A gun recoils back when the bullet is fired due to conservation of momentum. Conservation of momentum suggest that - the net change of momentum in a closed system must be zero.
Therefore in equation form -
[tex]mv - mu = 0[/tex]
Why does a gun recoil when a bullet is fired?
Explanation:When a bullet is fired, the gun exerts a force on the bullet through the trigger (this is action) in a particular direction. According to Newton's Third Law, we know to every action there is an equal and opposite reaction; action and reaction act on different bodies. So, the bullet will also exert an equal and opposite reaction on the gun in the opposite direction. That is why a gun recoils when a bullet is fired from it.
Hope it helps.ray4918 here to helpDefine relative motion??
Answer:
The laws of physics which apply when you are at rest on the earth also apply when you are in any reference frame which is moving at a constant velocity with respect to the earth. For example, you can toss and catch a ball in a moving bus if the motion is in a straight line at constant speed.
Explanation:
Hair sticking to a balloon after it
has been rubbed on a girl's head is
an example of
Answer:
Friction
Explanation:
Answer:
It causes electrostatic charging, which makes it stick to your hair,
Which two elements would most likely form an ionic compound?
a
From the Periodic Table of Elements - #6 carbon and #8 oxygen
b
From the Periodic Table of Elements - #3 lithium and #9 fluorine
c
From the Periodic Table of Elements - #1 hydrogen and #7 nitrogen
d
From the Periodic Table of Elements - #5 boron and #10 neon
Answer:
boron and neon.
Explanation:
two element compounds are usually ionic when one element is a metal and the other is a non-metal. boron is a metal, whereas neon is a non-metal. i might be wrong but hope it helps.
In the middle of a turn, you should _____________________. A. brake hard to maintain complete control over your vehicle B. accelerate abruptly and straighten the wheel C. avoid using your brake or clutch, unless it is necessary to do so in order to avoid a collision D. downshift to allow you to turn more sharply
Answer:
D. Downshift to allow you to turn more sharply
D. downshift to allow you to turn more sharply
What is the purpose of downshifting?The purpose of "downshifting", or shifting the transmission down from a higher gear to a lower gear, is so that a driver can accelerate their car as fast as possible when exiting a corner that they had to slow down for.
The reason for ensuring your downshift is completed before turning into the corner is because letting the clutch out and engaging a lower gear while turning can cause the car to spin.
Two special conditions where you should downshift
Before Starting Down a Hill.
Slow down and shift down to a speed that you can control without using the brakes hard. Otherwise the brakes can overheat and lose their braking power.
Before Entering a Curve.
Slow down to a safe speed, and downshift to the right gear before entering the curve. This lets you use some power through the curve to help the vehicle be more stable while turning. It also allows you speed up as soon as you are out of the curve.
Therefore,
In the middle of a turn, you should downshift to allow you to turn more sharply .
Learn more about Downshift here:
https://brainly.com/question/10600958
#SPJ2
What voltage is required to move 6A through 20?
Answer:
120V
Explanation:
Given parameters:
Current = 6A
Resistance = 20Ω
Unknown:
Voltage = ?
Solution:
According to ohms law;
V = IR
Where V is the voltage
I is the current
R is the resistance
Now, insert the parameters and solve;
V = 6 x 20 = 120V
11. A ball is thrown up and attains maximum height of 20 m. Calculate its initial speed. (g = 10 ms')
a) 19.7 m/s
b) 20 m/s
c) 25 m/s
d) 35m/s
final velocity = 0
acceleration = - 10 m/ s 2
distance. = 20 m
u = ?
v^2 - u ^2 = 2 a s
0^2 - u^ 2 = 2 * -10 * 20
-u^2 = -400
u = √ 400
u = 20m / s
Explain how to find the acceleration of an object that has one-dimensional horizontal motion.
A bus travels east for 3 km, then north for 4 km. What is its final displacement?
5 km southeast
5 km northwest
5 km northeast
5 km southwest
Un corp de masa m=5 kg se afla pe o suprafata orizontala pe care se poate deplasa cu frecare (µ = 0,02). Calculati forta de frecare.
