Answer:
Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)
Kinetic energy at the beginning: KEi =[tex]mv^{2}[/tex]/2
Potential energy at the end: Uf =(Qq/r)*k
Kinetic energy at the beginning: KEf = 0
From energy conservation:
r = 2*Qk/[tex]v^{2}[/tex]*q/m
Ranking of the closest approach to the gold nucleous comes from q/m:
- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest
- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking
- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4
Explanation:
Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)
Kinetic energy at the beginning: KEi =[tex]mv^{2}[/tex]/2
Potential energy at the end: Uf =(Qq/r)*k
Kinetic energy at the beginning: KEf = 0
From energy conservation:
r = 2*Qk/[tex]v^{2}[/tex]*q/m
Ranking of the closest approach to the gold nucleous comes from q/m:
- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest
- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking
- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4
The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1300 N/m that he will compress with a force of 6500 N. The inside of the gun barrel is coated with Teflon, so the average friction force will be only 50 N during the 5.0 mm he moves in the barrel.
Required:
At what speed will he emerge from the end of the barrel, 2.5 mabove his initial rest position?
Answer:
22m/s
Explanation:
Mass, m=60 kg
Force constant, k=1300N/m
Restoring force, Fx=6500 N
Average friction force, f=50 N
Length of barrel, l=5m
y=2.5 m
Initial velocity, u=0
[tex]F_x=kx[/tex]
Substitute the values
[tex]6500=1300x[/tex]
[tex]x=\frac{6500}{1300}=5[/tex]m
Work done due to friction force
[tex]W_f=fscos\theta[/tex]
We have [tex]\theta=180^{\circ}[/tex]
Substitute the values
[tex]W_f=50\times 5cos180^{\circ}[/tex]
[tex]W_f=-250J[/tex]
Initial kinetic energy, Ki=0
Initial gravitational energy, [tex]U_{grav,1}=0[/tex]\
Initial elastic potential energy
[tex]U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)[/tex]
[tex]U_{el,1}=16250J[/tex]
Final elastic energy,[tex]U_{el,2}=0[/tex]
Final kinetic energy, [tex]K_f=\frac{1}{2}(60)v^2=30v^2[/tex]
Final gravitational energy, [tex]U_{grav,2}=mgh=60\times 9.8\times 2.5[/tex]
Final gravitational energy, [tex]U_{grav,2}=1470J[/tex]
Using work-energy theorem
[tex]K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}[/tex]
Substitute the values
[tex]0+0+16250-250=30v^2+1470+0[/tex]
[tex]16000-1470=30v^2[/tex]
[tex]14530=30v^2[/tex]
[tex]v^2=\frac{14530}{30}[/tex]
[tex]v=\sqrt{\frac{14530}{30}}[/tex]
[tex]v=22m/s[/tex]
The picture to the right shows which wave behavior?
Answer:
It is refraction
Explanation:
State three factors affecting pressure in liquids
Answer:
Density of liquid
Depth of liquid
Acceleration due to gravity
Define personal health.
Answer:
Personal Health is the ability to take charge of your health by making conscious decisions to be healthy.
Answer:Personal Health is the ability to take charge of your health by making conscious decisions to be healthy. It not only refers to the physical well being of an individual but it also comprises the wellness of emotional, intellect, social, economical, spiritual and other areas of life.
Explanation:
How much force will a 5 kg rock hit the Earth with if it falls
for 1 second?
Answer:
f
Explanation:
f
Cesar and Jill went to a field to play soccer. As the ball downward toward Jill, Jill used her foot to kick the ball and keep it in play. Cesar realized he could apply scientific principles to a soccer game. Which of the following best describes the scientific principle that Cesar observes as Jill kicks the ball?
