ANSWER
3) elastic energy
EXPLANATION
When the gymnast reaches the trampolin, both the gymnast and the trampolin move down because of the elasticity of the trampolin, until the trampolin is at its maximum point of which it can stretch. At this point the gravitational energy of the gymnast is zero because of zero height and its kinetic energy is zero too, because of zero energy. Therefore all the energy is stored as elastic energy.
A syringe is filled with air of volume 10 cm at 100 kPa and then sealed. Itis compressed so that its volume is reduced by 45 %. Find the pressureinside it.
Answer:
[tex]P_2=222.22kPa[/tex]Explanation: By using Gas Law, the new pressure can be calculated as follows:
Gas law states:
[tex](P_1V_1=P_2V_2)\Rightarrow(1)[/tex]Using (1), and Identifying the knowns and unknowns, and plugging in (1) we get the following results:
[tex]\begin{gathered} V_1=10cm^3=1\cdot10^{-5}m^3 \\ P_1=100\text{kPa}=100000Pa \\ V_2=(0.45)\cdot V_1=(0.45)\cdot(1\cdot10^{-5}m^3)=4.5\cdot10^{-6}m^3 \\ V_2=4.5\cdot10^{-6}m^3 \end{gathered}[/tex]Finally, the New pressure is calculated as follows:
[tex]\begin{gathered} (P_1V_1=P_2V_2)\Rightarrow(1) \\ P_2=\frac{P_1V_1}{V_2} \\ \therefore\Rightarrow \\ P_2=\frac{P_1V_1}{V_2} \\ P_2=\frac{(100000Pa)\cdot(1\cdot10^{-5}m^3)}{(4.5\cdot10^{-6}m^3)} \\ P_2=2.22\cdot10^5Pa \\ P_2=222.22kPa \end{gathered}[/tex]An electron of an atom jumps down from one energy level to another and emits a photon that contains 8.83 x 10^-18joule of energy. What was the frequency of the emitted electromagnetic energy?(a) 1.33 x 10^16 Hz(b) 6.74 x 10^15 Hz(c) 3.27 x 10^-15 Hz(d) 7.5 x 10^-17 Hz
In order to determine the frequency of the emmited electromagnetic energy, use the following formula:
[tex]E=hf[/tex]where:
E: energy of the photon = 8.83*10^(-18) J
h: Planck's constant = 6.63*10^(-34)J*s
f: frequency of the photon = ?
solve the previous equation for f, and replace the values of the other parameters, as follow:
[tex]f=\frac{E}{h}=\frac{8.83\cdot10^{-18}J}{6.63\cdot10^{-34}J\cdot s}=1.33\cdot10^{16}Hz[/tex]I was able to find part A. But need help with part B & C
To find the magnitude, we have to use the following formula
[tex]|d|=\sqrt[]{x^2+y^2}[/tex]Where x = 1 and y = 3, let's use these values to find the magnitude.
[tex]\begin{gathered} |d|=\sqrt[]{1^2+3^2}=\sqrt[]{1+9}=\sqrt[]{10} \\ |d|\approx3.2 \end{gathered}[/tex]The magnitude of the resultant vector is 3.2 meters, approximately.To find the direction, we have to use the following formula
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{y}{x})=\tan ^{-1}(\frac{3}{1}) \\ \theta\approx71.6 \end{gathered}[/tex]The direction of the resultant vector is 71.6°.A. If a particular ligament has an effective spring constant of 167 N/mm as it is stretched, what is the tension in this ligament when it is stretched by 0.720 cm?B. If a particular ligament has an effective spring constant of 167 N/mm as it is stretched, what is the elastic energy stored in the ligament when stretched by 0.720 cm?
Part (A)
The tension in the ligament is equal to the force acting on ligament. The expression for the tension in ligament is,
[tex]F=kx[/tex]Substitute the given values,
[tex]\begin{gathered} F=(167\text{ N/mm)(}\frac{1\text{ mm}}{10^{-3}\text{ m}})\text{(0.720 cm)(}\frac{1\text{ m}}{100\text{ cm}}) \\ =1202.4\text{ N} \end{gathered}[/tex]Thus, the tension in the ligament is 1202.4 N.
