In order to compare the perimeters and areas, let's first find two adjacent sides of each rectangle.
From ABCD, let's calculate AB and BC:
A and B have the same y-coordinate, so the length is the difference in x-coordinate:
AB = -1 - (-5) = -1 + 5 = 4
B and C have the same x-coordinate, so the length is the difference in y-coordinate:
AB = -1 - (-4) = -1 + 4 = 3
Therefore the perimeter and area are:
[tex]\begin{gathered} P=4+3+4+3=14 \\ A=4\cdot3=12 \end{gathered}[/tex]Now, for rectangle WXYZ, let's use WX and XY:
W and X have the same y-coordinate, so the length is the difference in x-coordinate:
WX = 7 - 1 = 6
X and Y have the same x-coordinate, so the length is the difference in y-coordinate:
XY = 6 - (-2) = 6 + 2 = 8
So the perimeter and area are:
[tex]\begin{gathered} P=6+8+6+8=28 \\ A=6\cdot8=48 \end{gathered}[/tex]In order to check if the rectangles are similar, let's check the following relation:
[tex](\frac{P_1}{P_2})^2=\frac{A_1}{A_2}[/tex]So we have:
[tex]\begin{gathered} (\frac{14}{28})^2=\frac{12}{48} \\ (\frac{1}{2})^2=\frac{1}{4} \\ \frac{1}{4}=\frac{1}{4}\text{ (true)} \end{gathered}[/tex]Since the relation is true, so the rectangles are similar.
A mass of 10 kg, initially at rest on a horizontal frictionless surface, is acted upon by a horizontal force of 25 N. The speed of the mass after it has moved 5.0 m is:
We will have the following:
[tex]\begin{gathered} W\ast d=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \\ \\ \end{gathered}[/tex]Then, we will have that:
[tex]\begin{gathered} (25N)(5m)=\frac{1}{2}(10kg)v_f^2-\frac{1}{2}(10kg)(0m/s){}2 \\ \\ \Rightarrow125N\ast m=(5kg)v_f^2\Rightarrow v_f^2=\frac{25N\ast m}{kg} \\ \\ \Rightarrow v_f=5m/s \end{gathered}[/tex]So, the final velocity of the mass after it has moved 5 meters is 5m/s.
How do I solve this problem Hint: 1. Draw the forces -rear car: weight downwards, normal force upwards, tension right, drag leftmiddle car: weight downwards, normal force upwards, tension right and left, drag leftlocomotive: weight downwards, normal force upwards, tension left, drag left, thrust/applied right2. The system is moving with a constant speed, so the forces are balanced!3. Start by just looking at the rear car. The tension to the right must balance the drag. This will also be equal to the tension on the middle car. 4. Next, just look at the middle car. The tension to the right must balance the drag and tension to the left. This will also be equal to the tension on the locomotive. 5. Now, just look at the locomotive. The thrust/applied force must balance the drag plus the tension.
The free body diagram of locomotive can be shown as,
Therefore, according to free body diagram, four types of forces are acting on locomotive.
The free body diagram of middle car can be shown as,
Therefore, there are two forces acting on the middle car.
The free body diagram of the last car is shown as,
Therefore, there are three number of forces acting on the last car.
According to free body diagram of locomotive, the tension in cable 2 is equal to net force acting on locomotive. Therefore, the tension in cable 2 is 1000 N.
According to free body diagram of middle car, the tension in cable 1 is equal to tension in cable 2 therefore, the tension in cable 1 is also 1000 N.
The net force generated by locomotive can be given as,
[tex]F=F_{drag}-F_{middle\text{ car}}-F_{\text{rear car}}[/tex]Substitute the known values,
[tex]\begin{gathered} F=1000\text{ N-200 N-200 N} \\ =600\text{ N} \end{gathered}[/tex]Thus, the net force generated by locomotive is 600 N.
When you walk across a carpet during the winter you often get a shock when you touch a metal door knob. The reason is that you pick up electrons via friction between your feet and the carpet. If you pick up -97.37 microC of charge, how many electrons did you acquire ?
