Answers
The horizontal component of a force is defined as follows
[tex]F_x=F\cdot\cos \theta[/tex]Where F = 16 N, and theta = 68. Let's use these values to find the horizontal component of F.
[tex]F_x=16N\cdot\cos 68\approx6N[/tex]Therefore, the horizontal component of the pulling force is 6 Newtons.Related Questions
Skydivers accelerate downward when they first jump out of an airplane. Eventually they reach a terminal velocity, however, and their speed remains constant because two forces become equal in magnitude but opposite in direction. What are these two forces?
Answers
Take into account that the gravitational force implies an acceleration on the skydivers when they go downward. While they go down the drag force, due to the air, decelerates the motion of the skydiver. The drag force increases in a constant way (becasue the drag force depends on the speed of the skydiver, and the speed, due to the acceleration produced by the gravity, increases), until it equals the gravitational force and the speed of the skydiver becomes constant.
Hence, the two forces are the gravitational force and the drag force
A psychology professor testified as an expert witness in court on behalf of a man who was hit by a train. In his testimony, the professor noted that it is reasonable to believe that someone who lives next to train tracks might not even notice an approaching train. Which term does this situation exemplif
Answers
Habituation is a term does this situation exemplify.
In this case, the professor is talking about a situation where a person has become used to a particular stimulus. as a result, the rate at which the stimulus would be registering to the person would be much lesser as compared to someone who is fairly new in being exposed to the stimulus.
This process is called habituation. Sensitization is the situation where a person becomes extremely sensitive to even lower intensity of the stimulus.
Habituation is a form of non-associative studying in which an innate reaction to a stimulus decreases after repeated or prolonged presentations of that stimulus. Responses that habituate encompass those who involve the intact organism or those that contain simplest additives of the organism.
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Fill in the blank question.
To send information through verbal speech, ____
cords send information encoded in _____
waves through the air.
Answers
To send information through verbal speech, transmitter cords send information encoded in radio waves through the air.
The basic principle is simple. At one end a transmitter encodes or modulates the message by varying the amplitude or frequency of the waves. This is a bit like Morse code. Meanwhile, a receiver tuned to the same wavelength receives the signal and decodes it into the desired format sound image data, etc.
Communications and satellite transmissions. Radio waves travel easily through the air. Radio waves are used for the wireless transmission of sound messages and information for communications, and for maritime and air navigation. Information is applied to an electromagnetic carrier wave as amplitude modulation AM frequency modulation FM or digital form pulse modulation.
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A baseball player is dashing toward home plate with a speed of 5.6 m/s when she decides to hit the dirt. She slides for 1.5 s , just reaching the plate as she stops (safe, of course). Calculate the acceleration.
Answers
ANSWER:
-3.73 m/s^2
STEP-BY-STEP EXPLANATION:
Acceleration is the rate of change of velocity and we can calculate it as follows:
[tex]a=\frac{v_2-v_1}{\Delta t}[/tex]Replacing:
[tex]\begin{gathered} a=\frac{0-5.6}{1.5} \\ a=-3.73m/s^2 \end{gathered}[/tex]The baseball's player's deceleration is 3.73 m/s^2
Explain how electron microscopy has increased understanding of subcellular structures.
[2]
Answers
Answer:
The development of the electron microscopes therefore helped scientists to learn about the sub-cellular structures involved in aerobic respiration called mitochondria . The scientists developed their explanations about how the structure of the mitochondria allowed it to efficiently carry out aerobic respiration.
What is discharged body
Answers
A body is considered to be discharged when it loses charge to the ground or another body.
The human body is an excellent conductor. Therefore, that will fall to the earth as the Charge passes through our bodies. Discharge of the body is the process by which the excess energy leaves the body and flows through it before falling to the ground. The term "earthing" refers to this idea.
Allowing the electricity to discharge from your body into the ground is the quickest approach to eliminating static electricity from the body. Touch any conductive object that is not isolated from the ground to enable this, such as a metal light pole or the screw on a light switch panel.
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7/21/22, 7:37 AMProblem Set ThreeNotes: Use 9.8 m/s 2 for the acceleration due to gravity. Formust be expressed in m/sLaw calculations, mass must be expressed in kg and velocity.A man with a mass of 110kg slides down a vertical pole with an acceleration of 9 m/s2. What is the frictional force acting on the man?
