Answer:
[tex]\boxed {\boxed {\sf 36 \ meters}}[/tex]
Explanation:
We are asked to find the distance a body covers. We know the initial velocity, acceleration, and time, so we will use the following kinematic equation.
[tex]d= v_i t+ \frac {1}{2} \ at^2[/tex]
The body starts at rest with an initial velocity of 0 meters per second. The acceleration is 8 meters per second squared. The time is 3.0 seconds.
[tex]v_i[/tex]= 0 m/s a= 8 m/s²t= 3 sSubstitute the values into the formula.
[tex]d= (0 \ m/s)(3 \ s) + \frac{1}{2} (8 \ m/s^2)(3 \ s)^2[/tex]
Multiply the first set of parentheses.
[tex]d= ( 0 \ m/s * 3 \ s) + \frac{1}{2} ( 8 \ m/s^2)(3 \ s)^2[/tex]
[tex]d=0 \ m + \frac{1}{2} ( 8 \ m/s^2)(3 \ s)^2[/tex]
Solve the exponent.
(3 s)²= 3 s* 3 s= 9 s²[tex]d= 0 \ m + \frac{1}{2}( 8 \ m/s^2)(9 \ s^2)[/tex]
Multiply again.
[tex]d= 0 \ m + \frac{1}{2} ( 72 \ m)[/tex]
[tex]d= 36 \ m[/tex]
The body will cover a distance of 36 meters.
A 25.0kg boy is sliding on a frictionless frozen lake at 5.00m/s to the north when he is struck by a 1.00kg
snowball moving at 15.0m/s from the west. If the snowball sticks to him, how fast, and in what direction,
does the boy move after the collision?
The final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east
The given parameters;
Mass of the boy, m₁ = 25 kgSpeed of the boy, u₁ = 5 m/sMass of the snowball, m₂ = 1.0 kgSpeed of the snow ball, u₂ = 15 m/sThe initial momentum of the boy is calculated as follows;
[tex]P_y = m_1 u_1\\\\P_y = 25 \times 5\\\\P_y = 125 \ kgm/s \ \ north[/tex]
The initial momentum of the snowball is calculated as follows;
[tex]P_x = m_2 u_2\\\\P_x = 1 \times 15 \ \\\\P_x = 15 \ kgm/s \ \ west[/tex]
The resultant momentum of the boy and the snowball after collision is calculate as follows;
[tex]P_f = \sqrt{P_y^2 + P_x^2} \\\\P_f = \sqrt{125^2 + 15^2} \\\\P_f = 125 .9 \ kgm/s[/tex]
The final velocity of the system boy-snowball system is calculated as;
[tex]v_f(m_1 + m_2)= P_f\\\\v_f = \frac{P_f}{m_1 + m_2} \\\\v_f = \frac{125.9}{25 + 1} \\\\v_f = 4.84 \ m/s[/tex]
The direction of the boy after the collision is calculated as follows;
[tex]\theta = tan^{-1}(\frac{v_y}{v_x} )\\\\\theta = tan^{-1} (\frac{5}{15} )\\\\\theta = 18.4 \ ^0[/tex]
Thus, the final velocity of the boy after the collision with snowball is 4.84 m/s at 18.4⁰ north-east
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1. A roller coaster with a mass of 800 kg sits stationary at the top of a section of track, 75 m above
the ground as shown. When the brake is released, it starts to roll down the track
2. For each height represented in the diagram, calculate the gravitational potential energy using
Ep = mgh. Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for
each of the heights of the roller coaster. (6 marks)
3. Assuming there is no friction, determine the mechanical kinetic energy using Ek = Etotal - Ep.
Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for each of
the heights of the roller coaster. (6 marks)
4. For each height represented in the diagram, calculate the velocity using = �2
. Show ONE
SAMPLE calculation in the calculations section below and fill in Table 1 for each of the heights of
the roller coaster. (6 marks)
5. Use your answers to graph how gravitational potential energy, mechanical kinetic energy, and
velocity change as the roller coaster changes height. Use different colours for the three lines on
the graph. Graph paper is provided below. (3 marks)
6. Repeat steps 1 – 5 above for a roller coaster cart that has a mass of 300 kg and enter your
results in Table 2.
