When two particles collide and the masses of the particles are given, as well as the initial and final velocity of one particle and one of the velocities of the second particle, then the remaining velocity of the second particle is given by the expression:
[tex]v_2^{\prime}=\frac{m_1v_1+m_2v_2-m_1v_1}{m_2}[/tex]Which can be deduced from the Law of Conservation of Linear Momentum.
In the given problem, the initial and final velocities of the train car with mass 10kg are given, as well as the initial velocity of the 20kg car:
[tex]\begin{gathered} m_1=10kg \\ v_1=10\frac{m}{s} \\ v_1^{\prime}=-20\frac{m}{s} \\ \\ m_2=20kg \\ v_2=-40\frac{m}{s} \\ v_2^{\prime}=\text{ unknown} \end{gathered}[/tex]Replace those values into the given equation to find v₂':
[tex]\begin{gathered} v_2^{\prime}=\frac{(10kg)(10\frac{m}{s})+(20kg)(-40\frac{m}{s})-(10kg)(-20\frac{m}{s})}{20kg} \\ \\ \Rightarrow v_2^{\prime}=-25\frac{m}{s} \end{gathered}[/tex]Therefore, the velocity of the 20kg train car after the collision, is: -25 m/s.
Question 17 of 25A conductor is a material that:O A. allows easy movement of charge.B. is never made of metal,C. hinders the passage of electricity.O D. is made of glass.
From the given list, let's select the statement that best defines a conductor.
A conductor can be defined as any material that allows the flow of electric charge.
Examples of conductors are:
• Copper
,• Aluminium
,• Silver...
Therefore, the best statement that defines a conductor is that a conductor is a material that allows easy movement of charge.
• Option B is wrong because most conductors are made of metal.
,• Option C is wrong because it is an ,insulator ,hinders the passage of electricity.
,• Option D is wrong because a glass is an insulator not a conductor.
ANSWER:
A. allows easy movement of charge
carts, bricks, and bands
5. What acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick?
a. About 0.25 m/s2
b. About 0.50 m/s2
c. About 0.75 m/s2
d. About 1.00 m/s2
B. The acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
What is the applied force on an object?The force applied on object is obtained by multiplying the mass and acceleration of the object.
According to Newton's second law of motion, the force exerted on an object is directly proportional to the product of mass and acceleration of the object.
Also, the applied force is directly proportional to the change in the momentum of the object.
Mathematically, the force acting on object is given as;
F = ma
a = F/m
where;
a is the acceleration of the objectm is the mass of the objectF is the applied forceFrom the trials, the acceleration results when 2 rubber bands stretched to 20 cm are used pull a cart with one brick is 0.5 m/s².
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How long does it take to stop a 1000 kg object moving at 20 m/s with a force 5000N? 2500 N, 1000 N, 500 N, 400 N, 200 N, 100 N
ANSWER
[tex]\begin{gathered} 5000N\Rightarrow4s \\ 2500N=8s \end{gathered}[/tex]EXPLANATION
To find the time taken to stop the object, we first have to find the acceleration of the object.
Since the force is working to stop the object (slow down the object), it means that the object is decelerating (slowing down).
To find the acceleration (deceleration), apply Newton's second law of motion:
[tex]F=ma[/tex]where F = force; m = mass; a = acceleration
Therefore, for a force of 5000N, we have that:
[tex]\begin{gathered} 5000=1000\cdot a \\ \Rightarrow\frac{5000}{1000}=a \\ \Rightarrow a=5m\/s^2 \end{gathered}[/tex]Now, we can apply Newton's equation of motion to find the time taken to stop the object:
[tex]v=u-at[/tex]where v = final velocity = 0 m/s; u = initial velocity = 20 m/s; t = time taken
Note: the negative sign indicates deceleration
Hence, the time taken for a force of 5000N to stop the object is:
[tex]\begin{gathered} 0=20-5\cdot t \\ \Rightarrow5t=20 \\ \Rightarrow t=\frac{20}{5} \\ t=4s \end{gathered}[/tex]For a force of 2500N, the deceleration is:
[tex]\begin{gathered} 2500=1000\cdot a \\ a=\frac{2500}{1000} \\ a=2.5m\/s^2 \end{gathered}[/tex]Hence, the time taken for a force of 2500N to stop the object is:
[tex]\begin{gathered} 0=20-2.5\cdot t \\ \Rightarrow2.5t=20 \\ t=\frac{20}{2.5} \\ t=8s \end{gathered}[/tex]A 60.0 kg skier with an initial speed of 14 m/s coasts up a 2.50 m high rise as shown in the figure.