Răspuns:
0,98 N
Explicaţie:
Dat fiind:
µ = Coeficientul de frecare = 0,02
Masa (m) = 5 kg
Folosind relația:
Forța de frecare (F) = µ * N
N = reacție normală = masă * accelerație datorată gravitației
Prin urmare,
F = u * m * g
g = 9,8 m / s²
F = 0,02 * 5 kg * 9,8 m / s²
F = 0,98 kgm / s²
F = 0,98N
Which ball has larger Inertia before touching the ground when it falls freely from a
certain height (air friction is zero)
a. mass 50 kg b. mass 25 kg
c. mass 10 kg d. All of these
Answer:
[tex]Mass = 50kg[/tex]
Explanation:
Required
Which has the larger inertia
Inertia (I) is calculated as follows:
[tex]I = mr^2[/tex]
Where m represents the mass of the object and r represents the distance from the axis.
Take for instance m = 10 and r = 4
The inertia would be:
[tex]I = 10 * 4^2[/tex]
[tex]I = 160[/tex]
Assume r is constant and m is increased to 25
The inertia becomes:
[tex]I = 25 * 4^2[/tex]
[tex]I = 25 * 16[/tex]
[tex]I = 400[/tex]
Notice that there is an increment in inertia, when the mass increased.
Hence, mass of 50kg will have the larger inertia.
please help answer brainlest
each rock you need to tell where OK can you find them
like: your backyard or the road
Flakes of brick
Cracks in rocks
Lichen growing on rocks
Roots or plant growth making cracks in concrete
Worn away statues
Exposed rocks
Rust on rocks
Example You Found:
Answer:
Flakes of brick: Athens, Ohio
Cracks in rocks: Sandston, Mudcracks
Lichen growing on rocks: Antarctica living below the surface
Roots or plant growth making cracks in concrete: In the driveway, sidewalk, or road.
An object of mass m is traveling in a circle with centripetal force f c. If the velocity of the object is v, what is the radius of the circle?
Answer:
r = mv^2/ Fc
Explanation:
Got it right
Please answer:
Galileo, in order to measure the speed of light, performed this simple
experiment: He and his assistant each took a lantern that had a shutter in front of it,
and stood on hilltops one mile apart in the night.
Galileo was to flash his lantern, and the assistant was to open the shutter of his own
lantern as soon as he saw the light from Galileo's lantern. Galileo had planned to
measure the time taken for light to travel from one hill to the other and back, and
calculate the speed of light using the formula, Speed = distance traveled / time
taken But he could not measure the speed of light using this method. What could
have been the reason for this?
A. The distance between the two hills was too much for them to see the light from
the other person's lantern.
B. Light does not travel in straight lines and so the distance traveled by it cannot
be measured.
C. The time taken for the light to travel would have been too little to have been
measured by them.
D. Light does not travel because it is present everywhere, so its 'speed' cannot be
calculated
I asked this question before also please answer
Answer: C. The time taken for the light to travel would have been too little to have been measured by them.
Explanation:
Light travels at a speed of 300,000 km per second which is much too fast for the naked eye to even comprehend. This is why light shinning in a distance is picked up instantaneously by out eyes.
This experiment done by Galileo would therefore be unable to measure the speed of light because the light from either lantern would have traveled too fast for Galileo to be able to measure the time taken.
El motor de una licuadora gira a 3600 rpm, disminuye su velocidad angular hasta 2000 rpm realizando 120 vueltas. Calcular: a) La aceleración angular b) El tiempo transcurrido
Answer:
a) α = -65,2 rad/s².
b) t = 2,57 s.
Explanation:
a) La aceleración angular se puede calcular usando la siguiente ecuación:
[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \theta [/tex]
En donde:
[tex]\omega_{f}[/tex]: es la velocidad angular final = 2000 rpm = 209,4 rad/s
[tex]\omega_{0}[/tex]: es la velocidad angular inicial = 3600 rpm = 377,0 rad/s
α: es la aceleración angular=?
