Answer:
(B)
Explanation:
Time = change of velocity ÷ acceleration
= (6-0) ÷ 5
= 1.2
1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
To calculate the time it takes for an object to change its velocity from 0 to 6 m/s, we can use the formula:
time = change in velocity / acceleration
Given that the change in velocity (Δv) is 6 m/s and the acceleration (a) is 5 m/s², we can plug these values into the formula:
time = 6 m/s / 5 m/s²
time = 1.2 seconds
Therefore, 1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
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Can objects (aside from light/photons) travel the same speed or faster than the speed of light and still have Special Relativity work? Why or why not?
Answer:
who can apply for the award
Burns are a particular safety hazard when handling
Answer:
flammable liquids
Explanation:when it catches on fire your going to get burned.hope this helps.
What is the gravitational potential energy of a 2100 kg car at the top of an 18 m parking garage?
[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Gravitational Potential of an object with mass m, and height of h metre is :
[tex]\qquad \sf \dashrightarrow \:mgh[/tex]
Now, if the mass of object is 2100 kg, height of 18 m is placed in there, then it's gravitational potential energy is ~
[tex]\qquad \sf \dashrightarrow \:2100 \times 10 \times 18[/tex]
[tex]\qquad \sf \dashrightarrow \:378000 \: \: joules[/tex]
[tex]\qquad \sf \dashrightarrow \:3.78 \times 10 {}^{5} \: \: joules[/tex]
1 If electromagnetic radiation acted like particles in the double-slit experiment, what would be observed?
a The screen would remain dark because no radiation would reach the screen.
b One bright band would appear in the center of the screen.
c A series of light and dark bands would appear on the screen.
d Two bright bands would appear on the screen in line with the slits.
2 Which statement about the interference behavior of electromagnetic radiation seen in the double-slit test experiment is true?
a Waves that make up the radiation collide with each other so that they add together or cancel each other out.
b Particles that make up the radiation collide with each other and scatter randomly.
c Particles that make up the radiation collide with each other so that they add together or cancel each other out.
d Waves that make up the radiation do not interact with each other.
3 Which statement about the observed results of the double-slit experiment is true?
a Waves that are out of phase constructively interfere to create bright bands.
b Waves that are in phase destructively interfere to create bright bands.
c Waves that are out of phase constructively interfere to create bright bands.
d Waves that are in phase constructively interfere to create bright bands.
4 Which statement about the observed results of the double-slit experiment is true?
a Waves that are in phase constructively interfere to form dark bands.
b Waves that are out of phase constructively interfere to form dark bands.
c Waves that are in phase destructively interfere to form dark bands.
d Waves that are out of phase destructively interfere to form dark bands.
5 A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. What will be true about the new interference pattern seen on the screen compared to the original interference pattern?
a The spacing between the dark fringes will increase.
b The spacing between the bright fringes will increase.
c The spacing between the bright fringes will decrease.
d The spacing between the dark fringes will remain the same.
6 Consider the two-slit interference experiment. Electromagnetic radiation passes through the two slits that are a distance of 0.0170 nm apart. A fourth-order bright fringe forms at an angle of 8.0 degrees relative to the incident beam. What is the wavelength of the light?
a 789 nm
b 420 nm
c 581 nm
d 591 nm
Answer:
1. Two bright bands would appear on the screen in line with the slits.
2. Waves that make up the radiation collide with each other so that they add together or cancel each other out.
3. Waves that are in phase constructively interfere to create bright bands.
4. Waves that are out of phase destructively interfere to form dark bands.
5. The spacing between the bright fringes will decrease.
6. 581 nm
Explanation:
A proton in the nucleus of an atom has an electrical charge of:
neutral
-
+
zero
Answer:
proton is positively charged changechar
Explanation:
D. Calculate the electric force (F.) between an electron and
proton that are 5.29 x 10-11 meters apart.
Answer:
See below
Explanation:
F = C q1 q2 /r^2
8.988 x 10^9 * (1.60217 x 10^-19)^2 / ( 5.29 x 10^-11)^2 =
.00000008244 N
82.4 nano N
Select all the correct answers.
