}[/tex]
g ’= 3.44 10⁻⁵ m / s²
b) the acceleration created by the sun
mass of the sun M = 1,9991 10³⁰ ka
the distance from the sun wears down the Earth's surface
r = D -R_e
r = 1.496 10¹¹ -6.37 10⁶
r = 1.4959 10¹¹ m
let's calculate
g ’’ = [tex]6.67 \ 10^{-11} \frac{1.991 \ 10^{30}}{ (1.4959 \ 10^{11})^2 }[/tex]
g ‘’ = 5.934 10⁻³ m / s²
a crane lifts a 75 kg mass at a height of 8m. calculate the gravitational potential energy gained by the mass (g= 10N)
Answer:
6000 J
Explanation:
Potential energy = mass*gravity*height (U = m*g*h)
In this case g = 10 so
U = 75*10*8 = 6000 J (joules) [kg * m^2/s^2]
Simpson drives his car with an average velocity of 85 km/h eastward. How long will it take him to drive 560 km on a perfectly straight highway?
If the current in each wire is the same, which wire produces the strongest magnetic field?
-a wire that is 1 mm thick and not coiled
-a wire that is 2 mm thick and not coiled
-a 1-mm-thick coiled wire with ten loops
-a 2-mm-thick coiled wire with two loops
Answer:
C
Explanation:
I did this before
Answer:
it is c
Explanation:
cuz i know
What is the speed of a wave with a wavelength of 2 m and a frequency of 3 Hz?
Answer:
6m/s
Explanation:
speed=wavelength ×frequency
speed=2×3=6m/s
The speed of a wave with a wavelength of 2 m and a frequency of 3 Hz is 6m/s.
What is Speed?Speed is the distance travelled by an object in a given time period. It is a scalar quantity because it has only magnitude and no direction. The SI unit of speed is meter per second.
Wave speed is related to both the wavelength and wave frequency and it is represented by the equation:
Speed = Wavelength × Frequency.
Speed = 2m × 3Hz
Speed = 6m/s
The speed of the wave is 6 meter per second.
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A 25kg object accelerates at 5 m/s². Find the force.
Answer:
Force = 125 kg-m/s²
Explanation:
Given:
Mass = 25 kg
Acceleration = 5 m/s²
Find:
Force
Computation:
Force = Mass x Acceleration
Force = 25 x 5
Force = 125 kg-m/s²
1. Find the magnitude of the gravitational force a 66.5 kg person would experience
while standing on the surface of each of the following planets:
Gravit Force
Planet
Earth
Mars
Pluto
Mass
5.97 x 10^24 kg
6.42 x 10^23 kg
1.25 x 10^22 kg
Radius
6.38 x 10^6 m
3.40 x 10^6m
1.20 x 10^6m
Answer:
Gravitational forces: Earth: [tex]W = 650.969\,N[/tex], Mars: [tex]W = 246.449\,N[/tex], Pluto: [tex]W = 38.504\,N[/tex]
Explanation:
The weight ([tex]W[/tex]), measured in newtons, experimented by the person on the surface of the planet is defined by Newton's Laws of Motion:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
From Newton's Law of Gravitation we derive this expression for gravitational acceleration:
[tex]g = \frac{G\cdot M}{R^{2}}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, measured in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the planet, measured in kilograms.
[tex]R[/tex] - Radius of the planet, measured in meters.
By (2) we calculate the gravitational acceleration for each planet:
Earth ([tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.97\times 10^{24}\,kg[/tex], [tex]R = 6.38\times 10^{6}\,m[/tex])
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.97\times 10^{24}\,kg)}{(6.38\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 9.789\,\frac{m}{s^{2}}[/tex]
Mars ([tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 6.42\times 10^{23}\,kg[/tex], [tex]R = 3.40\times 10^{6}\,m[/tex])
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (6.42\times 10^{23}\,kg)}{(3.40\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 3.706\,\frac{m}{s^{2}}[/tex]
Pluto ([tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 1.25\times 10^{22}\,kg[/tex], [tex]R = 1.20\times 10^{6}\,m[/tex])
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1.25\times 10^{22}\,kg)}{(1.20\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 0.579\,\frac{m}{s^{2}}[/tex]
Lastly, we calculate the gravitational force for each case by (1):
Earth ([tex]m = 66.5\,kg[/tex], [tex]g = 9.789\,\frac{m}{s^{2}}[/tex])
[tex]W =(66.5\,kg)\cdot \left(9.789\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 650.969\,N[/tex]
Mars ([tex]m = 66.5\,kg[/tex], [tex]g = 3.706\,\frac{m}{s^{2}}[/tex])
[tex]W = (66.5\,kg)\cdot \left(3.706\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 246.449\,N[/tex]
Pluto ([tex]m = 66.5\,kg[/tex], [tex]g = 0.579\,\frac{m}{s^{2}}[/tex])
[tex]W = (66.5\,kg)\cdot \left(0.579\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 38.504\,N[/tex]
Which property allows objects that do not produce their own light
Answer:
Reflection
Explanation:
As light is scattered on items which do not generate illumination, they reflect it. This is attributed to the fact that light reflects off of themselves. The moon, for example, absorbs sunlight such that it can be seen at night.