θ: es el desplazamiento o número de vueltas = 120 rev = 754,0 rad
Las conversiones de unidades se hicieron sabiendo que 1 revolución = 2π radianes y que 1 minuto = 60 segundos.
Resolviendo la ecuación (1) para α, tenemos:
[tex]\alpha = \frac{\omega_{f}^{2} - \omega_{0}^{2}}{2\theta} = \frac{(209,4 rad/s)^{2} - (377,0 rad/s)^{2}}{2*754,0 rad} = -65,2 rad/s^{2}[/tex]
Entonces, la aceleración angular es -65,2 rad/s². El signo negativo se debe a que el motor está desacelerando.
b) El tiempo transcurrido se puede encontrar como sigue:
[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]
Resolviendo para t, tenemos:
[tex]t = \frac{\omega_{f} - \omega_{0}}{\alpha} = \frac{209,4 rad/s - 377,0 rad/s}{-65,3 rad/s^{2}} = 2,57 s[/tex]
Por lo tanto, el tiempo transcurrido fue 2,57 s.
Espero que te sea de utilidad!
PLZ HELP I DON'T UNDERSTAND!! a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the ball (2kg) is traveling 10m/s. Whats the ball's kinetic energy just as it leaves the boy's hand? What is the potential energy of the ball when it reaches the highest point?
Explanation:
The mass of a ball, m = 2 kg
It is traveling with a speed of 10 m/s
The ball's kinetic energy just as it leaves the boy's hand is calculated as follows :
[tex]K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 2\times (10)^2\\\\=100\ J[/tex]
The ball's kinetic energy just as it leaves the boy's hand is 100 J. The potential energy of the ball when it reaches the highest point is same as the kinetic energy as it leaves the boy's hand.
Hence, the required kinetic and potential energy is 100 J.
A 10 kg wagon is accelerated by a constant force of 60 N from an initial velocity of 5.0 m/s to a final velocity of 11 m/s. What is the impulse received by the wagon? a)15 N s b)60 N s c)17 N s d)70 N s
Anyone....................
Answer:
h₁ = 18 [cm]
Explanation:
This problem can be solved by applying the principles of the static pressure of a liquid column. The key to the solution is to know that the pressure on the horizontal line should be equal for both liquids, this must be met regardless of their density.
[tex]P_{2}=P_{1}\\[/tex]
where:
P₂ = Pressure exerted by the liquid 2 in the given point L₂
P₁ = Pressure exerted by the liquid 2 in the given point L₁
Now we can calculate the pressure in each point using the following expression:
[tex]P_{1}=Ro_{1}*g*h_{1}\\P_{2}=Ro_{2}*g*h_{2}[/tex]
Where:
Ro₁ = density of the liquid 1 = 0.8 [g/cm³]
Ro₂ = density of the liquid 2 = 1.8 [g/cm³]
g = gravity acceleration = 9.81 [m/s²]
h₁ = column of the liquid 1 [m]
h₂ = column of the liquid 2, = 8 [cm] = 0.08 [m]
But first we must convert the units from grams per cubic centimeter to kilograms per cubic meter.
[tex]0.8[\frac{g}{cm^{3}}]*[\frac{1kg}{1000g} ]*[\frac{100^{3}cm^{3} }{1m^{3} } ]=800[kg/m^{3} ]\\1.8[\frac{g}{cm^{3}}]*[\frac{1kg}{1000g} ]*[\frac{100^{3}cm^{3} }{1m^{3} } ]=1800[kg/m^{3} ][/tex]
Now replacing in the first equation:
[tex]800*9.81*h_{1}=1800*9.81*0.08\\h_{1}=\frac{1800}{800} *0.08\\h_{1}=0.18 [m] = 18 [cm][/tex]
Answer:
[tex]h_1 = 18 \: cm[/tex]
Explanation:
See image for explanation
I hope my working is correct.
Why does the Sun appear white at noon?
Un cuerpo se lanza hacia arriba con 50 m/s . ¿A qué altura se encuentra cuando su rapidez es 20 m/s por segunda vez? (g = 10 m/s ^ 2) .