A worker is holding a filled gas cylinder still. Which two sentences are true about the energy of the filled gas cylinder?
it has no energy because it's being held still
It has gravitational potential energy because of its height.
its atoms and molecules have thermal energy
It has motion energy because it will fall if let go
its kinetic energy is being converted to potential energy
ghts reserved
it has no energy because its being held still
it has motion energy because it will fall if let go
A car of mass 1200kg falls a vertical distance of 24m starting from rest what is the workdone by the force of gravity on the car?Use the work energy theorem to find the final velocity of the car just before it hits the water(Treat the as a point like object)
Answer:
PE (relative to water) = M g h
PE = 1200 kg * 9.8 m/s^2 * 24 m = 2.82E5 Joules
KE = PE when vehicle strikes water
1/2 M V^2 = 2.82E5
V = (2.82E5 * 2 / 1200)^1/2 = 21.7 m/s
Check:
M g h = 1/2 M V^2
V = (2 g h)^1/2 = (2 * 9.8 * 24)^1/2 = 21.7 m/s
What is the wavelength of the wave pictured above?
Answer:
its counting by 4 the multi 4,8,12,16
An output gear has 10 teeth and an input gear has 40 teeth. What is the mechanical advantage of this gear combination?
Answer:
4
Explanation:
We know that :
Mechanical Advantage = Input / OutputSolving
MA = Input / OutputMA = 40 / 10MA = 4BRAINLIEST
1. What does the percent difference between GPE and KE tell you about the efficiency of energy transformations in this experiment?
2. Use a computer or graphing calculator to graph the percent difference in data table 1 versus mass, and the percent difference in data table 2 versus change in height (Δh). Based on your results that you see plotted for the two rounds of experiments, how would you adjust the mass and height so that the design of the system leads to GPE and KE values that are as close as possible? Explain why you think this design would achieve the desired result.
btw are my datas in the graph right? i added the gpe ke and %diff
The percent difference between GPE and KE in the experiment illustrates that the higher the percentage, the more efficient the energy transfer.
What is energy transformation?It should be noted that energy transformation simply means the process of changing energy from one form to another form.
In this case, the percent difference between GPE and KE in the experiment illustrates that the higher the percentage, the more efficient the energy transfer.
Also, the mass and height should be adjusted in such a way that the design of the system leads to GPE and KE values that are as close as possible.
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The experiment's percent difference between GPE and KE shows that the efficiency of energy transfer increases with percentage.
Thus, It should be emphasized that the term "energy transformation" simply refers to the act of altering energy's form.
The experiment's percent difference between GPE and KE in this instance demonstrates that the more significant the percentage, the more effectively energy is transferred.
The experiment's percent difference between GPE and KE in this instance demonstrates that the more significant the percentage, the more effectively energy is transferred.
Thus, Energy is capable of changing its forms. For instance, your body stores chemical energy from the food you eat until you can use it as kinetic energy when working or playing.
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A uniformly charged insulating rod is bent into the shape of a semicircle of radius R = 5 cm. If the rod has a total charge of Q = 3.10-9C, find the magnitude and direction of the electric field at O, the center of the circle.
Hi there!
We can begin by using Coulomb's Law:
[tex]E = \frac{kq}{r^2}[/tex]
k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)
E = Electric field strength (N/C)
r = distance from point (m)
q = charge (C)
Since this is a continuous charge, we must use calculus.
We can express this as the following:
[tex]q = \lambda L[/tex]
λ = Linear charge density (C/m)
L = Length of rod (m)
Now, since this is an arc, L = s (arc length). Additionally, we must find the differential elements of each:
[tex]dq = \lambda ds\\\\dq = \lambda rd\theta[/tex]
Our new equation is:
[tex]dE = \frac{kdq}{r^2}\\\\dE = \frac{k\lambda rd\theta}{r^2}[/tex]
However, we will only take the cosine component of the electric field since the vertical components will cancel out. (Electric fields are a vector). Therefore:
[tex]dE = \frac{k\lambda rd\theta}{r^2}cos\theta\\\\dE = \frac{k\lambda}{r}cos\theta d\theta[/tex]
Integrate. For a semicircle, the bounds will be from -π/2 to π/2.