Part (B)
The elastic energy stored in the ligament can be given as,
[tex]U=\frac{1}{2}kx^2[/tex]Substitute the known values,
[tex]\begin{gathered} U=\frac{1}{2}(167N/mm)(\frac{1\text{ mm}}{10^{-3}\text{ m}})(0.720cm)^2(\frac{1\text{ m}}{100\text{ cm}})^2(\frac{1\text{ J}}{1\text{ Nm}}) \\ \approx4.33\text{ J} \end{gathered}[/tex]Thus, the elastic potential stored in the ligament is 4.33 J.
In the diagram, q₁ = +4.60 x 10-6 C,92 +3.75 x 10-6 C, and 93 = -8.30 x 10-5 C.Find the magnitude of the net force on 92.0.350 m9192↑0.155 m93(Make sure you know the direction of each force! Oppositesattract, similar repel.)
Given:
• q1 = +4.60 x 10⁻⁶ C
,• q2 = +3.75 x 10⁻⁶ C
,• q3 = -8.30 x 10⁻⁵ C.
,• d12 = 0.350 m
,• d23 = 0.155 m
Let's find the magnitude of the net force on q2.
• First find the force acting on charge 1 and 2:
[tex]\begin{gathered} F_{12}=-\frac{kq_1q_2}{(d_{12})^2} \\ \\ F_{12}=-\frac{9\times10^9*4.60\operatorname{\times}10^{-6}*3.75\operatorname{\times}10^{-6}}{0.350^2} \\ \\ F_{12}=-1.27\text{ N} \end{gathered}[/tex]• Let's find the force acting on charge 2 with respect to change 3:
[tex]\begin{gathered} F_{23}=\frac{kq_1q_2}{(d_{23})^2} \\ \\ F_{23}=\frac{9\times10^9*3.75\operatorname{\times}10^{-6}*8.30\operatorname{\times}10^{-5}}{0.155^2} \\ \\ F_{23}=116.60\text{ N} \end{gathered}[/tex]• Now, for the magnitude of the net force on q2, we have:
[tex]\begin{gathered} F_{net}=\sqrt{(F_{12})^2+(F_{23})^2} \\ \\ F_{net}=\sqrt{(-1.27)^2+(116.60)^2} \\ \\ F_{net}=\sqrt{1.6129+13595.56} \\ \\ F_{net}=\sqrt{13597.1729} \\ \\ F_{nrt}=116.60\text{ N} \end{gathered}[/tex]Therefore, the magnitude of the force on q2 is 116.6 N.
• ANSWER:
116.6 N
A green ball (ball 1) of mass M, collides with an orange (ball 2) of mass 1.26 m. The initial speed of the green ball is 5.4 m/s the final speed of the green ball is 2.6 m/s and theta = 36.9° A find the magnitude of the final speed of the orange ball? B. what is a direction of the final speed of the orange ball?
We know that in a collision the momentum is conserved, that is:
[tex]\vec{p}_0=\vec{p}_f[/tex]Since this is vector equation we can divide it in two scalar equations, one for x and for y. Then we have:
[tex]\begin{gathered} p_{0x}=p_{fx} \\ \text{and} \\ p_{0y}=p_{fy} \end{gathered}[/tex]Then we have for the x direction:
[tex]\begin{gathered} 5.4m=m(2.6)\cos 36.9+1.26mv_o\cos \phi \\ 5.4=2.6\cos 36.9+1.26v_o\cos \phi \end{gathered}[/tex]and for the y direction:
[tex]\begin{gathered} 0=-m2.6\sin 36.9+1.6mv_o\sin \phi \\ 2.6\sin 36.9=1.6v_o\sin \phi \end{gathered}[/tex]Hence, we have the system of equations:
[tex]\begin{gathered} 5.4=2.6\cos 36.9+1.26v_o\cos \phi \\ 2.6\sin 36.9=1.6v_o\sin \phi \end{gathered}[/tex]From the second equation we have:
[tex]v_o=\frac{2.6\sin 36.9}{1.6\sin \phi}[/tex]Plugging this in the first equation:
[tex]\begin{gathered} 5.4=2.6\cos 36.9+1.26(\frac{2.6\sin36.9}{1.6\sin\phi})\cos \phi \\ 1.26(\frac{2.6\sin36.9}{1.6})\tan \phi=5.4-2.6\cos 36.9 \\ \tan \phi=\frac{1.6(5.4-2.6\cos 36.9)}{(1.26)(2.6\sin 36.9)} \\ \phi=\tan ^{-1}(\frac{1.6(5.4-2.6\cos36.9)}{(1.26)(2.6\sin36.9)}) \\ \phi=69.69 \end{gathered}[/tex]Now that we know the value of the angle we plug it in the expression for the velocity, then we have:
[tex]\begin{gathered} v_o=\frac{2.6\sin 36.9}{1.6\sin 69.69} \\ v_0=1.04 \end{gathered}[/tex]Therefore, the magnitude of the final speed of the orange ball is 1.04 m/s and the direction is 69.69°
A 9300kg car traveling at a velocity of 5m/s strikes a second boxcar that weighs 5000kg at rest.At what speed will the second boxcar move?
