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.

Answers

Answer 1

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

Answer 2

(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 kg

The 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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Related Questions

A airplane accelerated from 59 m/s to 95 m/s in a distance of 123 meters what was its acceleration in m/s^2, assumed constant?

Answers

Answer:

22.54 m/s^2

Explanation:

vf = final velocity = 95 m/s

vi = initial velocity = 59 m/s

d = displacement = 123 m

a = acceleration, unknown

Use this kinematics equation to find a:

vf^2 = vi^2 + 2a*d

95^2 = 59^2 + 2*a*123

22.54 m/s^2 = a

Hope this helps!! :)

4. Draw conclusions: What is the minimum energy required to break the egg?
.

Answers

Answer:

0.25 J

Explanation:

The strength of the egg shell, the size of the egg, and the force used to break it are just some of the variables that affect how much energy is needed to crack an egg. When an object hits the egg with an impact energy of 12–26 mJ, cracks occur.

What is energy?

The capacity of a system or object to do work is called energy, which is a fundamental term in physics. Kinetic energy, potential energy, heat energy, electromagnetic energy, and nuclear energy are just a few of the different forms of energy.

While potential energy is the energy possessed by an object as a result of its position or position, kinetic energy is the energy of motion. While electromagnetic energy is energy carried by electromagnetic waves like light, thermal energy is energy related to the temperature of a substance. The energy stored in the nucleus of an atom is called nuclear energy.

Therefore, when an object hits the egg with an impact energy of 12–26 mJ, cracks occur.

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A squirrel drops an acorn onto the head of an unsuspecting dog. The acorn falls 20 m before it lands on the dog. Ignore air resistance. How many seconds did the acorn fall?

Answers

Answer:

2 seconds

Explanation:

From the question,

Applying the equation of motion for a body falling under gravity

s = ut+1/2gt²................................. Equation 1

Where, s = hieght of fall, u = initial velocity, t = time, g = acceleration due to gravity.

Given: s = 20 m, u = 0 m/s, t = ?, g = 10 m/s²

Substitute these values into equation 1

10 = 0(t²)+1/2(10×t)

10 = 5t

Solve for t

5t/5 = 10/5

t = 2 seconds

When talking about the products of photosynthesis, why is it a HUGE issue when a vast amount of the rainforests is dying along with vast amounts of forests from fires/deforestation? Remember, we know that this hurts the ecosystems of animals living there, but what else does it effect?
Question 5 options:


the creation of too much smoke


the creation of carbon dioxide


the creation of carbon monoxide


the creation of oxygen

Answers

Answer:

the creation of oxygen

Explanation: i just took the k12 test :)

Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces off a cement floor, and in Case 2, the ball bounces off a piece of stretchy rubber. Two cases of a ball dropping and bouncing off of a surface. In case one, the ball approaches a cement floor, comes in contact, momentarily comes to rest, and then bounces off with a velocity smaller than the approach velocity. In case 2, the ball approaches a piece of stretchy rubber. When the ball comes in contact with the rubber, it deforms the rubber and comes momentarily to rest. The ball bounces off again with a velocity smaller than the approach velocity, and the rubber regains its original shape. In both cases, the balls are dropped from the same initial height and reach the same final height. In which case is the magnitude of the ball's change in momentum the greatest

Answers

Answer:

the impulse must be the same in these two cases    F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

Explanation:

For this exercise we use the relationship between momentum and momentum

         I = Δp

         F t = m v_f - m v₀

To know the speed we use the conservation of energy

starting point. Highest point

       Em₀ = U = m g h

fincla point. Just before the crash

      Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

        m g h = ½ m v²

         v = [tex]\sqrt{2gh}[/tex]

we substitute in the impulse relation

     F t = m ([tex]\sqrt{2g h_f } - \sqrt{2g h_o}[/tex])

therefore we can see that as in case the initial and final heights are equal, the impulse must be the same in these two cases

A liquid fueled rocket is red on a test stand. The rocket nozzle has an exit diameter of 30 cm and the combustion gases leave the nozzle at a velocity of 3800 m/s and a pressure of 100 kPa, which is the same as the ambient pressure. The temperature of the gases in the combustion area is 2400 C. Find (a) the temperature of the gases at the nozzle exit plane, (b) the pressure in the combustion area, and (c) the thrust developed. Assume that the gases have a speci c heat ratio of 1.3, and a molar mass of 9. Assume that the ow in the nozzle is isentropic.