Calculations:
800 kg roller coaster cart:
Sample calculation for gravitational potential energy:
Sample calculation for Mechanical kinetic energy:
Sample calculation for velocity:
300 kg roller coaster cart:
Sample calculation for gravitational potential energy:
Sample calculation for mechanical kinetic energy:
Sample calculation for velocity:
Results:
Table 1: Potential energy, kinetic energy, total energy, and velocity of the 800 kg roller coaster cart
Table 2: Potential energy, kinetic energy, total energy, velocity of the 300 kg roller coaster cart.
Graphs:
It’s graphing time. These graphs are a bit different than the ones you did in the
data analysis assignment at the beginning of the course. In this case you have
three things to graph on each graph. (One graph for the 800 kg roller coaster cart
and one graph for the 300 kg roller coaster cart.) You need to graph the
gravitational potential energy with respect to height, the mechanical kinetic
energy vs height, and the velocity vs height.
Let’s look at the energy graphs first. In this case both kinetic energy and
mechanical energy cover the same range of values. This means they can use the
same scale on the y-axis. So, you will use the left y-axis and the x-axis to graph
the kinetic energy vs height and the potential energy vs height. You will need a
legend to explain which line is which. Colour coding is a nice way to highlight this.
The velocity values are much different than the energy values. This means you
need a totally different scale. So, your left y-axis won’t work. You need to make a
second scale on the right y-axis for your velocity values. You will plot the points
the same way as normal, but you will use the numbers on the right-hand scale
instead. Again, be sure to add your velocity line to the legend with a separate
colour code.
Discussion Questions:
1. Describe the relationship between the gravitational potential energy and the mechanical kinetic
energy of the roller coaster on your graph. (2 marks)
2. Describe the shapes of each of the three lines in the graph. Explain why the velocity is different.
(4 marks)
3. Describe how mass affects the speed at the bottom of the roller coaster. (2 marks)
4. Describe how mass affects the gravitational potential energy and the mechanical kinetic energy
of the roller coaster. (2 marks)
5. At what point does the roller coaster have a maximum value for the following? Justify your
answer by explaining why. (2 marks each)
a. Gravitational potential energy
b. Mechanical energy
c. Velocity
6. In your calculations, you assumed that the roller coaster was frictionless. All real roller coasters
encounter friction. Describe how the actual values of the variables would differ, or not differ,
from your calculated values for a real roller coaster. (Hint: what form of energy would some of
the total energy be converted to if there was friction in the system?) (4 marks)
How you will be graded:
Grades will be based on answering questions to demonstrate an understanding of the material covered
in this unit. Point form answers are okay if ideas are complete and use vocabulary (Word Bank)
provided. For questions out of 4 marks, there are 4 responses expected.
Answer:
Give me some hint please
Based on the calculations, potential energy of this roller coaster at a height of 75 meters is equal to 588,000 Joules.
How to calculate potential energy?Mathematically, potential energy is calculated by using this formula:
P.E = mgh
Where:
P.E represents potential energy.m is the mass.h is the height.g is acceleration due to gravity.Note: Acceleration due to gravity is equal to 9.8 m/s².
At a height of 75 m, we have:
P.E = 800 × 9.8 × 75
P.E = 588,000 Joules.
At a height of 60 m, we have:
P.E = 800 × 9.8 × 60
P.E = 470,400 Joules.
At a height of 45 m, we have:
P.E = 800 × 9.8 × 45
P.E = 352,800 Joules.
At a height of 30 m, we have:
P.E = 800 × 9.8 × 30
P.E = 235,200 Joules.
