1. A roller coaster with a mass of 800 kg sits stationary at the top of a section of track, 75 m above

the ground as shown. When the brake is released, it starts to roll down the track


2. For each height represented in the diagram, calculate the gravitational potential energy using

Ep = mgh. Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for

each of the heights of the roller coaster. (6 marks)


3. Assuming there is no friction, determine the mechanical kinetic energy using Ek = Etotal - Ep.

Show ONE SAMPLE calculation in the calculations section below and fill in Table 1 for each of

the heights of the roller coaster. (6 marks)


4. For each height represented in the diagram, calculate the velocity using = �2

. Show ONE

SAMPLE calculation in the calculations section below and fill in Table 1 for each of the heights of

the roller coaster. (6 marks)


5. Use your answers to graph how gravitational potential energy, mechanical kinetic energy, and

velocity change as the roller coaster changes height. Use different colours for the three lines on

the graph. Graph paper is provided below. (3 marks)


6. Repeat steps 1 – 5 above for a roller coaster cart that has a mass of 300 kg and enter your

results in Table 2.

Calculations:

800 kg roller coaster cart:

Sample calculation for gravitational potential energy:

Sample calculation for Mechanical kinetic energy:

Sample calculation for velocity:


300 kg roller coaster cart:

Sample calculation for gravitational potential energy:

Sample calculation for mechanical kinetic energy:

Sample calculation for velocity:


Results:

Table 1: Potential energy, kinetic energy, total energy, and velocity of the 800 kg roller coaster cart


Table 2: Potential energy, kinetic energy, total energy, velocity of the 300 kg roller coaster cart.


Graphs:

It’s graphing time. These graphs are a bit different than the ones you did in the

data analysis assignment at the beginning of the course. In this case you have

three things to graph on each graph. (One graph for the 800 kg roller coaster cart

and one graph for the 300 kg roller coaster cart.) You need to graph the

gravitational potential energy with respect to height, the mechanical kinetic

energy vs height, and the velocity vs height.

Let’s look at the energy graphs first. In this case both kinetic energy and

mechanical energy cover the same range of values. This means they can use the

same scale on the y-axis. So, you will use the left y-axis and the x-axis to graph

the kinetic energy vs height and the potential energy vs height. You will need a

legend to explain which line is which. Colour coding is a nice way to highlight this.

The velocity values are much different than the energy values. This means you

need a totally different scale. So, your left y-axis won’t work. You need to make a

second scale on the right y-axis for your velocity values. You will plot the points

the same way as normal, but you will use the numbers on the right-hand scale

instead. Again, be sure to add your velocity line to the legend with a separate

colour code.


Discussion Questions:

1. Describe the relationship between the gravitational potential energy and the mechanical kinetic

energy of the roller coaster on your graph. (2 marks)

2. Describe the shapes of each of the three lines in the graph. Explain why the velocity is different.

(4 marks)

3. Describe how mass affects the speed at the bottom of the roller coaster. (2 marks)

4. Describe how mass affects the gravitational potential energy and the mechanical kinetic energy

of the roller coaster. (2 marks)

5. At what point does the roller coaster have a maximum value for the following? Justify your

answer by explaining why. (2 marks each)

a. Gravitational potential energy

b. Mechanical energy

c. Velocity

6. In your calculations, you assumed that the roller coaster was frictionless. All real roller coasters

encounter friction. Describe how the actual values of the variables would differ, or not differ,

from your calculated values for a real roller coaster. (Hint: what form of energy would some of

the total energy be converted to if there was friction in the system?) (4 marks)

How you will be graded:

Grades will be based on answering questions to demonstrate an understanding of the material covered

in this unit. Point form answers are okay if ideas are complete and use vocabulary (Word Bank)

provided. For questions out of 4 marks, there are 4 responses expected.

Yes, it is possible to determine the final speed of the rollercoaster if the initial speed and the height to be reached are known.