Find her final speed right at the top, in meters per second, given that the coefficient of friction between her skis and the snow is 0.38?
The final speed of the skier at the top mountain is determined as 9.27 m/s.
What is the change in the energy of the skier?
The change in the energy of the skier due to frictional force is calculated as follows;
ΔP.E = Pi + Ef
where;
Pi is the initial potential at the topEf is the energy lost to frictionThe distance of the plane travelled is calculated as;
sin35 = 2.5/L
L = 2.5 / sin35
L = 4.36 m
ΔP.E = mghi - μmgcosθ(L)
where;
m is the masshi is the initial heightg is acceleration due to gravityμ is coefficient of frictionΔP.E = (60 x 9.8 x 2.5) - (0.38)(60)(9.8) cos(35) x (4.36)
ΔP.E = 671.98 1 J
The final speed of the skier at the top of the plane;
P.E = K.E
P.E = ¹/₂mv²
v² = 2P.E /m
v = √(2P.E /m)
v = √(2 x 671.98) / 60)
v = 4.73 m/s
Total speed = -4.73 m/s + 14 m/s = 9.27 m/s
Thus, due to frictional force opposing the upward motion of the skier, the final speed at the top will be smaller than the initial speed.
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If an astronaut weighs 148 N on the Moon and 893 N on Earth, then what is his mass on Earth?
_____ kg
The mass on Earth is 91.1 kg
The weight on the Moon is 148 N
The weight on the Earth is 893 N
We need to apply the concept of force.
Weight=massxacceleartion
893=mass of man x acceleration of gravity on earth
893= mass x 9.8
mass = 91.1 kg
Therefore, the mass on earth is 91.1 kg.
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Answer:
the mass on Earth is 91.1 kg
Explanation:
Block A in (Figure 1) has mass 0.900 kg , and block B has mass 3.00 kg . The blocks are forced together, compressing a spring S between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block B acquires a speed of 1.35 m/s .
Part A: What is the final speed of block A?
Part B: How much potential energy was stored in the compressed spring?
(a) The final speed of block A is determined as 4.5 m/s.
(b) The potential energy that was stored in the compressed spring is 11.85 J.
What is the final speed of block A?
The final speed of block A is determined by applying the principle of conservation of linear momentum as follows;
Pa = Pb
where;
Pa is the momentum of block APb is the momentum of block Bmv (block A) = mv (block B)
(0.9 kg)(v) = (3 kg)(1.35 m/s)
0.9v = 4.05
v = 4.05/0.9
v = 4.5 m/s
The potential energy stored in the compressed spring is calculated as follows;
Apply the principle of conservation of energy.
U = K.E
where;
K.E is the kinetic energy of the blocksU = ¹/₂mv² (A) + ¹/₂mv² (B)
U = ¹/₂(0.9)(4.5²) + ¹/₂(3)(1.35²)
U = 11.85 J
Thus, the potential energy that was stored in the compressed spring is determined by applying the principle of conservation of energy.
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The heating element of an iron operates at 110 V with a current of 11 A.(a) What is the resistance of the iron? Ω(b) What is the power dissipated by the iron? W
(a)
In order to calculate the resistance, we can use Ohm's Law:
[tex]\begin{gathered} R=\frac{V}{I}\\ \\ R=\frac{110}{11}\\ \\ R=10\text{ \Omega} \end{gathered}[/tex](b)
To calculate the power, we can use the formula below:
[tex]\begin{gathered} P=I\cdot V\\ \\ P=11\cdot110\\ \\ P=1210\text{ W} \end{gathered}[/tex]a car goes from 32 m/s to a complete stop in 4.8 seconds. calculate the average stopping force of the car if has a mass of 2500 kg
The average stopping force is 16,500 N
Initial velocity of car (v₁)= 32m/s
Final velocity (v₂) = 0m/s
Time to stop= 4.8 seconds
Mass of car= 2500 kg
we need to apply the concept of laws of motion
Acceleration of car (a)= Change in velocity/time
a= v₂-v₁/t
a= 0-32/4.8
a= -6.6 m/s² ( deceleration)
Force= mass x acceleration
Force= 2500x 6.6
Force= 16500 N
Therefore the average stopping force is 16,500 N.