[tex]E = \frac{k\lambda}{r}\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}} {cos\theta} \, d\theta\\\\E = \frac{k\lambda}{r}sin\theta\left \|{\frac{\pi}{2}} \atop {-\frac{\pi}{2}}} \right. \\\\E = \frac{k\lambda}{r}(1 - (-1)) = \frac{2k\lambda}{r}[/tex]
We need to solve for λ, which is Q/ L:
[tex]\lambda = \frac{3.10 \times 10^{-9} C}{\pi (0.05)} = 1.9735 \times 10^{-8} \frac{C}{m}[/tex]
Now, plug and solve for the electric field strength:
[tex]E = \frac{2(8.99\times 10^9)(1.9735\times 10^{-8})}{0.05} = \boxed{7096.783 \frac{N}{C}}[/tex]
**A diagram was not provided, but if the hemisphere's focus was to the right, the electric field would be to the right, and etcetera.
A turntable is switched from 11.3 rad/s to 1.9 rad/s, and the platter goes through an angle of 17.95 radian in reaching the new angular speed. What is the angular acceleration of the platter?
Hi there!
We can use the angular equivalent of the following kinematic equation:
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]
ωf = final angular velocity (1.9 rad/s)
ωi = initial angular velocity (11.3 rad/s)
α = angular acceleration (? rad/s²)
θ = angular displacement (17.95 rad)
We can rearrange the equation to solve for angular acceleration.
[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta\\\\\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}[/tex]
Plug in the given values and solve.
[tex]\alpha = \frac{1.9^2 - 11.3^2}{2(17.95)} = \boxed{3.456 \frac{rad}{s}}[/tex]
What are your ideas about how humans can reduce the amount of carbon dioxide and methane they add to the atmosphere?
Answer:
Find the heavy processes of CO2 and methane.
Explanation:
Step #1 Find a very reliable energy source.
Step #2 Find the heavy processes of CO2 and methane like in many factories and places with high carbon emissions.
Step #3 Shut down each of the heavy processes even 10 per day would do a lot.
Step #4 Use the renewable energy source as a replacement for the attempts to generate energy using carbon emissions.
Astronomers often detect stars that are rotating extremely rapidly, known as neutron stars. These stars are believed to have formed in the inner core of a larger star that collapsed, due to its own gravitation, to a star of a very small radius and very high density. Before collapse suppose the core of such star is the size of our Sun (R=7 x 10km) with mass 2.0 times as great as the Sun, and is rotating at a speed of 1 revolution every 10 days. If it were to undergo gravitational collapse to a neutron star of radius 10 km, what would its rotational speed be? Assume the star is a uniform solid sphere at all times. (1 = MR2) a
Based on law of conservation of angular momentum, the rotational speed of the star is equal to 6,000 rev/s.
Given the following data:
Radius of Sun = 7 × 10⁵ km.Mass of star = 2 Mass of Sun (M = 2M).Radius of star = 10 km.Time = 10 days.How to calculate the rotational speed.First of all, we would determine the initial angular speed of the neutron star as follows:
[tex]\omega_i = \frac{1 \;Rev}{10 \;days} \\\\\omega_i = \frac{1 \;Rev}{10 \times 24 \times 60 \times 60}\\\\\omega_i = 1.157 \times 10^{-6}\;rev/s[/tex]
Mathematically, the moment of inertia of a uniform solid sphere is given by this formula:
[tex]I=\frac{2}{5} mr^2[/tex]
Where:
I is the moment of inertia.m is the mass.r is the radius.In order to determine the rotational speed of this neutron star, we would apply the law of conservation of angular momentum:
[tex]L_1 = L_2\\\\I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} m_1r_1^2 \omega_1 = \frac{2}{5} m_2r_2^2\omega_2\\\\m_1r_1^2 \omega_1 = m_2r_2^2\omega_2\\\\\omega_2 =\frac{r_1^2 \omega_1}{r_2^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ (10 )^2}\\\\\\\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ 10 0}\\\\\omega_2 =\frac{566,930}{100}[/tex]
Final angular speed = 5,669 ≈ 6,000 rev/s.