A. An unbalanced force has no effect on the ball.
B. Gravity on the ball is equal to the force of friction.
C. A net force of zero changes the direction of the ball.
D. Unbalanced forces change the ball’s speed and the direction of motion.
Answer:
D
Explanation:
Two charges, one +Q and the other −Q, are held a distance d apart. Consider only points on the line passing through both charges and clearly explain your answers to the following: [You can answer this problem without any calculations]. Do not consider any points at infinite distance from the charges. [5 points](a) Find the location of all points, if any, where the electric potential is zero.(b) Find the location of all points, if any, where the electric field is zero.
Answer:
a. d/2 mid-way between the charges.
b. d/2 mid-way between the charges.
Explanation:
(a) Find the location of all points, if any, where the electric potential is zero.
Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric potential due to the +Q charge is V = kQ/x and that due to the -Q charge is V' = -kQ/(d - x) where x is the point of zero electric potential.
The potential is zero when V + V' = 0, and this can only be midway between the charges. This is shown below
So, kQ/x + [-kQ/(d - x)] = 0
kQ/x - kQ/(d - x) = 0
kQ/x = kQ/(d - x)
1/x = 1/(d - x)
(d - x) = x
d = x + x
d = 2x
x = d/2 which is mid-way between the charges.
(b) Find the location of all points, if any, where the electric field is zero.
Since the charges are of equal magnitude and opposite charge and separated by a distance, d, the electric field due to the +Q charge is E = kQ/x² and that due to the -Q charge is E' = -kQ/(d - x)² where x is the point of zero electric field.
The electric field is zero when E + E' = 0 and this can only be midway between the charges. This is shown below.
So, kQ/x² + [-kQ/(d - x)²] = 0
kQ/x² - kQ/(d - x)² = 0
kQ/x² = kQ/(d - x)²
1/x² = 1/(d - x)²
(d - x)² = x²
d - x = ± x
d = x ± x
d = x - x or x + x
d = 0 or 2x
d = 0 or d = 2x
Since d ≠ 0, d = 2x ⇒ x = d/2 which is midway between the charges.
A truck driver is attempting to deliver some furniture. First , he travels 8 km east, and then he turns around and travels 3 km west. Finally, he turns again and travels 12 km east to his destination. a- what distance has the driver traveled? b- what is the drivers total displacement?
Jeni walks 100 meters east and then 50 meters north. How big is Jeni's displacement from the starting point?
a. 100 meters
b. 150 meters
c. 50 meters
d. About 112 meters
Answer:
d. About 112 meters.
Explanation:
From the question, John's displacement forms a right angle triangle as below.
Using Pythagoras theorem,
a² = b²+c²....................... Equation 1
Where a = John's displacement from the starting point, b = 100 m, c = 50 m
Substitite these values into equation 1
a² = 100²+50²
a² = 10000+2500
a² = 12500
a = √12500
a = 111.8 meters.
a = about 112 meters.
The right answer is d. About 112 meters.
You and a friend each hold a lump of wet clay. Each lump has a mass of 25 grams. You each toss your lump of clay into the air, where the lumps collide and stick together. Just before the impact, the velocity of one lump was < 4, 4, -3 > m/s, and the velocity of the other lump was < -2, 0, -7 > m/s.
What was the the total momentum of the lumps just before the impact?
p(total) = ____kg·m/s.
What is the momentum of the stuck-together lump just after the collision?
p = ____kg·m/s.
What is the velocity of the stuck-together lump just after the collision?
v_f = ____m/s.
Answer:
a) p(total) = <0.05, 0.1, 0.1 > kg m/s
b) p = <0.05, 0.1, 0.1 > kg.m/s
c) v_f = < 1, 2, 2 > m/s
Explanation:
a.)
Mass of each lump = 25 g = 0.025 kg
Velocity of lump 1 = < -2, 0, -7 > m/s
Momentum of lump 1 = Mass×Velocity
= 0.025×< -2, 0, -7 >
= < -0.05, 0, 0.175> kg m/s
Velocity of lump 2 = < 4, 4, -3 > m/s
Momentum of lump 2 = Mass×Velocity
= 0.025×< 4, 4, -3 >
= < 0.1, 0.1, -0.075> kg m/s
Total momentum before impact = < -0.05, 0, 0.175 > + < 0.1, 0.1, -0.075>
= < 0.05, 0.1, 0.1 > kg m/s
⇒p(total) = <0.05, 0.1, 0.1 > kg m/s
b)
As we know that,
By the law of conservation of linear momentum,
The total momentum will be the same before and after the collision.