Answers

Answer:

1. Temperature= 869.35 K

2. Pressure of combustion = 12994.043 kpa

3. Thrust = 127x10⁶N

Explanation:

this problem has been fully explained in the attachment. please use it to get a clearer explanation of the answer.

1.

The temperature = (273+2400k) - (3800)²/2(4003)

= 2673 - 14440000/8006

= 2673 - 1803.65

= 869.35 K

Approximately 869.4K

2. We first get mach number

= 3800/√1.3(923.8)(869.35)

= 3800/1021.78

= 3.719

Pressure = 100kpa[1+2.07464415]^1.3/0.3

= 12995.043kpa

C. Thrust

Pi/4(3800)²(0.3)²(100x10³)/(923.8)(869.4)

= 12678.621

= 126.781 kN

Thrust is approximately 127kN = 127x10⁶N


How could being mindful in conversation be helpful?

Answers

You can see if the person your talking to and being mindfull they  will see respect towards you or they see you as a mindful person.It shows a person that you care.

Explanation:

what's the speed of the man who runs 10 miles in 2hrs?​

Answers

The man would be running at 5 miles per hour.
What you’re looking for is:
5 miles per hour

what is the relation of pressure of a liquid with its depth and density?​

Answers

Answer:

★ Pressure and depth have a directly proportional relationship. This is due to the greater column of water that pushes down on an object submerged. Conversely, as objects are lifted, and the depth decreases, the pressure is reduced.

Explanation:

Hope you have a great day :)

A makeshift sign hangs by a wire that is extended over an ideal pulley and is wrapped around a large potted plant on the
roof as shown in the figure below. When first set up by the shopkeeper on a sunny and dry day, the sign and the pot are in
equilibrium. The mass of the sign is 27.5 kg, and the mass of the potted plant is 67.5 kg.
Plant
sale
today!
(a) Assuming the objects are in equilibrium, determine the magnitude of the static friction force experienced by the
potted plant.
N
(b) What is the maximum value of the static friction force if the coefficient of static friction between the pot and the
roof is 0.707?
N

Answers

Answer:I know the answer for B cus I’m doing the same problem. For B, you would only take the coefficient of friction given and then multiply it by the Normal Force, which in this case is the same as the Gravitational Force.

Explanation:

We assume the foam plate has a positive charge when rubbed with paper towels.

Lift the pan away from the charged plate using the styrofoam cup. Briefly touch the rim of the pan to neutralize it. Place the neutralized pan on the plate and observe the tape rise. When the pan is on the plate, the rim of the plate has a _____________. This means that the pan base is ________________ charged because the net charge on the pan is __________. You know that this must be the case because as you lift the pan with the cup away from the plate, the tape on the rim goes down.

Answers

Answer:

POSITIVE CHARGE,  NEGATIVE CHARGE,  ZERO

Explanation:

To solve this completion exercise, we must remember that charges of the same sign repel each other and in a metallic object (frying pan) the charge is mobile.

Let's analyze the situation when we touch the pan, the charges are neutralized, therefore when we bring the pan to the plate that has a positive charge, it attracts the mobile negative charges in the pan, until it is neutralized, therefore on the opposite side of the pan. pan (edge ​​with a glued tape) is left with a positive charge; therefore the edge and the tape, which is very light, have positive charges and repel each other.

We must assume that the frying pan is insulated so that the net charge is zero, since the induction process.

Consequently the words to complete the sentence are

When the pan is on the plate, the edge of the plate has a _POSITIVE CHARGE_____.

This means that the base of the container is loaded NEGATIVE CHARGE_____ because the net charge of the container is ___ZERO_

Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.

Answers

Answer:It will take about 3000 years

Explanation:

PLEASE HELP ASAP WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST!!!

What 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right?
(Select 2 of the choices below)

A. gravity

B. normal force

C. air resistance

D. friction

Answers

Answer:

Friction and Air resistence

Explanation:

i already passed this grade years ago...