This is the answer tab
The total charge can be written as
[tex]Q=-97.37*10^{-6}C[/tex]And the charge of an individual electron is
[tex]q=-1.6*10^{-19}C[/tex]Thus, the number of electrons is the total charge acquired, divided by the amount of charge on each individual electron. This gives us:
[tex]n=\frac{Q}{q}=\frac{-97.37*10^{-6}}{-1.6*10^{-19}}=6.085625*10^{14}[/tex]So, we have acquired 6.085625*10^14 electrons
Consider an airless, non-rotating planet of mass M and radius R. and electromagnetic launcher standing on the surface of this planet shoots a projectile with initial velocity v0 directed straight up. Unfortunately, due to some error, v0 is less than the planet's escape velocity ve; specifically, v0 = 0.762ve. Unable to escape the planet's gravitational pull, the projectile rises to a maximal height h above the ground, then falls back to the ground. Calculate the ratio h/R of the projectile's maximum height to the planet's radius.
Given data:
The mass of planet is M.
The radius of planet is R.
The initial velocity of projectile is v₀=0.762ve.
The amount of kinetic and potential energy should be equal according to conservation of energy,
[tex]\begin{gathered} KE=PE \\ \frac{1}{2}mv^2_e=\frac{GMm}{R}^{}_{} \\ R=\frac{GM^{}_{}}{v^2_e} \end{gathered}[/tex]The escape velocity is given by,
[tex]v_e=\sqrt[]{\frac{2GR}{M}}[/tex]Here, G is the universal gravitational acceleration.
The time taken to reach the maximum height will be,
[tex]\begin{gathered} v_0=gt \\ t=\frac{v_0}{g} \end{gathered}[/tex]The maximum height reached by the projectile is given by,
[tex]\begin{gathered} h=v_0t+\frac{1}{2}gt^2 \\ h=v_0(\frac{v_0}{g})+\frac{1}{2}g(\frac{v_0}{g})^2 \\ h=\frac{v^2_0}{g}_{}+\frac{1}{2}\frac{v^2_0}{g}_{} \\ h=\frac{3}{2}\frac{v^2_0}{g}_{} \\ h=\frac{3}{2}R \\ \frac{h}{R}=1.5 \end{gathered}[/tex]Answer:
1.39
Explanation:
(1/2)mVe^2=(mMG)/R
Ve^2=2MG/R
(1/2)mVo^2=mMG(1/R-1/(R+h))=mMG(h/(R^2+Rh))
Vo^2=2MG(1/R)(h/(R+h))
Vo^2=Ve^2(h/(R+h))
Vo^2/Ve^2=h/(R+h)
Ve^2/Vo^2=(R+h)/h
1/.762^2=(R+h)/h
1.72h-h=R
.72h=R
h/R=h/.72h
h/R=1.39
Question 17 of 30Which of the following is the energy of atoms vibrating inside an object?A. RadiantB. ElectricC. ThermalD. MechanicalSUBMIT
ANSWER:
C. Thermal
STEP-BY-STEP EXPLANATION:
Thermal energy, or heat, is the internal energy of substances; is the vibration and movement of atoms and molecules within a substance.
Therefore, the correct answer is C. Thermal
The virtual image of a lens is always on the same side of the lens as the object. Is this true or false?
The lens follows the law of refraction to form an image of the object.
Thus, for the formation of real image, the image should be formed on the opposite side of the lens.
The virtual image is formed when the image is on the same side of the lens as where the object is present.
Thus, the given statement is true.
If it takes 299 calories to melt a mass of ice into water at a temperature of zero degrees Celsius then what was the mass of the ice in grams? A)0.0471 B)3.75 C)161000
Given:
The heat absorbed by ice to melt into water is Q = 299 calories.
To find the mass of the ice in grams.
Explanation:
When the ice starts melting into water, the temperature of the system does not change, all the heat provided to the system is used up to convert the ice to water. This is known as latent heat of fusion.