A 12-kg rotating disk of radius 0.75 m has an angular momentum of 0.85 kg.m^2/s. Calculate the angular speed of the disk.
The formula for determining angular momentum is expressed as
L = Iw
where
L = angular momentum
I = moment of inertia
w = angular speed
Also, the formula for finding moment of inertia is expressed as
I = 1/2MR^2
where
M = mass of body
R = radius of body
From the information given,
M = 12
R = 0.75
Thus,
I = 1/2 x 12 x 0.75^2 = 3.375
L = 0.85
From the first equation,
w = L/l
w = 0.85/3.375
w = 0.2519 rad/s
The current through a light bulb connected across the terminals of a 120 V outlet is 0.50 A. At what rate does the bulb convert electric energy to light?
The rate at which the bulb converts electric energy to light = 60 J/s
Explanation:The voltage across the terminals, v = 120 v
The current, I = 0.50 A
The rate at which the bulb convert electric energy to light is the power generated by the bulb
Power is given by the formula:
P = IV
Substitute I = 0.50A and v = 120 v into P = IV
P = 0.50 x 120
P = 60 J/s
The rate at which the bulb converts electric energy to light = 60 J/s
What is the distance between the Moon and the Earth if the mass of the moon is 7.34 x 1022 kg, the mass of the Earth is 5.98x1024 kg and the force of attraction between the two is 2.00 x 1020 N?
Answer:
[tex]r=3.83\times10^8m[/tex]Explanation: We need to find the distance between the moon and earth provided the force between them and their masses. the equation used to solve this problem is as follows:
[tex]\begin{gathered} F=\frac{m_1m_2}{r^2}G\Rightarrow(1) \\ G=6.674\times10^{-11}m^3kg^{-1}s^{-2}_{} \end{gathered}[/tex]Using the known values, and plugging in the equation (1) results in:
[tex]\begin{gathered} m_1=7.34\times10^{22}\operatorname{kg} \\ m_2=5.98\times10^{24}\operatorname{kg} \\ F=2.00\times10^{20}N \\ \end{gathered}[/tex]The final step is as follows:
[tex]\begin{gathered} (2.00\times10^{20}N)=\frac{(7.34\times10^{22}\operatorname{kg})\cdot(5.98\times10^{24}\operatorname{kg})}{r^2}\cdot(6.674\times10^{-11}m^3kg^{-1}s^{-2}_{}) \\ \text{ Rearranging} \\ r^2=\frac{(7.34\times10^{22}\operatorname{kg})\cdot(5.98\times10^{24}\operatorname{kg})\cdot(6.674\times10^{-11}m^3kg^{-1}s^{-2}_{})}{(2.00\times10^{20}N)} \\ r^2=1.4647\times10^{17}m^2 \\ r=\sqrt[]{1.4647\times10^{17}m^2}=3.83\times10^8m \\ r=3.83\times10^8m \end{gathered}[/tex]Therefore the distance between the moon and the earth is:
[tex]r=3.83\times10^8m[/tex]a radioactive substance, fermium-253, has a half-life of three days. how long will it take for this isotope to decay to one-eighth of its original amount?