The forces responsible for exerting forces to the left on a cyclist are -

Air resistanceFriction

What is air resistance?

Air resistance describes the forces that are in opposition to the relative motion of an object as it passes through the air. We can write air resistance as -

[tex]$F_{D}=\frac{1}{2} \rho v^{2} C_{D} A[/tex]

where -

F{D} = drag

{ρ} = density of fluid

{v} = speed of the object relative to the fluid

C{D}  = drag coefficient

{A} = cross sectional area

Given is to find what 2 forces would be responsible for exerting forces to the left on a cyclist that is already in motion moving to the right.

The forces responsible for exerting forces to the left on a cyclist are -

Air resistanceFriction

Therefore, the forces responsible for exerting forces to the left on a cyclist are -

Air resistanceFriction

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if a body of mass m is placed on earth ,what is the amount of potential energy possessed by it (g:-9.8m/s​

Answers

Answer:

mgh

Explanation:

Assume the height of the body is 1.8m.

The gravity?of the body is G=mg

the height of the gravity center is about 0.9m

E=mgh

=m*9.8m/s*0.9m

= 8.82mJ

Integrate your expressions for dEx and dEy from θ=0 to θ=π. The results will be the x-component and y-component of the electric field at P

.

Express your answers separated by a comma in terms of some, all, or none of the variables Q

and a and the constants k and π.

Answers

Answer:

hello your question is incomplete below is the missing part

Ex = 0

Ey = [tex]\frac{-2kQ}{\pi a^2}[/tex]

Explanation:

Attached below is a detailed solution showing the integration of the expression dEx and dEy from ∅ = 0 to ∅ =π

Ex = 0

Ey = [tex]\frac{-2kQ}{\pi a^2}[/tex]

A 450.0 kg roller coaster is traveling in a circle with radius 15.0m. Its speed at point A is 28.0m/s and its speed at point B is 14.0 m/s. At point A the cart is already moving with circular motion. a) Draw free bodydiagramsfor the cartatpointsAand B(two separate free body diagrams). b) Calculate the acceleration of the cartat pointsAandB(magnitude and direction). c) Calculate the magnitude of the normal force exerted by the trackson the cartat point A. d) Calculate the magnitude of the normal force exerted by the tracks on the cart at point B.

Answers

Answer:

b)  a = 52.26 m / s², a ’= 13.06 m / s², c) N = 2.79 10⁴ N, d) N = 1.89 10³ N

Explanation:

a) In the attached we can see the free body diagrams for the two positions, position A in the lower part of the circle and position B in the upper part of the circle

b) Let's start at point A

Let's use that the acceleration is centripetal

           a = v² / r

let's calculate

            a = 28² / 15.0

            a = 52.26 m / s²

as they relate it is centripetal it is directed towards the center of the circle, therefore for this point it is directed vertically upwards

Point B

           a ’= 142/15

           a ’= 13.06 m / s²

in this case the acceleration is vertical downwards

c) The values ​​of the normal force

point A

let's use Newton's second law

           ∑ F = m a

           N- W = m a

           N = mg + ma

           N = m (g + a)

           N = 450.0 (9.8 + 52.25)

           N = 2.79 10⁴ N

d) Point B

            -N -W = m (-a)

             N = ma -m g

             N = m (a-g)

             N = 450.0 (14.0 - 9.8)

             N = 1.89 10³ N

what are the importance of regulare
health examination.​

Answers

Answer:

The purpose of regular health examination is to evaluate health status, screen for risk factors and disease, and provide preventive counseling interventions. The major benefits of regular health examination is early detection of treatable disease.

Steve is planning his annual Spring Break road trip. He pulls out his map and draws out his route to visit the five locations that he has planned for this year. They go in a counterclockwise loop and he ends up at home, where he started, just in time to start classes again. Whenever he is on the road he travels a constant 60 miles/hour. When Steve adds up the total distance traveled, as measured by his odometer, and divides it by the time that his trip took, he has measured what quantity?

a. His average velocity.
b. His average speed.
c. His instantaneous velocity.
d. His instantaneous speed.