The latent heat of fusion for water is
[tex]L_f=79.8\text{ cal/g}[/tex]The formula to calculate mass is
[tex]Q=mL_f[/tex]Substituting the values, the mass will be
[tex]\begin{gathered} m=\frac{Q}{L_f} \\ =\frac{299}{79.8} \\ =3.75\text{ g} \end{gathered}[/tex]Final Answer: The mass of ice is 3.75 g
Each set of protons and electrons represents a different atom. Place theatoms in order of their overall charge. Order them from most positive to mostnegative.1 30 protons, 26 electrons11 17 protons, 10 electrons10 protons, 10 electrons12 protons, 11 electrons
Firstly, we need to remember how to find the liquid charge for a single atom. We can start by saying that
[tex]Q=n_pq_p+n_eq_e[/tex]Where Q is the total charge, np is the number of protons, qp is the charge of a proton, ne is the number of electrons and qe is the charge of an electron
As we know, the charge of an electron is the opposite of the proton's. If we consider it as -1 and 1, respectively, we're left with
[tex]Q=n_p-n_e[/tex]Then, let us find the charge for each atom.
So, for our atoms, we have
[tex]Q_1=30-26=4[/tex][tex]Q_2=17-10=7[/tex][tex]Q_3=10-10=0[/tex][tex]Q_4=12-11=1[/tex]Then, our answer is
[tex]Q_2>Q_1>Q_4>Q_3[/tex]Is it possible to get More work out of a machine than you put in?
If we were able to get more work out of a machine than we put in, energy would be created in the process. According to the Law of Conservation of Energy, this is not possible.
Therefore, the answer is:
[tex]\begin{gathered} \text{ No, it is not possible to get more work} \\ \text{ out of a machine than we put in.} \end{gathered}[/tex](a) How much power (in W) is dissipated in a short circuit of 225 V AC through a resistance of 0.280 Ω? W(b) What current (in A) flows? A
Let's name some variables:
V: voltage; V = 225V
R: resistance; R = 0.28 ohm
I: current
P: power
Let's use this equation first to solve for P:
P = V^2/R
Plugging in variables:
P = 225^2/0.28
P = 180803.57 W = 180.80357 kW
We can use the following to solve for I:
V = IR
225 = I*0.28
I = 803.571 A = 803571 mA
We can also use this method:
Now that we have P, we can use this to solve for I:
P = IV
Plugging in the known variables:
180803.57 = 225*I
I = 803.571 A = 803571 mA
A box weighing 100 newtons is pushed up an inclined plane that is 5 meters long. It takes a force of 75 newtons to push it to the top, which has a height of 3 meters. Work Output? Work Input? Efficiency?
The work output can be given as,
[tex]W_O=Fh[/tex]Substitute the known values,
[tex]\begin{gathered} W_O=(75\text{ N)(3 m)(}\frac{1\text{ J}}{1\text{ Nm}}) \\ =225\text{ J} \end{gathered}[/tex]Thus, the work output of the procedure is 225 J.
The work input of the procedure is,
[tex]W_i=wd[/tex]Substitute the known values,
[tex]\begin{gathered} W_i=(100\text{ N)(5 m)(}\frac{1\text{ J}}{1\text{ Nm}}) \\ =500\text{ J} \end{gathered}[/tex]Thus, the work input of the procedure is 225 J.
The efficiency of the work done is given as,
[tex]e=\frac{W_O}{W_i}[/tex]Substitute the known values,
[tex]\begin{gathered} e=\frac{225\text{ J}}{500\text{ J}} \\ =0.45 \end{gathered}[/tex]Thus, the efficiency of procedure is 0.45 or 45%.
I need help finding the area for these acceleration time graphs problem C.
We are given that an object starts with an initial velocity of 5 m/s and that we are required to determine its velocity after 2 seconds. We are given the area of acceleration vs time. In a graph of acceleration vs time, the area under the curve represents the velocity of the object under the given acceleration.
We can determine the velocity reached by the object between t = 0 and t = 2 finding the area under the curve. Since the acceleration is constant the area under the curve is a square, therefore:
[tex]v=(2\frac{m}{s^2})(2s)=4\frac{m}{s}[/tex]This is the velocity after the object starts accelerating the final velocity is determined by adding the initial velocity therefore, the final velocity is:
[tex]v_f=4\frac{m}{s}+5\frac{m}{s}=9\frac{m}{s}[/tex]
your calculation.11. What is the kinetic energy of a 2kg rabbit hopping at a speed of 1.25 m/s?
Kinetic energy (KE) = 1/2 m v^2
Where:
m= mass = 2 kg
v= velocity = 1.25 m/s
Replacing:
KE = 1/2 x 2 kg x (1.25 m/s)^2 = 1.5625 J
For the following calculation, give the answer to the correct number of significant figures.