⇒Momentum of the stuck together after the collision = Total momentum of the lumps just before impact.
⇒ p = <0.05, 0.1, 0.1 > kg m/s
c)
Let the final velocity = v_f
Total mass = 0.025 + 0.025 = 0.05 kg
As
Momentum = mass ×velocity
⇒ <0.05, 0.1, 0.1 > = 0.05 ×v_f
⇒ v_f = <0.05, 0.1, 0.1 > / 0.05
= < 1, 2, 2 > m/s
⇒v_f = < 1, 2, 2 > m/s
Assuming the speed of sound is 340 m/s, what is the most likely speed of the jet shown below?
Well we know it has to be greater than 300,000 km/s since we can't see it.
We can't calculate it any closer than that using the given information.
Draw a conclusion, based on the solubility curves shown above, of which compound would have the greatest
percentage recovered after cooling a saturated solution of that compound from 90°C to 30°C?
A) KCL
B) NaNO3
C) Nacl
D) KNO3
Answer: The answer is D. KNO3
Explanation:
The graph shows that the KN03 going straight up from the temperature sign so you reversed that so that it will make it to 90°C to 30°C
To solve this we must be knowing each and every concept related to solubility. Therefore, the correct option is option D among all the given options.
What is solubility?The greatest amount of one material that may be dissolved in the other is referred to as its solubility. It is the most solute that may be dissolved into a solvent near equilibrium, resulting in a saturated solution.
When specific circumstances are satisfied, more solute can be dissolved further than the solubility limit point, resulting in a supersaturated solution. Adding extra solute after saturation or supersaturation does not enhance the concentration in the solution. Rather, the excess solute begins to precipitated out of solution. KNO[tex]_3[/tex] is the compound that would have the greatest percentage recovered after cooling a saturated solution of that compound from 90°C to 30°C.
Therefore, the correct option is option D.
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A particle moves along the x-axis according to the equation (x=14-7t+t^2+t^3 ), where (x) in meter and (t) in seconds. At (t=7 sec) Find (a) The position of the particle (b) It’s velocity (c) It’s acceleration
Answer:
jjnn ok jjjmkkmmkijnnkko
Balanced forces acting on an object keeps it at ____or moving at _____in straight line.
Fill in the ___ spaces
Answer:
Please see below as the answer is self explanatory.
Explanation:
According to Newton's 2nd law, a net force acting on an object of mass m, causes the object to be accelerated.If the forces acting on the object are balanced, which means that the net force on the object is zero, just applying the same law, we find that the object is not accelerated.According to Newton's First law, an object that is not accelerated is at rest, or moves along a straight line at constant speed.So, if there are balanced forces acting on the object, if the object is at rest, will keep at rest, and if it is moving, it will keep moving at constant speed along a straight line.A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two balls after collision?
Answer:
We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.
Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.
Explanation:
Standing at a crosswalk, you hear a frequency of 530 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 424 Hz. Determine the ambulance's speed from these observations.
Answer:
_s = 37.77 m / s
Explanation:
This is an exercise of the Doppler effect that the change in the frequency of the sound due to the relative speed of the source and the observer, in this case the observer is still and the source is the one that moves closer to the observer, for which relation that describes the process is
f ’= f₀ [tex]\frac{v}{v - v_s}[/tex]
where d ’= 530 Make
when the ambulance passes away from the observer the relationship is
f ’’ = f₀ [tex]\frac{v}{v + v_s}[/tex]
where d ’’ = 424 beam
let's write the two expressions
f ’ (v-v_s) = fo v
f ’’ (v + v_s) = fo v
let's solve the system, subtract the two equations
v (f ’- f’ ’) - v_s (f’ + f ’’) = 0
v_s = v [tex]\frac{ f' - f''}{ f' + f''}[/tex]
the speed of sound is v = 340 m / s
let's calculate
v_s = 340 [tex](\frac{ 530 -424}{530+424} )[/tex]
v_s = 340 [tex](\frac{106}{954}[/tex])
v_s = 37.77 m / s
The variable ______________ describes how quickly something moves.