At a height of 15 m, we have:
P.E = 800 × 9.8 × 15
P.E = 117,600 Joules.
In conclusion, we can deduce that the potential energy of this roller coaster decreases with a decrease in height.
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f. Protons and neutrons
2. TRUE or FALSE: An object that is positively charged contains all protons and no electrons
3. TRUE or FALSE: An object that is negatively charged could contain only electrons with me
accompanying protons
4. TRUE or FALSE: An object that is electrically neutral contains only neutrons.
An object which has more electron than proton is negatively charged, otherwise positively charged. Every statement is false
What is atom?Atom is the smallest unit of the matter consist of the positive charged nucleus and the electrons which moves around it. The atom can not be divided further.
The atom of a matter is made by three elements-
1) Neutron-Neutron is the element of atom, which has zero charge.2) Proton-Proton is the element of atom, which has positive charge.3) Electron-Electron is the element of atom, which has negative charge.For the Protons and neutrons, lets check all the statement wheather they are true or false.
2. An object that is positively charged contains all protons and no electrons- An object which is positively charged, has more number of proton than electron. This is a false statement.3. An object that is negatively charged could contain only electrons with me accompanying protons- An object, which is negatively charged, has more number of electron than proton. This is a false statement.4. An object that is electrically neutral contains only neutrons-The number of electron and proton is equal in an electrically neutral object. This is a false statement.Thus, every statement is false as a object which has more electron than proton is negatively charged, otherwise positively charged.
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How is the A He related to the A Hof a reaction?
Answer:
by giving a person to person
What is an element?
1. A substance made of more than one type of atom.
2. A substance that is made of water.
3. A substance that is made of no atoms.
4. A substance made of only one type of atom.
Answer:
Hey Friend...
Explanation:
This is ur answer.....
4. A substance made of only one type of atom.Hope it helps!
Brainliest pls!
Follow me! :)
Answer:
4
Explanation:
element is a substance that cannot be broken down into any other substance which means every element is made up of its own type of atom
is xenon a pure substance
[tex]\large\huge\green{\sf{Yes}}[/tex]
In the following free body diagram, what is the net force on the object?
A.10n
B.5 N to the right
C.20 N to the right
D.7 N to the right
Answer:
B
Explanation:
Simply take all forces pointing to the right of the box as positive and all of the forces pointing to the left of the box as negative and add all values.
ΣF = 7 + 18 + (-20) = 5N to the right
Need help asap
A scientist has invented a robot to work on the seabed. According to his calculation, the armour of the robot can withstand a maximum pressure of 10⁵ Pa exerted by the sea water. If the density of the sea water is 1025 kg/m3, what is the maximum depth of the seabed that this robot can work? [Given g = 9.81 m/s2 and rho water = 1000 kg/m3]
Answer:
Explanation:
Well which is it ? ρ = 1000 kg/m³ or ρ = 1025 kg/m³?
Obviously the sea is salt water so we can ignore ρ = 1000 kg/m³
1025 kg/m³(d m)(9.81 N/kg) = 1 x 10⁵ N/m² = Pa
d = 9.9450535...
d = 10 meters
That's if we only account for the pressure due to the water. On top of that pressure would be atmospheric pressure which is about 101000 Pa
so the robot would be a hair above its pressure limit before it even got in the water.
a
A person throws a ball up into the air, and the ball falls back towar
would the kinetic energy be the lowest? (1 point)
at a point before the ball hits the ground
when the ball leaves the person's hand
o when the ball is at its highest point
at a point when the ball is still rising
Answer:
when the ball is at its highest point
Explanation:
Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.
Question: Self-test 3.12 Calculate the change in G for ice at -10°C, with density 917 kg mº, when the pressure is increased from 1.0 bar to 2.0 bar.
The change in the Gibb's free energy per mole (G) is 1.96 J.