Given,
[tex]\frac{(71.359\text{ m}-71.357\text{ m})}{(3.2\text{ s}\times3.67\text{ s})}[/tex]When calculating using the numbers with the different number of significant digits, the final answer should contain the same number of significant digits as the number with the least number of significant digits.
On simplifying the above equation,
[tex]\frac{0.002}{12}[/tex]In the above equation, the numerator (0.002) has only one significant digit.
Therefore the answer will have one significant digit
On further simplifying,
[tex]=2\times10^{-4}[/tex]Thus the correct answer is 2×10⁻⁴
A stone is dropped down a well. When it hits the bottom of the well it is travelling at 20 m/s. The mass of the stone is 0.02 kg.
a. Calculate the kinetic energy of the stone as it hits the bottom of the well.
b. What is the gravitational potential energy of the stone at the top of the well?
c. Calculate the height of the well.
Mass is the quantity of matter an object has. Is this true or false?
The mass of the object is the amount of matter present in the object.
Thus, the mass of the object does not change with the gravity or any other force unless the matter is removed from the object.
Hence, the given statement is true.
A ball is thrown upward from an initial height of 2 meters. The ball reaches a height of 5 meters then falls to the ground. What is the total distance traveled by the ball?
We will have the following:
First, we find the distance it traveled upward:
[tex]d_u=5m[/tex]Now, we find the distance it traveled to the ground:
[tex]d_d=5m+2m\Rightarrow d_d=7m[/tex]Now, the total distance is:
[tex]d_T=d_u+d_{\text{d}}\Rightarrow d_T=5m+7m[/tex][tex]\Rightarrow d_T=12m[/tex]So, the ball traveled a total of 12 meters.
What does the lower scale read? Answer in units of N
We will have the following:
First, we are given:
*Mass of the breaker: 1.1kg
*Mass of water: 3.3 kg
*Mass of metallic alloy: 4.2kg
*Density of the alloy: 5300kg/m^3
*Density of water: 1000kg/m^3
Now, we find the volume of water displaced by the alloy:
[tex]V_{\text{w}}=4.2\operatorname{kg}\cdot\frac{m^3}{5300\operatorname{kg}}\Rightarrow V_w=\frac{21}{26500}m^3\Rightarrow V_w\approx7.92\cdot10^{-4}m^3[/tex]Then, from the reading in the hanging scale we will have the force experienced by the alloy due to the upthrust when placed in water, that is:
[tex]R=mg-\rho Vg[/tex]So:
[tex]R=(4.2\operatorname{kg})(9.8m/s^2)-(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)[/tex][tex]\Rightarrow R=33.39396226\ldots N\Rightarrow R\approx33.4N[/tex]The reading on the lower scale is due to the weight of the water in the breaker and upthrust on the scale:
[tex]R=g(m_1+m_2)+\rho Vg[/tex]Finally:
[tex]R=(9.8m/s^2)(1.1\operatorname{kg}+3.3\operatorname{kg})+(1000\operatorname{kg}/m^3)(\frac{21}{26500}m^3)(9.8m/s^2)[/tex][tex]\Rightarrow R=50.886003774\ldots N\Rightarrow R\approx50.9N[/tex]So, the readin on the lower scale is approximately 50.9N.
A motorboat travels 408 kilometers in 6 hours going upstream and 882 kilometers in 9 hours going downstream. What is the rate of the boat, and what is the rate of the current?
We have the next variables
x = speed of the boat in still water
y = speed of the current
(x-y) = Upstream speed
(x+y) = downstream speed
So we have the next equations for the distance
[tex]\begin{gathered} 6\mleft(x-y\mright)=408 \\ \end{gathered}[/tex][tex]\begin{gathered} 9(x+y)=882 \\ \end{gathered}[/tex]We simplify each equation
[tex]\begin{gathered} x-y=68 \\ x+y=98 \end{gathered}[/tex]we sum both equations
[tex]\begin{gathered} 2x=166 \\ x=\frac{166}{2} \\ x=83 \end{gathered}[/tex]Then we calculate the y
[tex]\begin{gathered} y=98-x \\ y=15 \end{gathered}[/tex]x = speed of the boat in still water=83 km/hr
y = speed of the current 15 km/hr
A woman tries to throw a rock over a wall, releasing the rock at a height of 1.55m abovethe ground. If she throws the rock at 6 m/s, will it reach the top of the wall 3.75m aboveher?