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A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown.
The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
Angular Speed of the pulley
The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;
K.E = P.E
[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]
[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]
Substitute the given parameters and solve for the angular speed;
[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]
Linear speed of the blockThe linear speed of the block after travelling 0.7 m;
v = ωR₂
v = 35.39 x 0.03
v = 1.1 m/s
Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
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how to make sure a snow leopard does not escape?
build a fence.....................
explain the fleming left-hand rule with the diagram and what will the direction of the induced current in the figure if the magnet is going towards the coil.
Friction between our feet and the surface we walk on is desirable. True False
Answer:
True
Explanation:
(a) Calculate the capacitance of a parallel-plate capacitor whose plates are 30. cm x 5.0 cm and are separated by a 3.0-mm air gap. (b) What is the charge on each plate if a 6-V battery is connected across the two plates? (c) What is the electric field between the plates? (d) Estimate the area of the plates needed to achieve a capacitance of 5F, given the same air gap d.
The answers to your questions are as written below
A) capacitance of a parallel plate capacitor = 442.5 * 10⁻¹³ F
B) The charge on each plate is = 2655 * 10⁻¹³ C
C) The electric field between the plates is ; 2 * 10³ N/C
D) The estimated area of the plates is ; 1.695 * 10⁹ m²
A) Calculate the capacitance of a parallel plate capacitor
Plate dimension = 30 cm * 5 cm
Air gap ( d ) = 3 mm = 3 * 10⁻³ m
First step : calculate the area of the plate
Area of plate = 30 * 10⁻²m * 5 * 10⁻² m
= 150 * 10⁻⁴ m²
Next step : determine the capacitance
Capacitance = ε₀A / d
= ( 8.85 * 10⁻¹² * 150 * 10⁻⁴ ) / 3 * 10⁻³
= 442.5 * 10⁻¹³ F
B) Calculate the charge on each plate if a 6-V battery is connectedcharge on each plate can be calculated with the formula below
Q = CV
= 442.5 * 10⁻¹³ * 6
= 2655 * 10⁻¹³ C
C) Determine the electric field between the platesElectric field between plates ( E ) = V / d
= 6 / 3 * 10⁻³
= 2 * 10³ N/C
D) Estimate the area of the platesRequired capacitance = 5 F
Air gap ( d ) = 3 * 10⁻³ m
applying the equation below
C = ε₀A / d -----
5 = ( 8.85 * 10⁻¹² * A ) / 3 * 10⁻³
Therefore A :
A = ( 5 * 3 * 10⁻³ ) / 8.85 * 10⁻¹²
= 15 * 10⁻³ / 8.85 * 10⁻¹²
= 1.695 * 10⁹ m²
Hence we can conclude that the answers to your questions are as written above.
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What are some INDIRECT applications of centripetal force in real life?
Answer:
The circular turning of roads
Explanation:
please mark brainliest
Answer:
Explanation:
The circular turning of roads
Driving on Curves
above
3. DIRECTIONS: Read the following two statements carefully and choose the correct options.
Statement 1: In a room containing air, heat can go from one place to another by radiation only.
Statement 2: In convection, heat is transferred from one place to other by actual motion of the
molecules in the heated material.
A) Statement (1) is correct while statement (2) is incorrect.
B) Statement (2) is correct while statement (1) is incorrect.
C) Both statements are correct
D) Both statements are incorrect.