Answers
Given:
The mass of the man is m = 110 kg
The man slides down with acceleration,
[tex]a=9\text{ m/s}^2[/tex]The acceleration due to gravity is
[tex]g=\text{ 9.8 m/s}^2[/tex]Required: The frictional force acting on the man.
Explanation:
The equation of motion to calculate frictional force is given as
[tex]\begin{gathered} mg-f=ma \\ f=mg-ma \\ =m(g-a) \end{gathered}[/tex]On substituting the values, the frictional force will be
[tex]\begin{gathered} f=110(9.8-9) \\ =\text{ 88 N} \end{gathered}[/tex]Final Answer: The frictional force is 88 N
The SPH3U class designs a cannon able to shoot a human being out of it. . If the human is launched at a velocity of 40 m/s, 37° from the ground, how far from the cannon should you place a mattress to catch the human if the muzzle of the cannon is 0.75 m from the ground?
Answers
Given:
Velocity = 40 m/s
Angle = 37°
Let's find the distance from the cnnon where you should place a matress to catch the human if the muzzle is 0.75 meters from the ground.
Let's first draw the projectile motion diagram:
To find the distance, first apply the formula to solve for the time taken for the human to reach the matress.
[tex]y=y_o+v_{oy}t-\frac{1}{2}gt^2[/tex]Where:
Height from the ground to the muzzle = yo = 0.75
y = 0
Velocity = vo = 40 m/s
initial vertical velocity = voy = vo sinθ = 40 sin37
We have:
[tex]v_{oy}=v_o\sin \theta=40\sin 37=24.07\text{ m/s}[/tex]g is the acceleration due to gravity = 9.8 m/s²
Now, let's solve for the time t:
[tex]\begin{gathered} y=y_o+v_{oy}t-\frac{1}{2}gt^2 \\ \\ 0=0.75+24.07t-\frac{1}{2}\times9.8\times t^2 \\ \\ t=4.94\text{ seconds} \end{gathered}[/tex]To sove for the horizontal distance, x, apply the formula:
[tex]\begin{gathered} x=v_{ox}t \\ \\ \text{Where:} \\ v_{ox}=v_o\cos \theta=40\cos 37=31.94\text{ m/s} \end{gathered}[/tex]Hence, we have:
[tex]\begin{gathered} x=31.94\times4.94 \\ \\ x=157.8\text{ m} \end{gathered}[/tex]Therefore, the matress should be placed at 157.8 meters from the cannon in order to catch the human.
ANSWER:
157.8 meters
On July 19, 1969, the lunar orbit of Apollo 11 was adjusted to an average height of 172 kilometers above the Moon's surface. The radius of the Moon is 1840 kilometers, and the mass of the Moon is 7.3 x 1022 kilograms. At what speed did the spacecraft orbit the Moon? Include units in your answer. Answer must be in 3 significant digits.
Answers
In order to calculate the speed of the spacecraft, use the following formula;
[tex]v=\sqrt[]{\frac{GM}{r}}[/tex]where,
M: mass of the moon = 7.3*10^22 kg
G: Cavendish constant = 6.67*10^-11 Nm^2/kg^2
r: distance in between the spacecraft and the center of the Moon =
172 km + 1840 km = 2012 km
Replace the previous values of the parameters into the formula for v and simplify:
[tex]v=\sqrt[]{\frac{(6.67\cdot10^{-11}N\frac{m^2}{\operatorname{kg}^2})(7.3\cdot10^{22}kg)}{2012\operatorname{km}}}\approx4.91\cdot10^4\frac{m}{s}[/tex]Hence, the speed of the spacecraft is approximately 4.91*10^4 m/s
Two 3.09 cm by 3.09 cm plates that form a parallel-plate capacitor are charged to +/- 0.617 nC. What is the electric field strength inside the capacitor if the spacing between the plates is 1.784 mm?
Answers
Given:
The charge on the capacitor is Q = 0.617 nC
The distance between plates is d =1.784 mm
The area is
[tex]\begin{gathered} A=\text{ 3.09}\times3.09 \\ =9.5481\text{ cm}^2 \end{gathered}[/tex]To find the electric field strength.