Answer:
[tex]H_{\max }-H_{\text{wall}}=-0.3635m\Rightarrow Negative[/tex]Explanation: We need to find if the rock can reach the height of the wall, provided the initial velocity of 6m/s and it is thrown from the height of 1.55m above the ground., the equations used to solve this problem are as follows:
[tex]\begin{gathered} v(t)=v_o-gt\Rightarrow(1) \\ y(t)=y_o+v_ot-\frac{1}{2}gt^2\Rightarrow(2) \end{gathered}[/tex]Plugging in the known values in the equation (1) and (2) we get the following results:
[tex]\begin{gathered} v(t)=6ms^{-1}-(9.8ms^{-2})t\Rightarrow(3) \\ y(t)=1.55m+(6ms^{-1})t-\frac{1}{2}(9.8ms^{-2})t^2\Rightarrow(4) \\ \end{gathered}[/tex]Setting equation (3) equal to zero gives the time to reach the maximum height as follows:
[tex]\begin{gathered} v(t)=6ms^{-1}-(9.8ms^{-2})t=0 \\ t=\frac{(6ms^{-1})}{(9.8ms^{-2})}=0.61s \\ t=0.61s \end{gathered}[/tex]Substituting the time t calculated above in equation (4) gives the maximum height, the steps for the calculation are shown as follows:
[tex]\begin{gathered} y(0.62)=1.55m+(6ms^{-1})(0.62s)-\frac{1}{2}(9.8ms^{-2})(0.62s)^2 \\ y(0.62)=1.55m+3.72m-1.88356m=3.3865m \\ y(0.62)=3.3865m \\ H_{\max }=3.3865m \\ H_{\max }-H_{\text{wall}}=3.3865m-3.75m=-0.3635m\Rightarrow Negative \\ \\ \end{gathered}[/tex]Therefore, the ball can not reach the height of the wall!
If I am finding the x component of acceleration and it’s m/s^3, would the answer be m/s^3
No, the aceleration is alwas m/s^2
If you have something in m/s^3, probablly you are missing a multiplication times the time or maybe you have to integrated the function
[tex]y^{\prime}\left(t\right)=\frac{0.018m}{s^2}[/tex]
If the average kinetic energy of the molecules in an ideal gas initially at 26 degrees celsius doubles, which of the following is the final temperature of the gas?325 degrees Celsius, 554 degrees Celsius, 52
We will have the following:
The final termperature should then be 52 °C.
At which section does the object move with a positive velocity
The velocity of any objetc is defined in terms of initial and final position of the object is,
[tex]v=\frac{d_f-d_i}{t}[/tex]As in the graph during the motion of object, the final position of the object is decreasing with time. Thus, the final position is always less than the initial position.
[tex]d_f-d_i<0[/tex]Thus, the value of velocity at every point is,
[tex]v<0[/tex]Hence, first option (This object never has a positive velocity) is the correct answer.
There is a distance of 3.60 * 10 ^ 7 m between the Earth and a satellite. The magnitude of the gravitational force of attraction between the and the Earth is 400 N. What would be the magnitude of the gravitational force of attraction if the distance between them changed to 1.80 * 10 ^ 7 meters?
Given:
The initial distance between the earth and the satellite, R₁=3.60×10⁷ m
The initial force between the earth and the satellite, F₁=400 N
The changed distance between the earth and the satellite R₂=1.80×10⁷ m
To find:
The changed force between the earth and the satellite.
Explanation:
From Newton's gravitational law, the gravitational force between two objects is directly proportional to the product of the mass of the objects and inversely proportional to the square of the distance between them.
Let us assume that the mass of the earth is M and the mass of the satellite is m.