Answer:
105 m
Explanation:
Given that the initial velocity of the body, u= 50 m/s in upward direction.
Acceleration due to gravity, g= 10 m / [tex]s ^ 2[/tex] in the downward direction.
Let h be the height of the body when the velocity of the body is 20 m/s.
As the gravitational force acting in the downward direction, so taking it negative as
a= - 10 m/[tex]s ^ 2[/tex] in the upward direction.
By using the equation of motion [tex]v^2=u^2+2as[/tex] ...(i)
Here, u= 50 m/s
v= 20 m/s
[tex]a=- 10 m/s^2[/tex]
s=h
Putting all the values in the equation (i), we have
[tex]20^2=50^2+2(-10)h \\\\400=2500-20h \\\\h=\frac{2500-400}{20} \\\\h=105 m[/tex]
Hence, the height of the body is 105 m.
Billy throws a peach vertically upward. It takes 3 seconds for the peach to land back in his hand. How fast did Billy throw the peach?
Answer:
48 ft/s
Explanation:
Acceleration due to gravity decreased the upward velocity to zero in 1.5 seconds (half the total time), so the initial velocity must have been ...
(32 ft/s^2)(1.5 s) = 48 ft/s
If you have a liquid with a mass of 6g and a density of 25g/ml, what volume of liquid do you have?
Hello!!
For calculate the volume of the liquid, lets applicate the formula:
[tex]\boxed{V=m/ p }[/tex]
[tex]\textbf{Being:}[/tex]
[tex]\sqrt{}[/tex] V = Volume = ?
[tex]\sqrt{}[/tex] m = Mass = 6 g
[tex]\sqrt{}[/tex] p = Density = 25 g/ml
[tex]\text{Let's \textbf{replace} and resolve: }[/tex]
[tex]V = 6 \ g / 25 \ g / m l[/tex]
[tex]V = 0,24 \ m l[/tex]
[tex]\textbf{Result:}[/tex]
The volume of the liquid is 0,24 ml.
An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a rate of 24 m per year at a time when the region is 300 m wide. How fast is the area changing at that point in time
Answer:
[tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to RightEquality Properties
Geometry
Area of a Rectangle: A = lwAlgebra I
Exponential Property: [tex]w^n \cdot w^m = w^{n + m}[/tex]Calculus
Derivatives
Differentiating with respect to time
Basic Power Rule:
f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹Explanation:
Step 1: Define
Area is A = lw
2w = l
w = 300 m
[tex]\frac{dw}{dt} = 24 \ m/year[/tex]
Step 2: Rewrite Equation
Substitute in l: A = (2w)wMultiply: A = 2w²Step 3: Differentiate
Differentiate the new area formula with respect to time.
Differentiate [Basic Power Rule]: [tex]\frac{dA}{dt} = 2 \cdot 2w^{2-1}\frac{dw}{dt}[/tex]Simplify: [tex]\frac{dA}{dt} = 4w\frac{dw}{dt}[/tex]Step 4: Find Rate
Use defined variables
Substitute: [tex]\frac{dA}{dt} = 4(300 \ m)(24 \ m/year)[/tex]Multiply: [tex]\frac{dA}{dt} = (1200 \ m)(24 \ m/year)[/tex]Multiply: [tex]\frac{dA}{dt} = 28800 \ m^2/year[/tex]Answer:
28,800 m²/yr
Explanation:
This rectangle has dimensions such that:
width = wlength = 2wWe are given [tex]\displaystyle \frac{dw}{dt} = \frac{24 \ m}{yr}[/tex] and want to find [tex]\displaystyle \frac{dA}{dt} \Biggr | _{w \ = \ 300 \ m} = \ ?[/tex] when w = 300 m.
The area of a rectangle is denoted by Area = length * width.
Let's multiply the width and length (with respect to w) together to have an area equation in terms of w:
[tex]A=2w^2[/tex]Differentiate this equation with respect to time t.