Explanation:
The electric field strength can be calculated by the formula
[tex]E=\frac{Q}{\epsilon_oA}[/tex]The constant is
[tex]\epsilon_o=8.85\times10^{-12}\text{ C}^2\text{ / N m}^2[/tex]On substituting the values, the electric field strength will be
[tex]undefined[/tex]
A dog was running 10.8m/s for 61s. How far did it run in that time?
Answers
To solve using these three variables, we can use this equation:
distance = speed × timeSolving the QuestionWe're given:
Speed: 10.8 m/sTime: 61 sDistance: ?⇒ Plug the given values into the equation:
distance = speed × time
distance = 10.8 m/s × 61 s
distance = 658.8 m
AnswerThe dog ran 658.8 m in that time.
Earth has a radius R of about 6.37 × 106 m. How many watts of power does Earth receive from the sun? Remember that even though the Earth is a sphere, its image with respect to the sun is a circle — in other words, Earth's shadow is circular, meaning that it captures a ‘circular amount’ of light. Use the observed value of S = 1361 W/m2.
Answers
If the earth has a radius R of about 6.37 × 10⁶ m, then the power received by the earth would be 2.722 × 10¹⁰ watts.
What is power?
The rate of doing work is known as power. The Si unit of power is the watt.
Power =work/time
As given in the problem Earth has a radius R of about 6.37 × 10⁶ m.
Area of the earth assuming it is a circle = 3.14 × 6.37 × 10⁶
= 2 × 10⁷ meter²
The power received by the earth = 2 × 10⁷ × 1361
= 2.722 × 10¹⁰ watts
Thus, the power received by the earth would be 2.722 × 10¹⁰ watts.
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A car accelerates uniformly, passing a first checkpoint with velocity 20 m s-1, andpassing a second checkpoint with velocity 32 m s-1, after 4 seconds.a)Sketch a velocity-time graph to show this information.b)Use the velocity-time graph to find each of the following.C)The acceleration of the car during the 4 seconds.The total distance travelled by the car during the 4 seconds.
Answers
Part a: velocity-time graph
Where t is time in seconds after reaching the first checkpoint.
Part b:
Acceleration = change in velocity/time
a = (v2-v1)/t
v2: final velocity, v1: initial velocity, t: time
Let's keep track of all the variables we know:
v2 = 32 m/s, v1 = 20 m/s, t = 4 s
Now let's plug all the known variables in:
a = (32-20)/4 = 12/4
a = 3 m/s^2
d = (v1)*t + (a*t^2)/2
Let's keep track of all the variables we know:
v1 = 20 m/s, t = 4 s, a = 3 m/s^2
Now let's plug all the known variables in:
d = 20*4 + (3*4^2)/2 = 80 + 48/2 = 80 + 24
d = 104 m
/s
Given three electrically charged spheres A,B and C. Sphere A is brought near Sphere B and both are attracted to each other electrically. Sphere B is brought near Sphere C and both repel each other electrically. What happens when Sphere A is brought near Sphere C? A)electrical repulsion B)both attraction and repulsion C)electrical attraction D)neither attraction nor repulsion E)nothing
Answers
Given
Three electrically charged spheres A,B and C. Sphere A is brought near Sphere B and both are attracted to each other electrically. Sphere B is brought near Sphere C and both repel each other electrically.
To find
What happens when Sphere A is brought near Sphere C?
Explanation
Sphere A and sphere B are attracted to each other.
Thus, both A and B are of opposite charge.
Let A be a positive charge and B be a negative charge.
When sphere B is brought near sphere C both repel.
Thus, sphere C is also of negative charge.
So A and C are then of opposite charges and would both attract each other.
Conclusion
When sphere A is brought near sphere then C)electrical attraction occurs.
You are driving your 1800 kg car at 21 m/s over a circular hill that has a radius of 150 m. A deer running across the road causes you to hit the brakes hard while right at the summit of the hill, and you start to skid. The coefficient of kinetic friction between your tires and the road is 0.75.
What is the magnitude of your horizontal acceleration as you begin to slow?
Answers
The magnitude of your horizontal acceleration as you begin to slow is 7.35m/s².
Applying work-energy theorem,
Total work done on the body by all the forces = change in kinetic energy.
So, here,
Work done by friction = ∆KE
-uMgd = 1/2M(v²-u²)
u is the coefficient of friction,
M is the mass of the car,
g is the acceleration due to gravity,
v is the final velocity of car,
u is the final velocity of car,
d is the horizontal distance moved by car.