Steve’s average velocity for the whole trip is:______

a. greater than 60 miles/hour.
b. equal to 60 miles/hour.
c. less than 60 mile/hour, but greater than zero.
d. exactly zero.

Answers

Answer:

Part 1

Steve is measuring his average speed

Part 2

Average velocity is equal to 60 miles per hour

Explanation:

Part 1

Average velocity is equal to total distance travelled divided by total time taken. It also takes into consideration the change of direction through out the journey.

Hence, Steve is measuring his average speed

Option A is correct

Part 2

Average velocity is equal to 60 miles per hour only because velocity is a vector quantity

Option B is correct

reason why the center of gravity must not be at 50cm​

Answers

Answer:

hope this helps

hope this is what u want

A 3.0-kilogram mass is traveling in a circle of
0.20-meter radius with a speed of 2.0 meters per
second. What is its centripetal acceleration?
(1) 10. m/s
(3) 60. m/s2
(2) 20. m/s2
(4) 6.0 m/s2

Answers

Answer:

[tex]a=20\ m/s^2[/tex]

Explanation:

Given that,

The mass of an object, m = 3 kg

The radius of a circle, r = 0.2 m

The speed of the object, v = 2 m/s

We need to find the centripetal acceleration. Its formula is given by :

[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{0.2}\\\\a=20\ m/s^2[/tex]

So, the centripetal acceleration is [tex]20\ m/s^2[/tex].

You are standing on the bottom of a lake with your torso above water. Which statement is correct?

a. You feel a buoyant force only when you momentarily jump up from the bottom of the lake.
b. There is a buoyant force that is proportional to the weight of your body below the water level.
c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.
d. There is no buoyant force on you since you are supported by the lake bottom.

Answers

Answer:

c. There is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Explanation:

Buoyancy can be defined as a force which is created by the water displaced by an object.

Simply stated, buoyancy is directly proportional to the amount of water that is being displaced by an object.

Hence, the greater the amount of water an object displaces; the greater is the force of buoyancy pushing the object up.

The buoyancy of an object is given by the formula;

[tex] Fb = pgV [/tex]

[tex] But, \; V = Ah [/tex]

[tex] Hence, \; Fb = pgAh [/tex]

Where;

Fb = buoyant force of a liquid acting on an object.

g = acceleration due to gravity.

p = density of the liquid.

v = volume of the liquid displaced.

h = height of liquid (water) displaced by an object.

A = surface area of the floating object.

The unit of measurement for buoyancy is Newton (N).

In this scenario, you are standing on the bottom of a lake with your torso above water. Thus, there is a buoyant force that is proportional to the volume of your body that is below the level of the water.

Mr. Voytko wants to know how high in meters he can lift an 0.3 kg apple with 7.35 joules?

Answers

Answer:

the height above the ground through Mr. Voytko lifted the apple is 2.5 m.

Explanation:

Given;

energy of Mr. Voytko, E = 7.35 J

mass of the apple, m = 0.3 kg

Apply the principle of conservation of energy.

Energy of Mr. Voytko = Potential energy of the apple due to its height above the ground.

E = mgh

where;

h is the height above the ground through Mr. Voytko lifted the apple.

g is acceleration due to gravity = 9.8 m/s²

h = E / (mg)

h = 7.35 / (0.3 x 9.8)

h = 2.5 m

Therefore, the height above the ground through Mr. Voytko lifted the apple is 2.5 m.

231 91 Pa has__neutrons

Answers

Answer:

140

Explanation:

A 69.0 kg ice skater moving to the right with a velocity of 2.61 m/s throws a 0.22 kg snowball to the right with a velocity of 25.2 m/s relative to the ground. (a) What is the velocity of the ice skater after throwing the snowball

Answers

Answer:

0.08m/s

Explanation:

Given data

M1= 69kg

v1= 2.61m/s

M2= 0.22kg

v2= 25.2m/s

Before snowball is thrown:

Total mass of skater + snowball = 69+ 0.22 = 69.22kg

Total Momentum of skater + snowball = mv = 69.22 x 2.61 = 180.7 kgm/s

After snowball is thrown:

Let's call the velocity of the skater V.