it's up in Gogle trust me
A ball is thrown vertically downward from the top of a 37.4-m-tall building. The ball passes the top of a window that is 15.4 m above the ground 2.00 s after being thrown. What is the speed of the ball as it passes the top of the window?
Answer:
v= 20.8 m/s
Explanation:
Assuming no other forces acting on the ball, from the instant that is thrown vertically downward, it's only accelerated by gravity, in this same direction, with a constant value of -9.8 m/s2 (assuming the ground level as the zero reference level and the upward direction as positive).In order to find the final speed 2.00 s after being thrown, we can apply the definition of acceleration, rearranging terms, as follows:[tex]v_{f} = v_{o} + a*t = v_{o} + g*t (1)[/tex]
We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:[tex]\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)[/tex]
where Δy = yf - y₀ = 15.4 m - 37.4 m = -22 m (3)Replacing by the values of Δy, a and t, we can solve for v₀ as follows:[tex]v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)[/tex]
Replacing (4) , and the values of g and t in (1) we can find the value that we are looking for, vf:[tex]v_{f} = v_{o} + g*t = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)[/tex]
Therefore, the speed of the ball (the magnitude of the velocity) as it passes the top of the window is 20.8 m/s.Show two data points from your simulation that demonstrate this behavior.
I1 V1 I2= 2I1 V2=2V1 V1/ I1 =V2/I2
For the light bulb, why is it better to take more measurements in the range 20mA < I < 40mA, instead of just taking equally spaced measurements in the entire range of 0 mA < I< 55mA
Answer:
hello your question is incomplete attached below is the complete and the required circuit diagrams
answer :
Ai) This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well
B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature
hence At 0 mA current, there won't be any noticeable change
Explanation:
Ai) The voltage across the resistor will double when you double the current through the resistor
Given that : V = I*R.
lets assume : I = 2 amperes , R = 3 ohms
V = 2*3 = 6 v
secondly lets assume double the value of (I) i.e. I = 4 amperes
hence : V = 4*3 = 12 volts
This proves that when the current across the resistor is doubled the value of the voltage across the resistor doubles as well
Aii) Showing the two data points from simulation
I1 V1 I2= 2I1 V2=2V1 V1/ I1 =V2/I2
0.9*10^3 9 * 10^3 1.8*10^3 18*10^3 10 ohms
1.6 * 10^3 16 * 10^3 3.2*10^3 32*10^3 10 ohms
B) It is better to take more measurements in the range 20mA < I < 40mA because of the amount of temperature reached by the bulb and the change in resistance is affected by the temperature
hence At 0 mA current, there won't be any noticeable change
What force is needed to give a 4800.0 kg truck an acceleration of 6.2 m/s2 over a level road?
Answer:
the force needed to give the truck the acceleration is 29,760 N.
Explanation:
Given;
mass of truck, m = 4800 kg
acceleration of the truck, a = 6.2 m/s²
The force needed to give the truck the acceleration is calculated as;
F = ma
F = 4800 x 6.2
F = 29,760 N
Therefore, the force needed to give the truck the acceleration is 29,760 N.
Which graph best represents the greatest amount of work
The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8 in. to the final position x2 = 5 in. determine (a) the work done on the cart by the spring and (b) the work done on the cart by its weight.
This question is incomplete, the missing diagram is uploaded along this Answer below.