According the principle of conservation of energy, the total kinetic energy is equal to the total potential energy.

[tex]P.E _i + K.E_i = P.E_f + K.E_f \\\\mgh_i + \frac{1}{2} mv_i^2 = mgh_f + \frac{1}{2} mv_f^2\\\\gh_i + \frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\g(0) + \frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\\frac{1}{2} v_i^2 = gh_f + \frac{1}{2} v_f^2\\\\v_i^2 = 2gh_f + v_f^2\\\\v_i^2 - 2gh_f = v_f^2\\\\v_f = \sqrt{v_i^2 - 2gh_f}[/tex]

where;

Thus, it is possible to determine the final speed of the rollercoaster if the initial speed and the height to be reached are known.

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Answer:

a

Explanation:

i took the test 100%

Answer:

Allah asked Abraham(Ibrahim)

Answer:

c. Impulse

Explanation:

The total charge in this time is equal to 144 Coulomb.

Given the following data:

To calculate the total charge in this time:

First of all, we would convert the value of time in minutes to seconds.

Conversion:

1 minute = 60 seconds

3 minutes = 180 seconds

Mathematically, the quantity of charge is given by this formula:

[tex]Q=It[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q = 0.8 \times 180[/tex]

Quantity of charge, Q = 144 Coulomb.

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Answer:

960 watts

Explanation:

a) Sketches of all possible pv-diagrams for the cycle are attached below

b) The work W[tex]_{ac}[/tex] for the process Ca is : 2462.8  J

Given data :

Amount of heat flowing out = 800 J

Ta = 200 K

Tb = 300 K

R = 800

B) Determine the work W for the process Ca

Wₐs = -pdv

       = - [ pVb - pVa ] ---- ( 1 )

note : pVb = nRTb ,  pVa = nRTa

Equation ( 1 ) becomes

= -nR [ Tb - Ta ]

= - 2(8.314 ) [ 300 - 200 ]

= - 1662.87

given that W[tex]_{bs}[/tex] = 0 which is isochonic

dv = 0 ( cyclic process ) = d∅ - dw

∴ 0 = 800 - ( Wₐs  +  W[tex]_{ac}[/tex] )

Therefore : W[tex]_{ac}[/tex] = 800 + 1662.8 = 2462.8  J

Hence we can conclude that the work W for the process Ca = 2462.8  J

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Answer:

The astronomical unit is used primarily for measuring distances within the Solar System or around other stars.

Answer:

the plot is about a crocodile who lives in a river.....

Explanation:

It is about a crocodile who lives in a river. He becomes friends with a monkey that lives on a tree on the river bank. Consequently, the monkey shares the fruits of the tree with his friend, the crocodile. Further, the crocodile takes a share for his wife.

Answer:

strong nuclear force.

Explanation:

hope this helps you!!

Answer:

900N, charges repel

Explanation:

F = KQq/d²

K = 9 × 10^9Nm²/C²

Q= 10mC = 10 × 10^-3C

q = 1mc = 1 × 10^-3C

d = 10m

F = ?

Force = (9 × 10^9 × 10 × 10^-3 × 1 × 10^-3)/10²

= 9 × 10²

= 900N

It will be an electrostatic force of repulsion since like charges(two positive charges) repel

The magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.

Coulomb's law states that whenever the two charged particles are separated by a particular distance then there will be a force of attraction or repulsion acts between the charge.

The formula of coulomb's force will be given by

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

[tex]K=9\times 10^9\ \frac{Nm^2}{C^2}[/tex]

[tex]q_1= 10\times 10^{-3}\ C[/tex]

[tex]q_2=1\times 10^{-3}\ C[/tex]

[tex]\rm Distance \ d=10 \ m[/tex]

Now the force will be calculated as

[tex]F=(9\times 10^9)\dfrac{(10\times 10^{-3})\times (1\times 10^{-3})}{10}[/tex]

[tex]F= 900\ N[/tex]

The charges are positive in nature so they will repel each other

Hence the magnitude and the direction of the electrostatic force between the charges will be 900N and the charges will repel each other.