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What does a scientist mean when he or she says that an object is at rest
Answer:
I does not change position with respect to its surroundings with time
Analyze the collision of a baseball with a bat. During which portion of the collision does the baseball’s velocity reach zero?1) before the collision2) during the collision3) one second after the collision4) one-hundredth of a second after the collision
ANSWER:
2) during the collision
STEP-BY-STEP EXPLANATION:
We have that when a baseball ball collides with a bat, its velocity changes from positive to negative, that is, the ball when hit with an opposing force changes its direction exactly opposite to the initial one.
Therefore, during the collision, at some point in time, the velocity of the ball reaches zero and then finally changes its direction associated with an increase in the velocity of the ball.
The Diagram shows the forces involved as a student slides a water bottle across the desk in front of them to their friend. Based on the image, in which direction is there friction? (ignore the selected answer it’s random)
Answer:
Left.
Step-by-step explanation:
The force of friction opposes the motion of an object, causing moving objects to lose energy and slow down. Therefore, the friction goes to the left.
Jeff tosses a can of soda pop to Karen, who is standing on her 3rd floor balcony a distance of 8.5m above Jeff’s hand. Jeff gives the can an initial velocity of 16m/s, fast enough so that the can goes up past Karen, who catches the can on its way down. Calculate the velocity of the can the instant before Karen grabs the can. How long after Jeff tosses the can does Karen have to prepare to catch it?
ANSWER
9.51 m/s
EXPLANATION
We know that Jeff is 8.5m below Karen. He tosses the can up with initial velocity u = 16m/s and it passes where Karen is, so the maximum height of the can is 8.5m plus some more meters x. Then Karen catches the can in its way down, so when she does the can goes this distance x.
Let's find this distance. The height of an object thrown up with initial velocity u is:
[tex]y=ut-\frac{1}{2}gt^2[/tex]We know u = 16m/s but we don't know the time. This we can find from the final velocity of the can:
[tex]v=u-gt[/tex]At its maximum height the velocity is zero:
[tex]0=u-gt[/tex]Solving for t:
[tex]t=\frac{u}{g}[/tex]If we assume g = 9.8m/s²:
[tex]t=\frac{16m/s}{9.8m/s^2}=1.63s[/tex]We know that the can was in the air for 1.63 seconds until it reached its maximum height. The maximum height is:
[tex]y=16m/s\cdot1.63s-\frac{1}{2}\cdot9.8m/s^2\cdot1.63^2s^2[/tex][tex]y=26.08m-13.02m=13.06m[/tex]This is the maximum height of the can. The extra distance the can travelled above Karen is:
[tex]x=13.06m-8.5m=4.56m[/tex]In the can's way down, the initial velocity is 0, because it starts falling after stopping in its way up. The acceleration is still the acceleration of gravity and the height it falls is x. We can find the time it took to reach Karen's hand after it started falling:
[tex]y=\frac{1}{2}gt^2[/tex]Note that in this case we use the acceleration of gravity positive because it is in the same direction of the can's motion. Solving for t:
[tex]t=\sqrt[]{\frac{2y}{g}}[/tex][tex]t=\sqrt[]{\frac{2\cdot4.56m}{9.8m/s^2}}=\sqrt[]{0.93s^2}=0.97s[/tex]Knowing that the can was in the air for another 0.97 seconds after starting falling until it reached Karen's hand, we can find its velocity at that instant:
[tex]v=u+gt[/tex]Remember that in this case u = 0:
[tex]v=gt=9.8m/s^2\cdot0.97s=9.51m/s[/tex]The velocity of the can the instant before Karen grabs it is 9.51 m/s
Based on the circuit voltage and the wattage consumption,determine the approximate ampere rating of the followingappliances. Remember amps = watts divided by voltage.a = w÷ VRound to the nearest whole amp.1. AC Compressor on a 240 volt line and using 5,000 watts, amps =_____2. baseboard heater on a 120 volt line and using 1,200 watts, amps =_____3. vacuum cleaner on a 120 volt line and using 500 watts, amps =______4. blender on a 115 volt line and using 300 watts, amps5. toaster on a 120 volt line using 1,100 watts, amps =_____
Given:
1.