Thus,
[tex]\begin{gathered} F_1=\frac{GMm}{R_1^2} \\ \Rightarrow F_1R_1^2=GMm\text{ }\rightarrow\text{ \lparen i\rparen} \end{gathered}[/tex]And, the changed force is given by,
[tex]\begin{gathered} F_2=\frac{GMm}{R_2^2} \\ \Rightarrow F_2R_2^2=GMm\rightarrow\text{ \lparen ii\rparen} \end{gathered}[/tex]From equation (i) and equation (ii),
[tex]\begin{gathered} F_1R_1^2=F_2R_2^2 \\ \Rightarrow F_2=\frac{F_1R_1^2}{R_2^2} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} F_2=\frac{400\times\left(3.60\times10^7\right?^2}{\lparen1.80\times10^7)^2} \\ =1600\text{ N} \end{gathered}[/tex]Final answer:
The magnitude of the gravitational force of attraction between the earth and the satellite after the distance between them was changed is 1600 N.
3.00 m^3 of water is at 20.0°C.If you raise its temperature to60.0°C, by how much will itsvolume expand?WaterB = 207•10-6 0-1(Unit = m^3)
Given,
Initial volume of the water, V₁=3.00 m³
The initial temperature of the water, T₁=20.0 °C=293.15 K
The final temperature of the water, T₂=60 °C=333.15 K
From Charle's law, we have,
[tex]\frac{V_1}{T_1_{}}=\frac{V_2}{T_2}[/tex]On rearranging the above equation,
[tex]V_2=\frac{V_1T_2}{T_1}[/tex]On substituting the known values in the above equation,
[tex]V_2=\frac{3.00\times333.15}{293.15}=3.41m^3[/tex]Therefore the change in the volume is,
[tex]\Delta V=V_2-V_1[/tex]i.e.,
[tex]\Delta V=3.41-3.00=0.41m^3[/tex]Therefore, the volume of the water will expand by 0.41 m³
Questlon 7 of 10 Which two factors affect the amount of thermal energy an object has? O A. The directions in which the particles of the object are moving [ B. The amount of space between the particles of the object C. The mass of the object O D. The average kinetic energy of the particles of the object SUBM
C. The mass of the object
D. The average kinetic energy of the particles of the object
Thermal energy can be considered as kinetic energy, which involves the motions of the molecules of the object.
If an object has high kinetic energy, then the object has higher thermal energy.
If John takes 45 minutes to bicycle a total of four kilometers northwest to his grandmothers house , what is the velocity in km/hr?
time = 45 minutes
distance = 4 kilometers
direction = Northwest
Since the velocity is in km/hr, convert minutes to hours:
1 hour = 60 minutes
45 / 60 = 0.75
45 min = 0.75 hours
Velocity = distance / time = 4 km/ 0.75 h = 5.33 km/ h , northwest.
Describe how a convex lens could be used to make a magnifying lens.
Explanation and Answer:
when the object is placed between the optical center and the first focal point of the convex lens. its image is formed on the same side where the object is placed. the image is erect, magnified, and virtual.
we can see the ray diagram to understand it better,
as we can see from the ray diagram, a convex lens is used as a magnifying lens.
The image formed is larger than the original object.
if the object is placed between the first focal and the optical center. the image formed is larger than the object. the formed image is erect, virtual, and magnified. This way convex lens work as a magnifying lens.
Lola has a two-year lease that requires her to pay $295 rent per month.What is the total amount she will pay for rent during the term of herlease?a. $3,540b. $8,850c. $885d. $7,080
If the price that needs to be paid per month is $295, first let's convert the total amount of time (2 years) to months.
1 year has 12 months, therefore 2 years have 24 months.
Now, to find the total amount, let's multiply the cost per month by the number of months:
[tex]295\cdot24=7080[/tex]Therefore the total amount is $7,080. Correct option: d.
How much heat is needed to increase the internal energy of a gas in a piston by 4,258 J if the gas does 801 J of work on the environment by expanding if the process
According to the First Law of Thermodynamics, the change in internal energy (ΔU) of a system is equal to the sum of the heat (Q) added to the system and the work (W) exerted over the system:
[tex]\Delta U=Q+W[/tex]The sign of the heat and the work is positive when energy is transferred to the system and negative when energy is transferred from the system to its surroundings.
In this case, the system is formed by a gas on a piston, which gains 4,258 Joules in internal energy and releases 801 Joules as work. Then:
[tex]+4258J=Q-801J[/tex]Isolate Q from the equation to find the heat needed for this process:
[tex]\Rightarrow Q=4258J+801J=5059J[/tex]Therefore, the amount of heat needed to increase the internal energy of a system by 4,258J if it releases 801 of work to the environment, is 5059J.