(-0.75)(1800)(9.8)d = -1/2(1800)(21)²
d= 30 meters.
Applying Equation of motion,
V²-U²=2ad
Where a is the horizontal acceleration,
-(21×21) = 2(30)a
a = 7.35m/s².
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Two trains leave the same station at the same time, one traveling west at a constant speed of 60 miles per hour, the other traveling south at a constant speed of 80 miles per hour. After how long are the two trains exactly 300 miles apart?
Answers
If one train travels at constant speed 60 miles/h and other train at 80 miles/h south and west respectively then after 3 hours the two trains are exactly 300 miles apart.
What is speed?Speed is a scalar quantity that indicates how quickly an object is moving, regardless of its direction.
The formulas below can be used to determine an object's speed:
S = d/t
where
d is the distance an object travels and
where
The metre per second (m/s) is the unit of speed in the SI system.
As the train are traveling south and west the angle of direction between them is 90° hence we can use Pythagoras theorem to find displacement
c = √(a² + b²)
displacement of 1 hour = √(60 ² + 80²)
= √10000
= 100 miles
time taken 100 miles is = 1 hour
Then time taken for displacement of 300 miles = 3 × 1 hour
= 3 hours
Thus, After 3 hours the two trains are exactly 300 miles apart.
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A boy is sitting on the outside edge of a merry go round. If the angular velocity id doubled, then the linear speed of the boy will be
Answers
If the angular velocity id doubled, then the linear speed of the boy will be doubled.
As we know,
If a body is rotating in a circular path, it will have an angular as well as linear velocity.
The angular velocity is responsible for rotation and the linear velocity is responsible for changing the direction in a circular manner.
The angular velocity W is related to the linear velocity V as,
W = V/R
Let us say this is equation 1,
Where R is the radius of the merry go round,
Now if the angular velocity is doubled,
W' = V'/R
Let us say this is equation 2.
On dividing equation 1 and 2,
W/W' = V/V'
As we know, W' = 2W.
1/2 = V/V'
V' = 2V.
The linear velocity will increase by a factor of 2.
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This diagram shows an object with a mass of 3kg being pushed along africtionless surface. The object accelerates at 3m/s/s. What force was applied tothe object? *
Answers
We are asked to determine the force exerted on the object. To do that we will use Newton's second law:
[tex]F=ma[/tex]Where:
[tex]\begin{gathered} F=\text{ force} \\ m=\text{ mass} \\ a=\text{ acceleration} \end{gathered}[/tex]Now, we substitute the values:
[tex]F=(3kg)(3\frac{m}{s^2})[/tex]Solving the operations:
[tex]F=9N[/tex]Therefore, the force is 9N.
A projectile is launched at an angle of 30° above the horizontal with an initial velocity (vo) of 42.2 m/s. After some time passes, the projectile reaches the peak of its trajectory.Its horizontal velocity (vx) at the peak of its trajectory is ____ m/s.
Answers
First, for us to start, let us draw the problem:
When we analyse this problem, we can see that the only velocity that changes overtime is the vertical velocity (due to gravity), so the horizontal speed is the same throughout the whole movement. This gives us the following:
[tex]v_x=42.2*cos(30°)=36.546\frac{m}{s}[/tex]Then, the horizontal speed at the peak of the trajectory (and also during the whole movement) is v=36.546m/s
what is the center of mass?
a) the location that the object would concentrate to if we imagined it to shrink to a point.
b) the exact point of an object, measured from all its edges.
c) a point in space to which an object is attracted.
d) an arbitrarily chosen point in the object used to describe its motion.
e) a point in space that the object is moving toward.
Answers
The exact point of an object measured from all its edge center of mas.
What is center of mass?
the center of mass of a distribute of mass in space is the unique point where the weighted relatively position of the devided mass sums to 0. This is a point to which a force of may be applied to cause of a linear acceleration it without an angular acceleration.
Most of the time (not everytime)the centre of mass of an object lies within the object itself.
Eg-the center of mass of a ball in middle of a ball and the centre of mass of a copy is middle of a copy.
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A technician examines different electrical devices to determine the one that is the most energy efficient. While conducting a test, he notes that one of these devices consumes 550 000 J of energy and loses 315 000 J at the same time. What is the energy efficiency of this device?
Answers
The output of this device is equal to 550 000-315 000 = 235 000, or 42.7%, of its energy efficiency.