The given parameters:
Density of the ice, ρ = 917 kg/m³Initial pressure, P₁ = 1.0 barFinal pressure, P₂ = 2.0 barTemperature, T = - 10 CMass of water = 18 gThe change in the Gibb's free energy per mole (G) is calculated as follows;
[tex]\Delta G = V(P_2-P_1) \\\\[/tex]
where;
V is the volume of the ice
[tex]Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density} \\\\Volume = \frac{18 \times 10^{-3} \ kg}{917 \ m^3} \\\\Volume = 1.96 \times 10^{-5} \ m^3\\\\Volume = 1.96 \times 10^{-5} \ m^3 \times \frac{1000 \ L}{m^3} \\\\Volume = 0.0196 \ L[/tex]
Change in pressure;
[tex]P_2 - P_1 = 2.0 \ bar \ - \ 1.0 \ bar = 1.0 \ bar = 0.987 \ atm[/tex]
The change in the Gibb's free energy per mole (G);
[tex]\Delta G= V(P_2-P_1)\\\\\Delta G = 0.0196\ L \times 0.987\ atm \\\\\Delta G = 0.0193 \ L.atm\\\\1 \ L.atm = 101.325 \ J\\\\\Delta G = 0.0193 \ L.atm \times \frac{101.325 \ J}{1 \ L.atm} \\\\\Delta G = 1.96 \ J[/tex]
Thus, the change in the Gibb's free energy per mole (G) is 1.96 J.
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Two forces act on a block as shown in the picture. What is the net force of the block?
30 N to the right
30 N to the left
10 N to the left 10
10 N to the right
Answer:
10 N to the left.
Explanation:
Since the forces are acting in opposite directions, you need to calculate the difference.
20 N - 10 N = 10 N
More force is being exerted to the left. Therefore, the net force is 10 N to the left.
Which of the following is NOT a function of the lens in the eye?
W
It can perform minor adjustments for distance.
It flattens when light rays from distant objects are to be focused.
It is a light receptor that generates nerve signals that are sent to the brain
It maintains its spherical shape to view nearby objects.
Answer:
the last one
Explanation:
The average value of the load between A and B is 6.0 N. The spring has an unstretched length
increases from 4.0 cm to 6.0 cm.
The change in the length of the spring as it stretches from 4.0 cm to 6.0 cm is 2.0 cm.
The change in length of the spring can be calculated by subtracting the initial unstretched length (4.0 cm) from the final stretched length (6.0 cm).
Change in length = Final length - Initial length
Change in length = 6.0 cm - 4.0 cm
Change in length = 2.0 cm
Therefore, the change in the length of the spring as it stretches from 4.0 cm to 6.0 cm is 2.0 cm. This means the spring elongates by 2.0 cm under the applied load between points A and B.
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The complete question is:
The average value of the load between A and B is 6.0 N. The spring has an unstretched length increases from 4.0 cm to 6.0 cm. What is the change in the length of the spring as it stretches from 4.0 cm to 6.0 cm?
A. 4.0 cmB. 1.0 cmC. 2.0 cmD. 6.0 cmWhy is it important that we learn about our solar system, the sun, and planets in our solar system other than Earth?
Answer:
To understand all the different kind of elements and how they work. Nature is a beautiful thing and it is important we understand how they survive and how we can help. The solar system is a mandatory nature. Learning about other planets also helps us feel safe and more knowledgable.
How is Compression Force Measured?
don't just copy theanswer please
Compression force can be measured with a force gage or load cell.
the distance between student home and school is 1.5 km what is the distance traveld bythe students in a week
Answer:
10.5k
Explanation:
1.5x7=10.5
a bicycle with tires 68 cm in diameter travels 9.2 km. how many revolutions do the wheels make
Answer:
Explanation:
Circumference in meters is
C = πD = 0.68π
9200 m / 0.68π m = 4,306.545518...