The voltage of AC compressor is V = 250 V
The power of the AC compressor is P = 5000 W
2.
The voltage of the baseboard heater is V = 120 V
The power of the baseboard heater is P = 1200 W
3.
The voltage of the vacuum cleaner is V = 120 V
The power of the vacuum cleaner is P = 500 W
4.
The voltage of the blender is V = 115 V
The power of the blender is P = 300 W
5.
The voltage of the toaster is 120 V
The power of the toaster is P = 1100 W
Required:
1. The approximate ampere rating of the AC compressor.
2. The approximate ampere rating of the baseboard heater.
3. The approximate ampere rating of the vacuum cleaner.
4. The approximate ampere rating of the blender.
5. The approximate ampere rating of the toaster.
Explanation:
1. The approximate ampere rating of the AC compressor can be calculated as
[tex]\begin{gathered} I\text{ = }\frac{P}{V} \\ =\frac{5000}{240} \\ =20.833\text{ A} \\ \approx21\text{ A} \end{gathered}[/tex]2. The approximate ampere rating of the baseboard heater can be calculated as
[tex]\begin{gathered} I=\frac{1200}{120} \\ =\text{ 10 A} \end{gathered}[/tex]3. The approximate ampere rating of the vacuum cleaner can be calculated as
[tex]\begin{gathered} I\text{ = }\frac{500}{120} \\ =4.2\text{ A} \\ \approx4\text{ A} \end{gathered}[/tex]4. The approximate ampere rating of the blender can be calculated as
[tex]\begin{gathered} I\text{ =}\frac{300}{115} \\ =2.6\text{ A} \\ \approx3\text{ A} \end{gathered}[/tex]5. The approximate ampere rating of the toaster can be calculated as
[tex]\begin{gathered} I\text{ =}\frac{1100}{120} \\ =9.2\text{ A} \\ \approx\text{ 9 A} \end{gathered}[/tex]Final Answer:
1. The approximate ampere rating of the AC compressor is 21 A.
2. The approximate ampere rating of the baseboard heater is 10 A.
3. The approximate ampere rating of the vacuum cleaner is 4 A.
4. The approximate ampere rating of the blender is 3 A.
5. The approximate ampere rating of the toaster is 9 A.
How much power is created when you perform 55 Joule of work with a time 20 sec?
Answer:
2.75 watts
Explanation:
The power is equal to the work divided by time, so
P = W/t
Then, replacing W = 55 J and t = 20 sec, we get:
P = 55 J / 20 s = 2.75 Watts
Therefore, the power created is 2.75 watts
Which of the following terms represents the number of waves passing a given point each second?Frequency⊝Wavelength⊝Amplitude⊝Velocity⊝CLEAR ALL
Frequency is defined as number of waves or vibration per unit time.
Therefore, option (a), frequency is the number of waves passing a given point each second, is the correct choice.
Which statement best describes Earth's oceans?A. The waters of Earth's five major oceans rarely mix with oneanother.B. More than 95% of the water in the hydrosphere is found in oceans.C. The oceans contain almost all of Earth's freshwaterD. Oceans cover about 25% of Earth's surface.
From the given list, let's select the best statement that describes Earth's oceans.
The oceans of the earth are the principal component of the Earth's hydrosphere. The major oceans are the pacific ocean, atlantic ocean, indian ocean, souther ocean and the Arctic ocean.
The ocean covers more than 70% of the surface of the Earth and 97% of the Earth's water.
More than 95% of the Earth's water are found in the oceans.
The oceans contain only about 3% of the Earth's freshwater.
The waters of the Earth's five major oceans are connected with one another.
Therefore, the best statement which best describes the Earth's oceans is:
More than 95% of the water in the hydrosphere is found in oceans.
ANSWER:
B. More than 95% of the water in the hydrosphere is found in oceans.
Two ropes support a 15.0 kg load between them. One rope points NW at an angle of 15.0 degrees to the horizontal. Second rope points NE at an angle of 20.0 degrees to the vertical. Determine the magnitude of the tension force in each of the ropes.