What does "machine efficiency" mean?Efficiency is the percentage of input work that is actually completed by the machine (output work). The fact that some of the input work is used to reduce friction is the main cause of the output work always being less than the input labor. Efficiency is never 100% and is therefore always less than 100%..
How effective is a basic machine?Due to internal friction, simple machines are always less efficient than 1.0. conserving energy When friction is ignored, the work performed on a simple machine is equivalent to the work performed by the machine to complete a task.
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A 55 kg motorcyclist is flying through the air at 72 km/h at the apex of the jump 8.3 m above the ground. Find his landing speed assuming energy is conserved.
Answers
Answer: 85.4 km/h
Explanation:
We would apply the law of conservation of kinetic energy. The formula is expressed as
KE1 + PE1 = KE2 + PE2
where
KE1 and KE2 are the initial and final kinetic energies
PE1 and PE2 are the initial and final potential energies
The formula for calculating kinetic energy is
KE = 1/2mv^2
where
m is the mass of the object
v is the velocity
The formula for calculating potential energy is
PE = mgh
where
g is the acceleration due to gravity = 9.8m/s^2
h is the height of the object
From the information given,
m = 55
h = 8.3
Potential energy at the top = 55 x 9.8 x 8.3 = 4473.7
At the top, v = 72km/h
Recall, 1 km = 1000m
1 hour = 3600s
72km/h = 72 x 1000/3600 = 20m/s
kinetic energy at the top = 1/2 x 55 x 20^2 = 11000
KE1 + PE1 = 4473.7 + 11000 = 15473.7
At the point of landing,
PE2 =0
KE = 1/2 x 55 x v^2 = 27.5v^2
Thus,
27.5v^2 = 15473.7
v^2 = 15473.7/27.5
v = √(15473.7/27.5)
v = 23.72 m/s
his landing speed is 23.72 m/s
Converting to km/h, we would multiply 23.72 by 18/5
Landing speed = 85.4 km/h
Find the magnitude of the force F exerted by the leg on the small pulley. (By Newton's third law, the small pulley exerts an equal and opposite force on the leg.) Let the mass m be 2.27 kg.
Answers
The magnitude of the force F exerted by the leg on the small pulley is, F = 38.53 N
We can calvulate the magnitude of the force by a mathematical calculation, which is as follows. As the small pulley and mass are not moving, they are stationary.
t2=t1 = mg = (2.27)(9.8)
= 22.246 N
Now resolve the force in the opposite direction.
F = 2Tcos30°
F = (2)(22.246)(cos30°)
F = 38.53 N
Therefore, the magnitude of the force is 38.53 N.
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A force of magnitude Fx acting in the x-direction on a 2.35-kg particle varies in time as shown in the figure below. (Indicate the direction with the sign of your answer.)
Answers
The impulse is mathematically given as
v_f=3.61m/s
This is further explained below.
What is impulse?Generally, In classical mechanics, the integral of a force F over the time period t during which it operates is referred to as the impulse of the system.
It should come as no surprise that impulse is likewise a vector quantity given that force is one.
When an impulse is applied to an item, a corresponding vector change in the object's linear momentum occurs, also in the direction that the impulse causes.
In conclusion, Impulse = Area Under Curve
=2*4+1*4
[tex]m\left(v_f-v_i\right)=12 \\\\v_f=\frac{12}{2.35}=5.11 \mathrm{~m} / \mathrm{s}\\\\m\left(v_f-(-1.5)\right)=12 \\v_f=5.11-1.5 \\\\ =3.61 \mathrm{~m} / \mathrm{s} \\v_f=3.61 \mathrm{~m} / \mathrm{s}\end{gathered}$$[/tex]
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CQ
A force of magnitude Fx acting in the x-direction on a 2.35-kg particle varies in time as shown in the figure below.
(a) Find the impulse of the force.
____________kg
Which is true about a step-down transformer?Voltage out is larger than the voltage in.Voltage out is smaller than voltage in.It is when a robot turns into a car.Voltage out is the same as the voltage in.
Answers
Answer:
Explanation:
A step down transformer decreases the incoming voltage. It converts high voltage and and low current from the primary side to low voltage and high current. Thus, the correct option is
Voltage out is smaller than voltage in.