4,306.5 revolutions
A bicycle with tires 68 cm in diameter travels 9.2 km. the number of revolutions made by the wheel would be 4309
What is Velocity?The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. It can also be represented by the infinitesimal rate of change of displacement with respect to time. The unit of velocity is meter/second.
The mathematical expression for velocity is given by
velocity= displacement / time taken
As given in the problem a bicycle with tires 68 cm in diameter travels 9.2 km
Diameter = 68 cm
radius = diameter /2
= 68/2
= 34 cm
The distance covered by the tire in one cycle would be
distance in one revolution = 2π×Radius
= 2×3.14× 34
= 213.52 cm
=2.13 m
The number of revolutions by wheel = total distance/distance in one revolution
number of revolutions = 9200/2.135
=4309 revolutions
Thus, the number of revolutions made by the wheel would be 4309
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A textbook weighs 34 N on the surface of the Earth. What is the book’s mass on Earth’s surface?
Answer:
About 3.47kg
Explanation:
Recall that weight is equal to mass times acceleration.
In this case, our acceleration is due to gravity which on earth is about 9.8m/s/s
So we have 34N=9.8 *mass, divide both sides by 9.8 we get mass is equal to about 3.47kg.
I need help with parts a and B of this question
Answer:
Explanation:
Let x be the leaf spring compression distance
F = kx
5.20 x 10⁵ = (5.45 x 10⁵)x + (3.80 x 10⁵)(x - 0.500)
5.20 x 10⁵ = (5.45 x 10⁵)x + (3.80 x 10⁵)(x) - 1.90 x 10⁵
7.10 x 10⁵ = (9.25 x 10⁵)x
x = 0.76756...
x = 0.768 m
W = ½(5.45 x 10⁵)0.768² + ½(3.80 x 10⁵)(0.768 - 0.500)²
W = 174,148.648648...
W = 174 KJ
What are the two types of force
Answer:
Forces can be divided into primarily into two types of forces:
Contact Forces. Non-contact Forces.Explanation:
This is physics and it says collision and elastic/inelastic i need help
The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
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A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building. Part A If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?
Answer:
H = Vy t - 1,2 g t^2 formula for height of ball after t sec
H = 10 - 1.3 = 8.7 m
Vy = 27 sin 47 = 19.7 m/s vertical speed of ball
8.7 = 19.7 t - 9.8/2 t^2 height of ball after t sec
4.9 t^2 - 19.7 + 8.7 = 0 rearranging
[19.7 ± (388 - 170)^1/2] / 2 *4.9 = [19.7 ± 14.7] / 9.8 = .51 ,3.5 sec
.51 sec would be on the way up and 3.5 sec on the way down
Sx = 27 * cos 47 * 3.5 = 64.4 m around 200 ft seems reasonable
PLEASE HELP ASP WILL GIVE 50 POINT AND BRAINLIEST!!!!!!!!!!!!!
A. In the Bohr model of the hydrogen atom, the speed of the electron is approximately
2.16 × 10⁶ m/s. Find the central force acting on the electron as it revolves in a circular orbit of radius 5.17 × 10⁻¹¹ m.
Answer in units of N.
B. Find the centripetal acceleration of the electron
Explanation:
A. The centripetal force experienced by an electron as it goes around a hydrogen nucleus is given by
[tex]F_c = m_e\dfrac{v^2}{r}[/tex]
where [tex]m_e = \text{electron\:mass} = 9.11×10^{-31}\:\text{kg}[/tex]
[tex]r = 5.17×10^{-11}\:\text{m}[/tex] = orbital radius
[tex]v = 2.16×10^6\;\text{m/s}[/tex] = orbital velocity
so the centripetal force is
[tex]F_c = (9.11×10^{-31}\:\text{kg})\dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=8.22×10^{-8}\:\text{N}[/tex]
B. The electron's centripetal acceleration is given by
[tex]a_c = \dfrac{v^2}{r}[/tex]
Using the values from (A), we get
[tex]a_c = \dfrac{(2.16×10^6\;\text{m/s})^2}{5.17×10^{-11}\:\text{m}}[/tex]
[tex]\;\;\;=9.02×10^{22}\:\text{m/s}^2[/tex]
Where do hyperbolic comets originate?