Radioactive decay is characterized by the formula
[tex]N=(\frac{1}{2})^{\frac{t}{t_h}}N_0[/tex]Where N is the amount of the element remaining,
[tex]N_0\text{ is initial amount of the element}[/tex]t is the time taken to reduce the amount of the element to N and th is the half life of the element
In this case
[tex]N=\frac{N_0}{8}[/tex]And
[tex]th=3days=3\times24\times60\times60=2.6\times10^5\text{ s}[/tex]We need to find t
So we substitute the known values in the above equation
[tex]\frac{N_0}{8}=N_0\times(\frac{1}{2})^{\frac{t}{2.6\times10^5}}[/tex]Which is
[tex]\frac{1}{8}=(\frac{1}{2})^{\frac{t}{2.6\times10^5}}[/tex]Taking the log on both sides,
[tex]\log (\frac{1}{8})=-\log (8)=\frac{t}{2.6\times10^5}\log (\frac{1}{2})[/tex]Simplifying,
[tex]-0.9=\frac{-0.3\times t}{2.6\times10^5}[/tex]On rearranging and further simplifying we get,
[tex]t=1.15\times10^{-5}\text{ s}[/tex]Thus the time taken for fermium 253 to reduce to one-eight of its initial amount is
[tex]1.15\times10^{-5}\text{ s}[/tex]As a steel beam is winched up to the top floor of a new building, the mechanical energy is being converted into what?Light energyKinetic energyThermal energyPotential energy
Given that a steel beam is winched up to the top floor of a new buiding, let's determine the new foem of energy.
Mechanical energy is the sum of kinetic energy and potential energy.
The steel beam currently has a mechanical energy. Since the steel beam is now at the top of a new building, the mechanical energy is being converted to potential energy.
Potential energy is the energy held by an object due to its position relative to other objects. It is the energy stored from an object.
Since the steel beam is now at the top of the building, it has the potential to fall.
Therefore, the mechanical energy is now being converted to potential energy.
ANSWER:
Potential energy
A student is conducting an experiment todetermine what will happen to the gravitationforce of two boxes, with different masses, if theboxes are pushed together. One box has a-mass of 100 g and the other box has a mass of 150 g.What will happen if these two boxes were movedcoser together?A gravitational force would remain constantB. gravitational force would not be createdC. gravitational force would decreaseD. gravitational force would Increase
as the distance between the objects is decreasing
the force will be incresed
as the gravitational force is inversely proportional to the square of the distance between the objects.
option D is the correct answer
while on a luge run in the olympics the sled (mass=157 kg) rounds a horizontal curve of raidus 5.8 m with a speed of 18.5 m/s assuming there is no friction what is the normal force exerted by the seat on the rider if he has a mass of 83 kg?
The normal force is the centripetal force which is given by:
[tex]F=\frac{mv^2}{r}[/tex]plugging the values given we have:
[tex]\begin{gathered} F=\frac{(157)(18.5)^2}{5.8} \\ F=9264.35 \end{gathered}[/tex]Therefore, the normal force is 9264.35 N
Help with homework not graded I learn best by watching you do problem Option D = 3.19 s
Given data:
* The height of the stone is h = 50 m.
* The initial horizontal speed of the stone is v = 4 m/s.
Solution:
By the kinematics equation along the vertical direction, the time taken by the stone to reach the ground is,
[tex]h=ut+\frac{1}{2}gt^2[/tex]where u is the initial vertical speed of the stone,
The initial vertical velocity of the stone is zero, thus,
[tex]h=\frac{1}{2}gt^2[/tex]where g is the acceleration due to gravity,
Substituting the known values,
[tex]\begin{gathered} 50=\frac{1}{2}\times9.8\times t^2 \\ t^2=\frac{2\times50}{9.8} \\ t^2=10.2 \\ t=\pm3.19\text{ s} \end{gathered}[/tex]Neglecting the negative value of time,
Thus, the time taken by the stone to reach the ground is 3.19 s.
Hence, option D is the correct answer.
Question 7 please. Asking for what horizontal force is necessary to start the box into motion
Given,
The weight of the box, W=300 N
The coefficient of static friction, μ=0.30
(a) In order to move the box, the force applied must be greater than the static friction that exists between the box and the floor. The static friction is given by,
[tex]\begin{gathered} f=N\mu \\ =W\mu \end{gathered}[/tex]Where N is the normal force which is the same as the weight of the box.
On substituting the known values,
[tex]\begin{gathered} f=300\times0.3 \\ =90\text{ N} \end{gathered}[/tex]Thus the horizontal force required to start the box into motion is 90 N.