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Answer: The answer is A) 49 N(Newtons).

Answer:

P = 0

Explanation:

Momentum is defined as the quentity of motion contained in a body. Mathematically, it can be defined as the product of mass and velocity. So, in order to determine the Volkswagen Beetle, at rest, we can use the simple formula, as follows:

where,

P = Momentum of Volkswagen Beetle = ?

m = Mass of Volkswagen Beetle = 750 kg

v = Velocity of Volkswagen Beetle = 0 m/s (since, it is at rest)

Therefore,

P =

Answer:

Explanation:

Not in Genesis. The proposed sacrifice of Isaac was all Abraham's doing. Sarah was not present and could not, therefore offer any input. It might have been covered in other Jewish writings, but it is not recorded in Genesis.

Since the book most widely used by Everyone is the Bible, I would answer false.

Answer:

Explanation:

A) 215 + 45.0 = 260 km/h West

B) 215 - 45.0 = 170 km/h West

C) √(215² + 45.0²) = 219.6588... 220 km/h

θ = arctan(45.0/215) = 11.8214... 11.8° S of W

The airplane's ground velocity is westward at 170 km/h regardless of wind direction. The south wind does not affect the plane's westward velocity in component C.

Vector addition will determine each scenario's plane velocity relative to the ground. We'll simplify by assuming a consistent horizontal breeze.

Let's divide the airplane's velocity and wind velocity into westward (negative) and eastward (positive) components.

Data: Plane airspeed = 215 km/h.

45 km/h wind.

A. West wind:

The airplane's westward (negative) ground velocity:

West aeroplane velocity = -215 km/h

-45 km/h west wind.

Find Vg when the wind blows west:

Vg (west) = Aeroplane velocity + Wind velocity (west)

Vg(west) = -215 km/h - (-45 km/h)

Vg (west)=-170 km/h

B. East wind: 2. Westward (negative) aeroplane velocity relative to the ground:

West aeroplane velocity = -215 km/h

Wind velocity (east) = 45 km/h (because the wind is blowing eastward)

Find Vg when the wind blows east:

Vg (west) = Aeroplane velocity (west) + Wind velocity (east).

Vg (west) = -215 + 45 km/h

Vg (west)=-170 km/h

C. Southerly wind: 3. Westward (negative) aeroplane velocity:

West aeroplane velocity = -215 km/h

Wind velocity (south) = 0 km/h (no effect since wind is not moving westward)

Find Vg when the wind blows south:

Vg (west) = Aeroplane velocity (west) + Wind velocity (south).

Vg (west) = -215 + 0 km/h.

Vg (west)=-215 km/h

All resultant velocities are westward and 170 km/h.

Part C's plane's westward velocity is unaffected by the south wind. The plane's ground velocity remains -215 km/h (westward). Since the heading is westward, no angle is needed.

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Answer:

I think some drawbacks are that since there are no solid walls meaning it is weak and if murphy's law is in place, the water will destroy the substance. Tsunami waves also happen very quickly so even if the water can travel thru the substance quickly, it probably won't be quick enough. This design could help if the wave is smaller because less destruction would occur.

Explanation:

yeah

Answer: push force gravity tension and reverse force

Explanation: sense they are pushing the box there is a push force gravity because this is likely on earth tension because it is the reverse of gravity and the reverse force because you have to have the reverse of push

Answer:

Explanation:

If we ASSUME that a CCW torque is positive

τ = 10sin135[8.0] - 16sin90[4.0] + 19sin160[3.0] = 12 N•m

as the result is positive, the net torque is CCW.

The conventions for the current allow to find the result for which is the direction of the current the is circuit is:

Current is the movement of electrons in a circuit per unit time.