[tex]\displaystyle \frac{dA}{dt} =4w \cdot \frac{dw}{dt}[/tex]Let's plug known values into the equation:
[tex]\displaystyle \frac{dA}{dt} =4(300) \cdot (24)[/tex]Simplify this equation.
[tex]\displaystyle \frac{dA}{dt} =1200 \cdot 24[/tex] [tex]\displaystyle \frac{dA}{dt} =28800[/tex]The area is changing at a rate of 28,800 m²/yr at this point in time.
a tiger moving with constant accleration covers the distance between two points 70 meter apart in 7 seconds . its speed as it pass the second point 15 meter per second . 1) what is the speed at the first point. 2) what us its accleration
Answer:
a = 1.428 [m/s²]
v₀ = 5 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}[/tex]
where:
x = final point [m]
x₀ = initial point [m]
v₀ = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
But we need to use this additional equation.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
vf = final velocity = 15 [m/s]
Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]
[tex]x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ][/tex]
Now replacing this value in the second equation, we can find the initial velocity.
[tex]15=v_{o}+1.428*7\\v_{o}=5[m/s][/tex]
I got a new kitten and I dont know how to train them to use a leash. Can anyone help?
You used 100 N of force to push a shopping cart 5m down the aisle in the grocery store. How much work did you do? *
Answer:
[tex]\boxed {\boxed {\sf 500 \ Joules}}[/tex]
Explanation:
Work can be found by multiplying the force by the distance.
[tex]W=F*d[/tex]
The force is 100 Newtons and the distance is 5 meters. Therefore,
[tex]F= 100 \ N \\d= 5 \ m[/tex]
We can substitute the values into the formula.
[tex]W= 100 \ N * 5 \ m[/tex]
Multiply.
[tex]W= 500 \ N*m[/tex]
1 Newton meter is equal to 1 Joule. Thus, our answer of 500 Newton meters is equal to 500 Joules.[tex]W= 500 \ J[/tex]
You did 500 Joules of work to push a shopping cart 5 meters with a force of 100 Newtons.
Answer:
Given :-Force (F) = 100 NDisplacement (D) = 5 mTo Find :-Work done
Solution :-As we know that Work done is the product of Force and displacement
[tex] \tt \: w = force \: \times displacement[/tex]
[tex] \tt \: w = 100 \times 5[/tex]
[tex] \tt \: w = 500 \: j[/tex]
Therefore, Work done is 500 Joule
Give three examples of how a cell's shape makes possible the cell's function
Answer:
A) a nerve cells long extensions enable it to transmit electrical impulses from one body part to another
B) the sheet like organization of epithelial cells enables them to protect underlying cells
C) the alignment of contractile proteins within muscle cells enables them to contract, pulling closer together the structures to which they attach
Explanation:
What is a joule? Pls only answer if you know the answer.
Answer:
the SI unit of work or energy, equal to the work done by a force of one newton when its point of application moves one meter in the direction of action of the force, equivalent to one 3600th of a watt-hour.
Explanation:
Joule, unit of work or energy in the International System of Units (SI); it is equal to the work done by a force of one newton acting through one metre. Named in honour of the English physicist James Prescott Joule, it equals 107 ergs, or approximately 0.7377 foot-pounds.
A horizontal disc of radius 45 cm rotates about a vertical axis through its centre. The disc makes one full
revolution in 1.40 s. A particle of mass 0.054 kg is placed at a distance of 22 cm from the centre of the disc.
The particle does not move relative to the disc.
a On a copy of the diagram draw arrows to represent the velocity and acceleration of the particle. [2]
b Calculate the angular speed and the linear speed of the particle. [2]
c The coe cient of static friction between the disc and the particle is 0.82. Determine the largest distance
from the centre of the disc where the particle can be placed and still not move relative to the disc. [3]
d The particle is to remain at its original distance of 22cm from the centre of the disc.