A class of 10 students taking an exam has a power output per student of about 200W. Assume the initial temperature of the room is 20 C and that its dimensions are 6.0 m by 15.0 m by 3.0 m. What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source? The specific heat is 837 J/kg C, and its density is about 1.3 x 10^-3 g/cm^3
For a class of 10 students taking an exam has a power output per student of about 200W, the temperature is mathematically given as
dT=25C
What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source?Generally, the equation for the temperature is mathematically given as
dT=Q/mc
Where
Q=Pt
Q=2000*3600
Q=7200000J
And m
m=pv
m=1.3*270
m=351kg
Therefore
dT=7200000J/351*837
dT=25C
In conclusion, The temperature of the room at the end of 1.0 h
dT=25C
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What is refractive index?
[tex] \huge \mathbb \red{HEY \: THERE ♡}[/tex]
Refractive Index:The ratio of the sine of angle of incidence to the sine of angle of refraction in case of lens for light of a given colour and given pair of media is constant. Note** It is also called Snell's Law of Refraction. [tex] \mathsf \orange{\frac{sine \: i}{sine \: r} = constant}[/tex][tex] \huge \mathbb\pink{HOPE \: IT \: HELPS}[/tex]
A wave is sent back and forth along a rope 4 m long with a mass of 0.6 kg by exerting a force a force of 30 N. Calculate the linear mass density of the rope (in kg/m).
The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.
Linear mass density of the ropeThe linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.
The linear mass density of the rope is calculated as follows;
μ = m/L
μ = 0.6 kg / 4 m
μ = 0.15 kg/m
Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.
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give an example of the 4 steps of the scientific method
Answer:
1) asking a question about something you observe, 2) doing background research to learn what is already known about the topic, 3) constructing a hypothesis, 4) experimenting to test the hypothesis
Answer:
1) make an observation that describes a problem - see something you can fix or improve and try to describe the problem like which type of vegetables does rabbit like
2) create a hypothesis - What do you think will happen - If the rabbit eat the lettuce than .........
3) test the hypothesis - Do your experiment with the variables and follow the procedures
4) draw conclusions and refine the hypothesis - see if your hypothesis was correct - In conclusion my hypothesis was not correct because.......
If a person is putting in 500 J of work in a 25 second period, how much power are they producing?
A body is piloted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. Find the amount of force about the pivot .
Question :-
A Body is Pivoted at a Point. A Force of 10 N is Applied at a Distance of 30 cm from the Pivot. Find the Amount of Force about the Pivot.Answer :-
Amount of Force is 3 Nm .Explanation :-
As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .
For calculating the Force , we will use the Formula :-
[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]
Therefore , by Substituting the given values in the above Formula :-
[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]
[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]
[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]
Hence :-
Amount of Force = 0.3 Nm .[tex] \underline {\rule {185pt}{4pt}} [/tex]
A rocket weighing 220,000 N is taking off from Earth with a total thrust of
500,000 N at an angle of 20 degrees, as shown in the image below. What is
the approximate vertical component of the net force that is moving the rocket
away from Earth?
20°
Vertical
Component
of Net Force
Thrust
500,000 N
Weight
220,000 N
The vertical component of the net force that is moving the rocket away from Earth is determined as 171,010.1 N.
Vertical component of the lift force
The vertical component of the net force that is moving the rocket away from Earth is determined as follows;
Fy = Fsinθ
where;
F is applied forceθ is the angle of inclinationFy = 500,000 x sin(20)
Fy = 171,010.1 N
Thus, the vertical component of the net force that is moving the rocket away from Earth is determined as 171,010.1 N.
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Answer:
250,000
Explanation:
A garage hoist lifts a truck up 2 m above the ground in 15seconds.Find the power delivered to the truck (given 1000kg as the mass of the truck and gas 9.81m/s
The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.
What is power?
Power can be deifned as the rate at which work is done.
To calculate the power delivered to the truck, we use the formula below.
Formula:
P = mgh/t.............. Equation 1Where:
P = Power delivered to the truckm = Mass of the truckh = Heightg = Acceleration due to gravityt = timeFrom the question,
Given:
m = 1000 kgh = 2 mt = 15 secondsg = 9.81 m/s²Subsitute these values into equation 1
P = (1000×2×9.81)/15P = 19620/15P = 1308 WHence, The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.
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