A. the Oort cloud
B. the asteroid belt
C. the Kuiper belt
D. interstellar space
Answer:A.the Oort cloud
Explanation:
Answer:
try answer A..the Ootor cloud
a uniform thin rod of length l and mass m is allowed to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. a.) What is the speed of the center of gravity when the rod reaches its lowest position? b.) What is the tangential speed of the lowest point of the rod when the rod reaches its lowest position?
Answer:
Explanation:
Potential energy gets converted to rotational kinetic energy
a) ½Iω² = mgh
½(mL²/3)ω² = mgL/2
(L/3)ω² = g
ω = [tex]\sqrt{3g/L}[/tex]
v(CG) = (L/2) [tex]\sqrt{3g/L}[/tex]
Not sure if you wanted angular speed or tangential speed of the CG so I gave both.
b) v = Rω = L [tex]\sqrt{3g/L}[/tex]
True or False. What makes faith genuine is its object.
Answer:true
Explanation:
n earthquake emits both P-waves and S-waves that travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave at 4 000 m/s. If P-waves are received at a seismic station 34.0 s before an S-wave arrives, how far is the station from the earthquake center
Answer:
Yes , i also need answer
Explanation:
n earthquake emits both P-waves and S-waves that travel at different speeds through the Earth. A P-wave travels at 9 000 m/s and an S-wave at 4 000 m/s. If P-waves are received at a seismic station 34.0 s before an S-wave arrives, how far is the station from the earthquake center
Hello I am absolutely stumped on these six physics problems. Please help me on them.
Answer:
Explanation:
20° from the normal = 110° from parallel
1a) τ = (200sin90)[6] + (75sin110)[3] - (100sin110)[3]
τ = 1,129.52305... = 1100 N•m CCW
1b) τ = 200(6)(sin90) + 75(-3)(sin(360-110)) + 100(3)(sin(270 + 20)
τ = 1,129.52305... = 1100 N•m CCW
1c) My directions agree, both are positive z by right hand rule.
1d) Moment of inertia for a thin rod about an axis perpendicular to its center is
I = (1/12)mL²
τ = Iα
α = τ/I = 1129.523 / ((1/12)(200)(12²)) = 0.4706 rad/s²
θ = ½αt²
θ = ½(0.4706)(2•60)² = 3,388.56916... radians
θ = 3400 radians
at which time is is spinning about 9 revolutions per second
An LED is useful because when a current passes through it, it gives out... what?
Enter your answer
An LED is useful because when a current passes through it, it gives out light.
An LED is useful because when a current passes through it, it gives out Light.
What is an LED?
LED, in full light-emitting diode, in electronics, a semiconductor device that emits infrared or visible light when charged with an electric current.
A light-emitting diode (LED) emits light by applying a forward current to the pn junction of a compound semiconductor.
When forward current is passed through the light-emitting diode, carriers (electrons and holes) move. The holes in the p-type region move to the n-type region and the electrons in the n-type region move to the p-type region. The injected carriers recombine, and the energy difference before and after recombination is released as light. The emitted light depends on the energy band gap (Eg) of the compound semiconductor.
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A change
1. Ricardo has purchased a forklift for his business. The forklift can put out 4950 W of power. If the forklift is operating at full capacity, how much work can it do in 2.40 seconds?
O 2060)
O 4950)
O 6490)
O 11,900
The work done if the forklift is operating at full capacity is 11,900 J.
We have to recall that power is defined as the rate of doing work. The rate of doing work is defined as;
Power = Work done/time taken
When;
Power = 4950 W
Time taken = 2.40 s
Work done = Power × time taken
Work done = 4950 W × 2.40 s
Work done = 11,900 J
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