(b) The applied force is F= 50.0 N
The coefficient of kinetic friction μ=0.3
Thus, the kinetic friction offered by the floor is,
[tex]f_k=W\mu[/tex]On substituting the known values,
[tex]f_k=300\times0.3=90\text{ N}[/tex]Thus the net force acting on the body will be,
[tex]\begin{gathered} F_n=ma \\ =F-f_k\text{ }\rightarrow(i) \end{gathered}[/tex]Where m is the mass of the box and a is the acceleration of the box
The mass is given by
[tex]m=\frac{W}{g}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} m=\frac{300}{9.8} \\ =30.6\text{ kg} \end{gathered}[/tex]Thus, from equation (i) the acceleration is calculated as
[tex]\begin{gathered} a=\frac{F-f_k}{m} \\ =\frac{50-90}{30.6} \\ =-1.31m/s^2 \end{gathered}[/tex]Thus the acceleration of the box will be -1.31 m/s². The negative sign indicates that it is a deceleration. Thus, the box will eventually slow down and will come to rest.
A stretched string vibrates with a fundamental frequency of 200 Hz. What is the frequency of the second overtone?
Given that the fundamental frequency of the string is f = 200 Hz
We have to find the second overtone frequency.
The second overtone frequency can be calculated by the formula
[tex]f_2=3f[/tex]Substituting the value, the frequency of the second overtone is
[tex]\begin{gathered} f_2=3\times200 \\ =600\text{ Hz} \end{gathered}[/tex]Thus, the second overtone frequency is 600 Hz.
Determine the mass of the moon if you are given the following data (Given: Mass of Earth, Force acting on the Moon due to Earth, and the distance between the centers of Earth and the Moon).
Answer:
Explanation:
If we call
M_E = mass of the earth
M_0 = mass of the moon
R = distance between the earth and the moon
Then the gravitational force between the earth and the moon is given by
[tex]F=G\frac{M_EM_O}{R^2}[/tex]Solving for M_O gives
[tex]\boxed{M_O=\frac{FR^2}{GM_E}\text{.}}[/tex]Now, when F = force acting on the Moon due to Earth, R = the distance between the centres of Earth and the Moon, and M_E = mass of the earth are put into the above equation, we can determine the M_O the mass of the moon.
Given a horizontal uncharged conducting rod. A positively charged rod touches the left end of the horizontal conducting rod. Which of these best describes what happens? the rod becomes positively charged on its left end and negatively charged on its right end the rod becomes negatively charged on its left end and positively charged on its right end the rod becomes positively charged all over its surface the rod becomes negatively charged all over its surface the rod remains uncharged
Given
A horizontal uncharged conducting rod. A positively charged rod touches the left end of the horizontal conducting rod.
To find
Which of these best describes what happens?
Explanation
A positive charge attracts the negative charge.
Since a positive charged rod is brought near the left end so the left end will be negatively charged.
The right end of the uncharged conducting rod will be of positive charge as like charges repeal.
Conclusion
The correct statement is
the rod becomes negatively charged on its left end and positively charged on its right end
which of the following statement is correct about the uses of nuclear energy?
Fossil fuels are produced from the dead plants and animals present under the ground. These fossil fuels are burned to release the energy.
The examples of the fossil fuels are coal, petroleum and natural gas.
These fossil fuels are present in limited amount, thus, the use of fossil fuel should be avoided.
Thus, option B is the correct answer.
How is charge transferred from one object to another? What dictates which way the charge moves?
A charge can transfer from one object to another in one of the three ways: Conduction, friction and polarization.
The potential difference causes the charges to move. The direction in which a charge moves depends on the sign of the charge. A negative charge moves from negative potential towards positive potential. A positive charge moves from positive potential towards negative potential.
Question 43?Find the resistance of the same type of wire with a diameter of 0.01 in and a length of 5 ft.
An AM radio listener is located 5.0 km from the radio station.Part AIf he is listening to the radio at the frequency of 660 kHz, how many wavelengths fit in the distance from the station to his house?