The tension in the first rope is 38.05 N and the tension in the second rope is 138.1 N.
What is the weight of the load?
The weight of the load due to the force of gravity is calculated as follows;
W = mg
where;
m is mass of the loadg is acceleration due to gravityW = 15 kg x 9.8 m/s²
W = 147 N
Since the weight of the load is acting downwards, the tension in each rope is calculated as follows;
The tension in the first rope, T1 = W sinθ
where;
θ is the angle of inclination above the horizontalT1 = 147 sin(15)
T1 = 38.05 N
The tension in the first rope, T2 = W sinθ
where;
θ is the angle of inclination above the horizontal = 90 - 20 = 70⁰T2 = 147 x sin(70)
T2 = 138.1 N
Thus, the tension in each rope is determined by calculating the vertical component of force in each rope.
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Hey! I really need help with this question please :)
Answer: B
Explanation:
The formula for calculating the efficiency of a heat engine is expressed as
Efficiency = useful work done/Heat energy supplied x 100
From the information given,
Heat energy supplied = 500
useful work done = 50
Efficiency = 50/500 x 100
Efficiency = 10%
What is the frequency of a photon of EMR with a wavelength of 2.55x10*³m?1.18x1011 Hz8.50x10 12 Hz7.65x105 Hz1.18x105 Hz
In order to solve this equation, we will need to use the formula
[tex]f=\frac{c}{\lambda}[/tex]where f = frequency, c is the speed of light and lamda is wavelength
c = 3x10^8 m/s
lamda = 2.55x10^-3 m
f = (3x10^8)/(2.55x10^-3) = 1.18x10^11 1/s
A sound wave of wavelength 1.66m at a temperature of 23 C is produced for 2.5 seconds. How far does this wave travel? How many complete waves are emitted in this time interval?
To find how far it travels, we just have to multiply
[tex]1.66\times2.5=4.15[/tex]It travels 4.15 meters.
During this interval, there are emitted 2.5 waves.
You observe waves on the beach and measure that a wave hits the beach every 5 seconds. What is the period and the frequency of the waves ?
Answer:
Period = 5 seconds
Frequency = 0.2 Hz
Explanation:
The period is the time per cycle. So, if a wave hits the beach every 5 seconds, the period will be 5 seconds.
Additionally, the frequency is the inverse of the period, so the frequency of the waves can be calculated as:
[tex]\text{frequency = }\frac{1}{Period}=\frac{1}{5}=0.2\text{ Hz}[/tex]So, the answers are:
Period = 5 seconds
Frequency = 0.2 Hz
An interesting question is “In what direction is the dot (representing a particle in the medium)moving at the instant shown above?” The correct answer is “_______”. The way to understandthat is to imagine the pulse an instant later. The wave will have moved a bit to the right.Therefore, since the particle is still on the wave, and can only move up or down, it must be______.
We will have the following:
The correct answer is vertically.
The way to understand that is to image the pulse an instant later. The wave will have moved a bit to the right. Therefore, since the particle is still on the wav, and can only move up or down, it must be lower.
Giving a test to a group of students, the grades and gender are summarized belowGrades vs. Gender ABCMale10316Female465If one student was chosen at random, find the probability that the student got a B.
Answer:
20.45%
Explanation:
The probability that the student got a B is
[tex]\frac{total\#of\text{ students that got B}}{\text{total students }}\times100\%[/tex]Now, how many students are there in total?
The answer is
[tex]10+3+16+4+6+5=44\; \text{students}[/tex]How many students got a B?
The answer is
[tex]3+6=9\; \text{students}[/tex]therefore, the probability that the student has got a B is
[tex]\frac{9\text{ students }}{44\text{ students }}\times100\%=20.45\%[/tex]Hence, the probability that a student has got a B is 20.45%
Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially elastic. Car B is stopped at a light when it is struck. Car A has mass m and speed v before the collision. After the collision.. A)each car has half the impulse of before B)each car has double the impulse of before C)each car have the same impulse D)each car has an impulse in ratio to its mass.
We are given the following information.
The collision is elastic.
Car B has twice the mass of car A.
Recall that in an elastic collision, the momentum and the kinetic energy are conserved.