i Determine the maximum angular speed of the disc so that the particle does not move relative to
the disc. [2]
ii The disc now begins to rotate at an angular speed that is greater than the answer in d i. Describe
qualitatively what happens to the particle. [2]
Answer:
a. Please find attached the diagram of the disc, having arrows that represent the velocity and the acceleration of the particle placed on it
b. The angular speed is approximately 4.488 rad/s
The linear speed is approximately 0.987 m/s
c. The largest distance from the center of the disc where the particle can be placed and still not move is approximately 0.399 m from the center of the disc
d. i The maximum angular speed of the disk so that the particle does not move relative to the disk is approximately 6.044 rad/sec
ii When the angular speed with which the disc rotates is more than the the answer of question d i above, the particle slips on the disc, and the disc begins to rotate faster than the particle, while the particle is swung in an outward radial direction off the disc due to the centrifugal forces
Explanation:
The given parameters are;
The radius of the horizontal disc, r = 45 cm = 0.45 m
The time the disc makes one full revolution, T = 1.40 s
The mass of the particle placed on the disc = 0.054 kg
The location on the disc the particle is placed = 22 cm from the disc's center
a. Please find attached the diagram of the disc created with Microsoft Visio, with arrows representing the velocity and the acceleration of the particle placed on the disk
b. The angular speed, ω = 2·π/T = 2 × π/1.4 ≈ 4.488 rad/s
The linear speed, v = ω × r = 4.488 rad/s × 0.22 m ≈ 0.987 m/s
The linear speed, v ≈ 0.987 m/s
c. The given coefficient of static friction = 0.82
Therefore;
The frictional force that prevents motion = Weight of the particle × The coefficient of static friction
The frictional force that prevents motion is [tex]F_f[/tex] = 0.054 × 9.8 × 0.82 ≈ 0.434 N
[tex]F_f[/tex] ≈ 0.434 N
Therefore, for the largest distance from the center of the disc where the particle can be placed and still not move, r, is given by the formula for the centripetal force, [tex]F_c[/tex], acting on the particle as follows;
For static equilibrium, no movement of the particle relative to the disc, we have;
[tex]F_f[/tex] = [tex]F_c[/tex]
Where;
[tex]F_c = \dfrac{m \times v^2}{r} = m \times \omega ^2 \times r[/tex]
Which gives;
[tex]F_c = {0.054 \ kg \times (4.488 \ rad/s)^2} \times r = F_f = 0.434 \ N[/tex]
r = 0.434 N/(0.054 kg × (4.488 rad/s)²) ≈ 0.399 m
The largest distance from the center of the disc where the particle can be placed and still not move, r = 0.399 m from the center of the disc
d. i From the static equilibrium equation where r = 0.22 m, we have;
[tex]F_c = {0.054 \ kg \times \omega ^2} \times 0.22 \ m = F_f = 0.434 \ N[/tex]
ω = √(0.434 N/(0.054 kg × (0.22 m))) ≈ 6.044 rad/sec
The maximum angular speed of the disk so that the particle does not move relative to the disk, ω ≈ 6.044 rad/sec
ii When the angular speed with which the disc rotates is more than the the answer of question d i above, we have
The particle begins to slip on the disc such that the disc rotates faster than the particle and the particle tends to rotate slower than the speed pf the disc and is swung off the disc by centripetal force acting on the particle due to the rotational motion of the disc.
Current in a wire continusly increasing what will be the effect on the magnitude of the displacement current
Explanation:
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If you drive your 1,000 kg car from sea level up to the mountain, which is 366 m above sea level, how much will you have increased your car's potential energy?(consider : g=10m/s2) *
1 point
a) 366,000 J
b) 3,660,000 J
c) 2.73 J
d) 20.73 J
Answer:
Option B. 3660000 J
Explanation:
At the sea level, we'll assume that the height is 0 m. Hence, the potential energy at the sea level is zero.
Next, we shall determine the potential energy at a height of 366 m above the sea level. This can be obtained as follow:
Mass (m) = 1000 kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 366 m
Potential energy (PE) =?
PE = mgh
PE = 1000 × 10 × 366
PE = 3660000 J
From the calculations made above, we can see clearly that the potential energy of the car at a height of 366 m above sea level is 3660000 J.
Hence, the potential energy of the car increases from 0 at the sea level to 3660000 J at 366 m above the sea level.