Total momentum = momentum of skater + momentum of snowball

=69.22V + (5.544)

= 69.22V + 5.544

So:

180.7  = 69.22V+5.544

180.7- 5.544= 69.22V

175.156= 69.22V

V= 175.156/69.22

V = 2.53m/s

The total momentum after catching the snowball is mV or:

(69.0 + 0.22) x V

So:

5.544= 69.22V

V= 5.544/69.22

V=0.08m/s

The velocity of the ice skater after throwing the snowball is 0.08m/s

A 10-kg rock falls from a height of 8-m above the ground. What is the kinetic energy of the rock just before it hits the ground?

Answers

Answer: 800

Explanation:

1/2 x m x v^2 = m x g x h

KE = 10 x 10 x 8

KE= 800

The energy of the body by the virtue of its motion is known as the kinetic energy of the body. The kinetic energy of the rock just before it hits the ground will be 784.8 J.

What is kinetic energy?

The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.

According to the law of conservation of energy, energy can not be created nor be destroyed can be transferred from one form to another form.

Kinetic energy= potential energy

Kinetic energy= mgh

Kinetic energy= 10×9.81×8

Kinetic energy=784.8 J

Hence the kinetic energy of the rock just before it hits the ground will be 784.8 J.

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HELP PLS!
(LOOK AT THE PICTURE)

Answers

The answer is b !!!! Hope it helps

a. Using the ideas of electric field and force, explain what would happen to an electron if released from rest at r=2.0m?
b. Would the electron released from rest move to a region of higher electrical potential or lower electrical potential?
c. Would the electron released from rest move such that the system would have higher potential energy or lower potential energy?

Answers

I’m pretty sure it’s C

What is any push or pull on an object called?

Answers

Answer:

A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects.

Explanation:

man is walking due east at the rate of of 4kmph and the rain is falling 30° east of vertical with a velocity of 6kmph the velocity of rain relative to the man will be?

Answers

Answer:

No answer

Explanation:

no explanation

A cylinder containing the air comprises the systemm. Cycle is completed as follows : (i) 82000 N-m of work is done by the piston on the air during compression stroke and 45 kJ of heat are rejected to the surroundings. (ii) During expansion stroke 100000 N-m of work is done by the air on the piston. Calculate the quantity of heat added to the system؟?​

Answers

Answer & Explanation:

1 N-m = 1 Joule

So 82 kJ of energy put into the system during (i).

45 kJ of heat leaves the system, so 82 kJ - 45 kJ  = 37 kJ is remaining.

(ii) requires 100 kJ of energy but only 37 kJ is available, so 100 kJ - 37 kJ = 63 kJ of heat energy must be added to the system.

The heat given would be equal to the heat emitted from the system and by providing some external source of energy the volume or temperature of the system may increase.

The amount of heat added to the system is 63kJ.

The energy can be estimated as:

Given,

Work done by piston = 82000 Nm

Heat rejected in surrounding = 45 kJ

Work done during expansion stroke = 100000 Nm

Quantity of added heat = ?

During compression stroke:

Work done by the piston [tex]\rm (W_{1-2})[/tex] = - 82000Nm or - 82kJ

Heat rejected to the system [tex]\rm (Q_{1-2})[/tex] = - 45kJ

We know that,

[tex]\rm Q_{1-2} = (U_{2} - U_{1}) + W[/tex]

Therefore,

[tex]\begin{aligned}-45 &= \rm (U_{2} - \rm U_{1}) + (-82)\\\\\rm (U_{2}-U_{1}) &= 37\rm (U_{2} - U_{1}) = 37\; kJ \end{aligned}[/tex]          (equation 1 )

During Expansion system:

Work done by the piston [tex]\rm (W_{2-1})[/tex] = 100000 Nm or 100 kJ

Now putting values in the equation:

[tex]\begin{aligned}\rm Q_{2-1} &= \rm U_{1} - U_{2} + W\\\\&=\rm (U_{1} - U_{2}) + W\end{aligned}[/tex]

Substituting value from equation 1:

[tex]\begin{aligned}\rm Q_{2-1} &= - 37+100\rm \;kJ\\&= 63\rm \; kJ\end{aligned}[/tex]

Therefore, 63kJ of energy is added to the system.

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