ANSWER
11 wavelengths
EXPLANATION
Given:
• The distance between the radio station and the listener, 5 km = 5000 m
,• The broadcasting frequency, f = 660 kHz = 660000 Hz
Unknown:
• The number of wavelengths that fit in the distance from the station to the listener's house
First, we have to find the wavelength,
[tex]\lambda=\frac{c}{f}[/tex]Where c is the speed of light in m/s and f is the frequency in Hz,
[tex]\lambda=\frac{3\cdot10^8m/s}{660000Hz}\approx454.55m[/tex]To find how many wavelengths fit in the distance from the station to the listener's house, we have to divide the distance between them by the wavelength of the wave,
[tex]n=\frac{5000m}{454.55m}\approx11[/tex]Hence, a total of 11 wavelengths fit in the distance from the station to his house.
Energy has the same MKS Unit as _________. *1 pointA:AccelerationB:WorkC:PowerD:Speed
Unit of acceleration is m/s^2
Unit of work is joule
Unit of power is watt
Unit of speed is m/s
Unit of energy is joule
So the correct option is work.
How was Lighting created?
It's created by electrical discharge as a result of the interaction between the clouds and the ground. In other words, a lightning is an 'electric current' that goes from cloud to the ground, the charges attract and create the phenomena.
What are the essential elements of a roller coaster and how is it related to energy
Some of the essential elements are:
Launch track: It is the section where the train is accelerated to its full speed in a matter of seconds.
Lift hill: It transports the roller coaster train to an elevated point.
Train: It is the vehicle which transports passengers around a roller coaster's circuit.
In roller coasters, the two forms of energy that are most important are gravitational potential energy and kinetic energy.
The gravitational potential energy is given by:
[tex]\begin{gathered} U=mgh \\ _{\text{ }}where\colon \\ m=_{\text{ }}mass_{\text{ }}of_{\text{ }}the_{\text{ }}train \\ g=_{\text{ }}gravitational_{\text{ }}acceleration \\ h=_{\text{ }}height \end{gathered}[/tex]And the kinetic energy is given by:
[tex]\begin{gathered} K=\frac{1}{2}mv^2 \\ _{\text{ }}where\colon \\ v=_{\text{ }}velocity_{\text{ }}of_{\text{ }}the_{\text{ }}train \end{gathered}[/tex]From these equations we can conclude that they are important since they influence on the mass, the speed and the height. Which are part of the two forms of energy.
Mac and Tosh are arguing in the cafeteria. Tosh says that if he clings the jello with a greater speed it will have a greater inertia. Mac argues that inertia does NOT depend upon speed, but rather upon mass. Who is correct?
ANSWER
Mac is correct
EXPLANATION
Inertia only depends upon the mass of the object, not its speed. The more mass, the more inertia.
A man throws a stone vertically down from a height with an initial velocity = -5 m / s . After 3 seconds , the stone reaches the ground ( at point A ) . What is the height h ?
Given:
The initial velocity of the stone, v₀=-5 m/s
The time it takes for the stone to reach the ground, t=3 s
To find:
The height.
Explanation:
From the equation of motion,
[tex]-h=v_0t-\frac{1}{2}gt^2[/tex]On substituting the known values,
[tex]\begin{gathered} -h=-5\times3-\frac{1}{2}\times9.8\times3^2 \\ \Rightarrow-h=-60 \\ \Rightarrow h=60\text{ m} \end{gathered}[/tex]Final answer:
The height from which the
2. Mass and Weight CalculationsHow much (in Newtons) does a 20.0 kg dog weigh?A student weighs 600 N. What is the mass of the student?
The weight of the dog = 196N
The mass of the student = 61.2 kg
Explanation:The weight of an object is the product of the mass and acceleration due to gravity
Let the weight be represented by W
Let the mass be represented by m
The acceleration due to gravity, g = 9.8 m/s^2
W = mg
How much (in Newtons) does a 20.0 kg dog weigh?
m = 20.0 kg
g = 9.8 m/s^2
W = mg
W = 20 x 9.8
W = 196 N
The weight of the dog = 196N
A student weighs 600 N. What is the mass of the student?
W = 600 N
g = 9.8 m/s^2
W = mg
600 = m x 9.8
m = 600 / 9.8
m = 61.2 kg
The mass of the student = 61.2 kg
What power (in kW) is supplied to the starter motor of a large truck that draws 240 A of current from a 22.5 V battery hookup?
Let's name some variables:
I: current; I = 240A
V: voltage; V = 22.5V
P: power; this is what we need to solve for
Given these variables, we can use the following equation to find P:
P = IV
Now, plugging in the known variables,
P = 240*22.5
P = 5400W = 5.4kW