Do we know which has more potential energy? object A or B? Best answer with reasoning gets brainliest.
Answer:
Object C has the most potential energy.
Between A and B, we do not know which has more potential energy.
Explanation:
We know the object with the most potential energy and this is the object at C.
Potential energy is the energy due to the position of a body above the ground surface.
The higher a body is above ground, the more its potential energy.
Potential energy = mass x acceleration due to gravity x height
So;
Object C has the most potential energy.
Between A and B, we do not know which has more potential energy.
This is because, the height and mass of the objects are not quantified using numbers.
Potential energy is a function of mass and height and acceleration due to gravity but acceleration due gravity is a constant.
llustration 2: Aman can run a distance of 100 m in 20 seconds. Find the speed of Aman in m/s.
Answer:
[tex]\boxed {\boxed {\sf 5 \ meters/second}}[/tex]
Explanation:
Speed is equal to distance over time.
[tex]s=\frac{d}{t}[/tex]
The distance is 100 meters and the time is 20 seconds.
[tex]d= 100 \ m \\t= 20 \ s[/tex]
Substitute the values into the formula.
[tex]s=\frac{100 \ m }{20 \ s}[/tex]
Divide.
[tex]s= 5 \ m/s[/tex]
Aman's speed is 5 meters per second.
In economics what goods and services will be produced?
Answer:A market economy is a system where the laws of supply and those of demand direct the production of goods and services. 1 Supply includes natural resources, capital, and labor. Demand includes purchases by consumers, businesses, and the government. Businesses sell their wares at the highest price consumers will pay.
Explanation:
which form of energy is an example of kinetic energy
Answer:
1. realizing of arrow
2. kicking of ball
3. punching the punching bag
why is the compass needle oriented as it is in the diagram
Answer:
its north pole is attracted to the south pole of the magnet
Explanation:
just did the quiz :)
A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner and outer surfaces of the shell are:
Answer:
1) The charge on the outer shell is +4·Q
2) The charge on the inner shell is +Q
Explanation:
1) The given parameters of the spherical shell are;
The net charge on the spherical shell = 3·Q
The point charge surrounded by the spherical shell = -Q
Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;
The net charge, 3·Q = -q + x
∴ x = 3·Q + Q = 4·Q
The charge on the outer shell, x = 4·Q
2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;
-Q + y = 0
∴ y = +Q
The charge on the inner shell, y = +Q
uniform acceleration means the velocity of the body is either increasing or decreasing changing at a constant rate TRUE OR FALSE? thank youu
Answer:
False.
Explanation:
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.
Hence, if we subtract the final speed from the initial speed and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{initial speed - final speed}{time}[/tex]
[tex]a = \frac{v - u}{t}[/tex]
Where,
a is acceleration measured in [tex]ms^{-2}[/tex]
v and u is initial and final speed respectively, measured in [tex]ms^{-1}[/tex]
t is time measured in seconds.
Additionally, acceleration is a vector quantity because it has both magnitude and direction.
Hence, uniform acceleration means the velocity of the body is increasing at a constant rate. For example, an object or body that is experiencing a free fall in a uniform gravitational field is said to be in uniform acceleration.
Which of the following terms best describes a group of galaxies?
Cluster
Constellation
Solar system
Universe
Answer:
Galaxy groups or clusters.
SO CLUSTERS
(Now i want me some culvers)
fasttt plsss!!!!
Energy A car engine drives a generator, which produces and stores electric charge in the car's battery. The headlamps use the electric charge stored in the car battery. List the forms of energy in these three operations
What is the ratio P1/P2P1/P2 of their power dissipations if the same current passes through each resistor
This question is incomplete, the complete question is;
Resistors 1 and 2 are connected to identical batteries, resistor 1 dissipates 4.0 times more power than resistor 2
What is the ratio P1/P2of their power dissipations if the same current passes through each resistor
Answer:
the ratio P₁/P₂ of their power dissipations is 0.25
Explanation:
Given the data in the question;
[tex]P_{i}[/tex] = [tex]I^{2} . R_i[/tex] --------------let this be equation 1
[tex]P_{i}[/tex] = [tex]\frac{V^{2} }{R_i}[/tex] -----------------let this be equation 2
P₁ = V²/R₁
P₁ = 4P₂
so
P₁ = V²/R₁ = 4( V²/R² )
now, form the above equation;
R₁ = R₂/4
using equation 1, we get;
P₁ = [tex]I^{2} . R_1[/tex]
P₂ = [tex]I^{2} . R_2[/tex]
Hence;
P₁ = [tex]I^{2}[/tex]( R₂/4)
P₁ = [tex]I^{2}[/tex] × ( R₂/4)
P₁ = (1/4) P₂ or 0.25
Therefore; the ratio P₁/P₂ of their power dissipations is 0.25
Calculate the force between two objects that have a masses of
1000 kg and 4000 kg that are separated by a distance of 25 m.
1.1 x 10-5 N
4.3 x 1015 N
О
4.3 x 10-7N
Explane Information-processing approach