Answer:
a) the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight is - 3.935 lb-ft
Explanation:
Given the data in the question;
(a) determine the work done on the cart by the spring
we calculate the work done on the cart by the spring as follows;
[tex]W_{spring}[/tex] = 1/2×k( [tex]x^{2} _{1}[/tex] - [tex]x^{2} _{2}[/tex] )
where k is spring constant ( 3 lb/in )
we substitute
[tex]W_{spring}[/tex] = 1/2 × 3( (-8)² - (5)² )
[tex]W_{spring}[/tex] = 1/2 × 3( 64 - 25 )
[tex]W_{spring}[/tex] = 1/2 × 3( 39 )
[tex]W_{spring}[/tex] = 58.5 lb-in
we convert to pound force-foot
[tex]W_{spring}[/tex] = 58.5 × 0.0833333 lb-ft
[tex]W_{spring}[/tex] = 4.875 lb-ft
Therefore, the work done on the cart by the spring is 4.875 lb-ft
b) the work done on the cart by its weight
work done by its weight;
[tex]W_{gravity}[/tex] = -mgsin∅( x₂ - x₁ )
we substitute in of values from the image below;
[tex]W_{gravity}[/tex] = -14 × sin(15°)( 5 - (-8) )
[tex]W_{gravity}[/tex] = -14 × 0.2588 × 13
[tex]W_{gravity}[/tex] = -47.1 lb-in
we convert to pound force-foot
[tex]W_{gravity}[/tex] = -47.1 × 0.0833333 lb-ft
[tex]W_{gravity}[/tex] = - 3.935 lb-ft
Therefore, the work done on the cart by its weight is - 3.935 lb-ft
a) the work done on the cart by the spring is 4.875 lb-ft.
b) the work done on the cart by its weight is - 3.935 lb-ft.
Calculation of the work done:a. The work done on the cart by the spring is
= 1/2 × 3( (-8)² - (5)² )
= 1/2 × 3( 64 - 25 )
= 1/2 × 3( 39 )
= 58.5 lb-in
Now we have to convert to pound force-foot
So,
= 58.5 × 0.0833333 lb-ft
= 4.875 lb-ft
b) Now
work done by its weight;
= -mgsin∅( x₂ - x₁ )
So,
= -14 × sin(15°)( 5 - (-8) )
= -14 × 0.2588 × 13
= -47.1 lb-in
Now we convert to pound force-foot
= -47.1 × 0.0833333 lb-ft
= - 3.935 lb-ft
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Describe effective communication strategies for gathering information, educating patients
Explanation:
Effective communication with patients will enable one to know the needs of the patient better as well as reducing the barriers to understanding each other for both parties.
To be an effective communicator while educating patients, the person must:
It is important to establish good rapport with the patient. By so doing they can trust you and let you in. Show empathy. Do not make them feel like you are judging themUse proper body language. Make eye contacts and try to be on the same level as the patient so you can be face to face with them.make the interaction easier for them. You have to keep questions as well as your sentences short and moderate. Stay on topic and always make sure that concepts are clear to them.show respect. try not to speak with commands. Give the patient opportunity to make choices.be patient with them. Due to age or the nature of their illnesses, the patient may be slow in speech or movement. help them to move at their own pace by not rushing them.give them time to respond and ask questions. this will make communication more effective.you cause graphics where necessary or written instructions for the patient.How would you feel if everyone hated you?
Answer:
awful
Explanation:
can you help me with my question Which of the themes of Hawthorne's "Dr. Heidegger's Experiment" is illustrated by this passage? Paragraph 36: The most singular effect of their gayety was an impulse to mock the infirmity and decrepitude of which they had so lately been the victims. A. Gayety produces an impulse to mock. B. It is a great release to look back on our problems happily C It is human nature to mock or make light of problems we have been delivered from
What did people assume Katherine was when she entered the room?
Answer: custodian,
Explanation: they never saw any colored women in the division before
toy car A drives with a steady force of 35N and covers 2000 m with fully charged battery. toy car B drives with a steady force of 80 N. how far would it be able to drive using the same fully charged battery as car A.