It is now 10:29 a.m., but when the bell rings at 10:30 a.m. Suzette will be latefor French class for the third time this week. She must get from one side of theschool to the other by hurrying down three different hallways. She runs downthe first hallway, a distance of 35.0 m, at a speed of 3.50 m/s. The secondhallway is filled with students, and she covers its 48.0-m length at an averagespeed of 1.20 m/s. The final hallway is empty, and Suzette sprints its 60.0-mlength at a speed of 5.00 m/s. a) Does Suzette make it to class on time or doesshe get detention for being late again? b) Draw a distance vs. time graph ofthe situation.
Answers
Answer:
a) She get detention for being late again
Explanation:
First, we need to identify how much time does she take on each hallway.
With the distance and the speed, we can calculate the time as:
t = distance/speed
So, for each hallway, we get:
First hallway:
distance = 35 m
speed = 3.5 m/s
time = 35/3.5 = 10s
Second Hallway
distance = 48 m
speed = 1.2 m/s
time = 40s
Third Hallway
distance = 60 m
speed = 5 m/s
time = 60/5 = 12 s
Therefore, the total time that she takes was
10s + 40s + 12s = 62s
Since she takes more than 60 seconds, she will be late again.
Finally, we know that she takes 10s to run a distance of 35m, then another 40s to run a distance of 48 m, and another 12s to run a distance of 60 m. Therefore, the distance vs. time graph for this situation is
Which phrase is an example of kinetic energy?
Answers
Have an AWESOME day! :)
How old is an artifact if four half-lives have occurred and the half-live of carbon-14 is 5730 years?
Answers
22920 years old is an artifact if four half-lives have occurred and the half-live of carbon-14 is 5730 years.
Why does carbon-14 have a short half-life?Due to radioactive decay to nitrogen-14, carbon-14 has a very short half-life of 5,730 years, which means that over this time, the proportion of carbon-14 in a sample will have decreased by 50%.
Why is the term "half-life" used?A half-life is the amount of time it takes for something to go from 100% to 50%. The phrase is most frequently used in reference to radioactive decay, which happens when energetic atomic particles that are unstable lose momentum.
Briefing:The half-life of carbon-14 is 5730 years. Accordingly, a sample of 100 carbon-14 atoms has a half-life of 5730 years, or 50 carbon-14 and 50 carbon-12 (i.e., 100 carbon-14, one half-life = 50 carbon-14, 50 carbon-12).
4 half lives
=5730×4
=22,920 years old.
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A track and field athlete is competing in the hammer throw event. The athlete is 181cm tall, with an arm length of 90 cm and is using a standard ball and chain for the event which is 121cm long. As the athlete begins to spin , their body becomes the center of the motion, with both arms holding on to the ball and chain handleIf the initial linear ve locity of the ball and chain is 29m/s upon its release, how fast (total angular velocity rads) was the athlete spinning it.
Answers
The angular velocity is obtained from the calculation as 13.7 rad/s.
What is the angular velocity?Let us recall that the term circular motion is applied to the kind of motion that is carried out by an object that is moving along a circular path. Given the fact that we have an object that is spinning around in a circular path, we can say that a circular motion is actually taking place in the system that involves the athlete, the ball and the rope.
We are told that the athlete is 181cm tall, with an arm length of 90 cm and is using a standard ball and chain for the event which is 121cm long. As the athlete begins to spin , their body becomes the center of the motion, with both arms holding on to the ball and chain handle then the initial linear velocity of the ball and chain is 29m/s upon its release.
We can now see that the radius of the path is 90 cm + 121 cm = 211cm or 2.11 m
Using;
V = rω
ω = V/r
ω = 29m/s/2.11 m
ω = 13.7 rad/s
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Question 2 of 25If the law of conservation of energy applies to a situation, then:A. the system is an open one.B. the output work being done is greater than the work being put into the system.C. the total (PE + KE) before an event is equal to the total (PE+KE) after the event.D. matter is entering and leaving the system.
Answers
ANSWER
C. The total (PE + KE) before an event is equal to the total (PE + KE) after the event.
EXPLANATION
We want to identify the correct option.
The law of conservation of energy is a law that applies only to closed systems and it states that:
In other words, energy cannot be created nor destroyed.
This implies that the total energy (sum of potential and kinetic energy) before an event occurs must be equal to the total energy after the event has occurred.
Therefore, the correct option is option C.