Impulse is basically the change in momentum.
Since car B has 2 times the mass of car A, the momentum of car B will be 4 times the momentum of car A.
This means that the impulse of car B will be greater than the impulse of car A.
Option D says that each car has an impulse in ratio to its mass meaning that a car with a larger mass will have a larger impulse and vice versa.
Therefore, we can conclude that after the collision, each car has an impulse in ratio to its mass.
On which of the following does the speed of a falling object depend?a.) v ∝ mb.)v ∝ mc.)v ∝ Δh
Given:
The falling object
To find:
The dependence on the speed of the falling object
Explanation:
For an object falling freely, the total mechanical energy remains always constant. So, we can write, that the decrease in potential energy will be equal to the increase in the kinetic energy that is
[tex]\begin{gathered} \frac{1}{2}mv^2=mgh \\ v^2=2gh \end{gathered}[/tex]Hence, the speed of the falling object does not depend on the mass it depends on the height difference.
What is absolute zero?0k0c273-100
The absolute zero is measured in Kelvin scale
Thus absolute zero is 0K.
A sample of unknown material weight 900N In air and and 400N when submerged in an alcohol solution with a density of 0.7 g/cm³.What is the density of the material ?
1.26 g/cm³
ExplanationStep 1
given
[tex]\begin{gathered} F_{g(air)}=900\text{ N} \\ F_{g(alchodol)}=400\text{ N} \\ \rho_{alcohol}=0.7\text{ }\frac{g}{cm^3} \end{gathered}[/tex]unknown; the density of the material, so
[tex]\begin{gathered} F_B=F_{g(air)}-F_{g(alcohol)} \\ F_B=900\text{ N-400 N=500 N} \end{gathered}[/tex]so, the proportion is
the ratio of the force equals the ratio of the density ,so
[tex]\begin{gathered} \frac{F_{g(air)}}{F_B}=\frac{\rho_{material}}{\rho_{alcholol}} \\ replace \\ \frac{900\text{ N}}{500\text{ N}}=\frac{\rho_{material}}{0.7\text{ }\frac{g}{cm^3}} \\ mutliply\text{ both sides by 0.7}\frac{g}{cm^3} \\ \frac{900\text{N}}{500\text{N}}*0.7\text{ }\frac{g}{cm^3}=\frac{\rho_{mater\imaginaryI al}}{0.7\text{\frac{g}{cm^{3}}}}*0.7\frac{g}{cm^3} \\ 1.26\frac{g}{cm^3}=\text{ density of the material } \\ \end{gathered}[/tex]so, the density of the material is
1.26 g/cm³
I hope this helps you
Timmy walks 5 m North, 3m West, and finally 1 m South. What is his displacement from his starting point?
Timmy walks 5 m North, 3m West, and finally 1 m South then his displacement from the starting point would be 5 meters in the northwest direction.
What is displacement?Displacement describes this shift in location and it is calculated with the help of the initial and the final position of the object.
As given in the problem If Timmy walks 5 m North, 3m West, and finally 1 m South ,
The resultant displacement of the Timmy = √(4² + 3²)
= 5 meters
Thus, the resultant displacement of the Timmy would be 5 meters
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a ball starts from rest. It rolls down a ramp and reaches the ground after 8 seconds. It's final velocity when it reaches the ground is 14.0 meters/second. What is the initial velocity and acceleration?
A ball starts from rest such that initial velocity, u=0, and final velocity, v = 14 m/s
and the time duration, t = 8 seconds.
To find initial velocity and acceleration, a.
As the ball is at rest, thus initial velocity is zero.
Acceleration is given by the formula,
[tex]a=\frac{v-u}{t}[/tex]Substituting the values in the above equation, we get
[tex]\begin{gathered} a=\frac{14-0}{8} \\ =1.75m/s^2 \end{gathered}[/tex]Hence the acceleration is 1.75 m/s^2
A worker pushes horizontally on a large crate with a force of 245 N, and the crate is moved 3.5 m. How much work was done? answer in : ___J
The amount of work done by a force can be written as the following:
[tex]W=F.\Delta x[/tex]For our case, we can replace our values and we'll get:
[tex]W=245*3.5=857.5J[/tex]Thus, the amount of work done is 857.5J