The distance travelled by toy car B using the same fully charged battery as car A is 875 m
How to determine the energy of car AForce (F) = 35 NDistance of car A (d) = 2000 mEnergy (E) = ?E = fd
E = 35 × 2000
E = 70000 J
How to determine the distance travelled by car BEnergy (E) = 70000 JForce (F) = 80 NDistance of car B =?E = fd
70000 = 80 × Distance of car B
Divide both sides by 80
Distance of car B = 70000 / 80
Distance of car B = 875 m
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if a car travels 200 m to the east in 8.0 s what is the cars average velocity?
Answer:
25 m/s
Explanation:
200/8 = 25
.................,,,,,,,,,,,
Answer:
B
Explanation:
Motion is movement, the teacher's movement is motion
what is diffrence between damping and undamping?
Answer:
Oscillation whose amplitude reduce with time are called damped oscillation. This happen because of the friction. In oscillation if its amplitude doesn't change with time then they are called Undamped oscillation
Damped and undamped vibration refer to two different types of vibrations. The main difference between damped and undamped vibration is that undamped vibration refer to vibrations where energy of the vibrating object does not get dissipated to surroundings over time, whereas damped vibration refers to vibrations where the vibrating object loses its energy to the surroundings.
A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back up 1.50 m before falling to the ground. If the mass of the coconut is
2.00 kg, calculate the potential energy of the coconut relative to the ground at each of the following sites:
(a) while it is still in the tree,
(b) when it hits the bystander on the head,
(c) when it bounces up to its maximum height,
(d) when it lands on the ground,
(e) when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole.
Answer:
A. 240 J
B. 60 J
C. 90 J
D. 0 J
E. 50 J
Explanation:
A. Determination of the potential energy of the coconut while it is still in the tree
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 12 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 12
PE = 240 J
B. Determination of the potential energy of the coconut when it hits the bystander on the head,
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 3
PE = 60 J
C. Determination of the potential energy of the coconut when it bounces up to its maximum height,
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 3 + 1.5 = 4.5 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 4.5
PE = 90 J
D. Determination of the potential energy of the coconut when it lands on the ground,
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 0 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 0
PE = 0 J
E. Determination of the potential energy of the coconut when it rolls into a ground hole, and falls 2.50 m to the bottom of the hole.
Mass (m) = 2 Kg
Acceleration due to gravity (g) = 10 m/s²
Height (h) = 2.50 m
Potential energy (PE) =.?
PE = mgh
PE = 2 × 10 × 2.50
PE = 50 J
(a) The potential energy of the coconut relative to the ground while it is still in the tree is 235.2 J.
(b) The potential energy of the coconut relative to the ground when it hits the bystander on the head is 58.8 J.
(c) The potential energy of the coconut relative to the ground when it bounces up to its maximum height is 88.2 J.
(d) The potential energy of the coconut relative to the ground when it lands on the ground is 0 J.
(e) The potential energy of the coconut when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole is 49 J.
The given parameters;
height of the tree, h = 12 mheight of the bystander, h' = 3 mheight it bounced back = 1.5 mmass of the coconut, m = 2.0 kgThe potential energy of the coconut relative to the ground while it is still in the tree;
[tex]P.E = mgh\\\\P.E = 2 \times 9.8 \times 12\\\\P.E = 235.2 \ J[/tex]
The potential energy of the coconut relative to the ground when it hits the bystander on the head;
[tex]P.E = 2 \times 9.8 \times 3 \\\\P.E = 58.8 \ J[/tex]
The potential energy of the coconut relative to the ground when it bounces up to its maximum height;
[tex]P.E = 2 \times 9.8 (1.5 + 3)\\\\P.E = 88.2 \ J[/tex]
The potential energy of the coconut relative to the ground when it lands on the ground;
[tex]P.E = 2 \times 9.8 \times 0\\\\P.E = 0 \ J[/tex]
The potential energy of the coconut relative to the ground when it rolls into a groundhog hole, and falls 2.50 m to the bottom of the hole;
[tex]P.E = 2\times 9.8 \times 2.5 \\\\P.E = 49 \ J[/tex]
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