In an electrical circuit the electrons that have a negative charge leave the negative electrode towards the positive electrode, by convention the current goes in the opposite direction from the positive electrode towards the negative electrode.

When analyzing the given circuit, the current leaves the positive pole at the top of the battery and moves through the blue wire, reaches the light bulb and moves the red wire and reaches the negative pole of the battery.

In conclusion using the conventions for the current we can find the result for which is the direction of the current the is circuit is:

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Answer:

net force would be 50 N right

Explanation:

Answer:

The altitude of the satellite above the surface of the Earth is 1,528.48 km.

F=rmv2=R2GmMrv2=R2GMr=GMv2R2r=(6.67259×10−11)×(5.98×1024)(6900)2×(6,370,000)2r=4,841,516.6 mr=4,841.52 km

Answer:

Option C, increases and decreases

Explanation:

When an object making noise approaches you, the wave frequency increases leading to a higher pitch. Conversely, when it moves away from you or retreats, the wave frequency decreases leading to a lower pitch. This can be observed in ambulance sirens.

The muzzle speed of the bullet before the collision is 415.3 m/s.

The given parameters:

Apply the principle of work-energy theorem to determine the final velocity of the block-bullet system;

[tex]F_f \times d = \frac{1}{2} mv^2\\\\\mu_k F_n \times d = \frac{1}{2} mv^2\\\\\mu_ k (m_1 + m_2)g \times d = \frac{1}{2} (m_1 + m_2)v^2\\\\\mu_k g \times d= \frac{1}{2} v^2\\\\2\mu _k gd = v^2\\\\v= \sqrt{2\mu _k gd } \\\\v = \sqrt{2 \times 0.23 \times 9.8 \times 9.5} \\\\v = 6.54 \ m/s[/tex]

Apply the principle of conservation of linear momentum to determine the muzzle speed of the bullet;

[tex]m_1 u_1 \ + \ m_2u_2 = v(m_1 + m_2)\\\\0.024(u_1) \ + \ 1.5(0) = 6.54(0.024 + 1.5)\\\\0.024u_1 = 9.967\\\\u_1 = \frac{9.967}{0.024} \\\\u_1 = 415.3 \ m/s[/tex]

Thus, the muzzle speed of the bullet before the collision is 415.3 m/s.

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Answer:

Explanation:

Answer:

Explanation:

The escape velocity of the dwarf planet is 1,721.8 m/s.

The given parameters:

The mass of the of the dwarf planet is calculated as follows;

[tex]m = 0.0045 \times 5.98 \times 10^{24} \ kg\\\\m = 2.69\times 10^{22} \ kg[/tex]

The radius of the dwarf planet is calculated as follows;

[tex]r = \frac{0.19 D}{2} \\\\r = \frac{0.19 \times 12, 742, 000}{2} \\\\r = 1,210,490 \ m[/tex]

The escape velocity of the dwarf planet is calculated as follows;

[tex]v _e = \sqrt{\frac{2GM}{r} } \\\\v_e = \sqrt{\frac{2\times 6.67 \times 10^{-11} \times 2.69 \times 10^{22}}{1,210,490}}\\\\v_e = 1,721.8 \ m/s[/tex]

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Answer:

for reasoning you could put -> the line is at the top at the left handed sided of the chart (also the beginning) slopes down towards the very bottom right handed side of the chart, symbolising less force as distance goes.

The charge moved by this electric force is [tex]5.4 \times 10^{-5} \; Coulombs[/tex].

Given the following data:

To determine the charge:

Mathematically, the work done by an electric force in moving a charge from one point to another is given by the formula:

[tex]W = PQ[/tex]

Where:

Making Q the subject of formula, we have:

[tex]Q =\frac{W}{P}[/tex]

Substituting the given parameters into the formula, we have;

[tex]Q =\frac{2.70 \times 10^{-3}}{50}\\\\Q =5.4 \times 10^{-5} \; Coulombs[/tex]

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