Answer:
1. 0.574 kJ/kg
2. 315.7 MW
Explanation:
1. The mechanical energy per unit mass of the river is given by:
[tex] E_{m} = E_{k} + E_{p} [/tex]
[tex] E_{m} = \frac{1}{2}v^{2} + gh [/tex]
Where:
Ek is the kinetic energy
Ep is the potential energy
v is the speed of the river = 3 m/s
g is the gravity = 9.81 m/s²
h is the height = 58 m
[tex] E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg [/tex]
Hence, the total mechanical energy of the river is 0.574 kJ/kg.
2. The power generation potential on the river is:
[tex] P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW [/tex]
Therefore, the power generation potential of the entire river is 315.7 MW.
I hope it helps you!
ionic bonds form when electrons?
Answer:
when the electron transferred permanently to another atom
2. An archer shoots an arrow. The action force is the bowstring against the arrow, The reaction force is...
a. Air resistance against the bow
b. Arrow's push against the bowstring
c. Grip of the archer's hand on the bow
Answer:b the arrows push against the bow string
Explanation:
The reaction force will be Arrow's push against the bowstring.
Option b is correct.
From Newton's Third Law of Action & Reaction, It states that for every action in nature there is an equal and opposite reaction.
It means that If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A.Therefore, If an archer shoots an arrow. The action force is the bowstring against the arrow, The reaction force will be Arrow's push against the bowstring
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Question #4
4. Anthony walks to the pizza place for lunch. He walk 4 km
east, he realized he passed it and then walked 1 km west.
What distance did he cover? What was his displacement?
Answer:
Distance 5 km, Displacement 3 km east
Explanation:
The distance covered by Anthony is 5 km, while his displacement is 3 km.
Distance and displacement:The distance is defined as the total length of space covered during motion between the starting point and end point, not necessarily a straight line. Whereas displacement is defined as the minimum distance between two points in space, that is a straight line.
Suppose, if you start from one point and walk for 100 meters then come back at the same point that you started your walk, you would have traveled 2 times 100 meters that is 200 meters. But your displacement will be zero because your starting point and the end point are the same.
The given question clarifies the difference between distance and displacement:
Since Anthony walks 4 km east and then walks 1 km west, the distance covered by him is:
distance = 4 + 1 = 5 km
but as he walks west, he comes closer to the starting point, and his displacement is the distance between the starting point and the end point.
displacement = 4 - 1 = 3 km
Referto the figure below for more information.
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Which of the following is an example of benefits?
Travel
Physical activity
Salary
Insurance
Insurance is an example of benefits therefore the corrct answer is option D.
What is exercise?The exercise has a beneficial effect on the body and differs the type of death a person undergoes. If a person exercises but sits for too long can result in many diseases that ultimately lead to death. Along with exercise, the person should do mobile activities so that it increases their fitness.
It is among the most notable and significant advantages of insurance. According to insurance plans, the insured person or companies are protected from liabilities. The correct kind of insurance coverage might help you protect yourself from losses brought on by various life uncertainty.
Insurance is a good example of a benefit because, unlike salaries, which are things you earn as a base when you start a job, benefits are things you only receive as a result of working in certain jobs.
Thus,the correct answer is option D.
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Weight of a body becomes greater at the pole than that at the equator . why ?
Which statement best describes an atom? (2 points)
оа
Protons and neutrons grouped in a specific pattern
Ob
Protons and electrons spread around randomly
ос
A group of protons and neutrons that are surrounded by electrons
Od
A ball of electrons and neutrons surrounded by protons
Answer:
A group of protons and neutrons that are surrounded by electrons I think that's the answer...
Explanation:
Help me answer this please!
Answer:
Option iv
Explanation:
The acceleration due to gravity (g) of the earth is constant (i.e 9.81 ms¯²). As long as the stone remains within the earth, the acceleration due to gravity experienced by the stone will be 9.81 ms¯².
The maximum height reached by an object is the height where the object no longer have velocity to move further hence it begins to fall back. At this point (i.e at the maximum height), the velocity of the object is zero.
Summary:
At maximum height:
Acceleration due to gravity of the stone = 9.81 ms¯²
Velocity = 0
Thus, option iv gives the correct answer.
Donald Duck Matthews wanted to show up all his friends with his newfound dancing skills he left his house to go to dressers house 20 m north from where he was he's so dresser his moves went off to the next place and next stop was 10 m west at choirboys house luckily he saw Eddie there too. After showing off his skills agai he was satisfied, but he still had to show one last person up. His brother who was 15 m South. After making it there he showed hos moves and smiled. what was Ducks displacement?
A. 12 m SW
B. 11 m NW
C. 45 m NW
D. 45 m NE
The final displacement of the duck would be 11.18 meters in NW, therefore the correct answer is option B.
What is displacementDisplacement describes this shift in location.
As given in the problem Donald Duck Matthews wanted to show up all his friends with the newfound dancing skills he left his house to go to dressers house 20 m north from where he was he's so dresser his moves went off to the next place and next stop was 10 m west at choirboys house luckily he saw Eddie there too.
After showing off his skills again he was satisfied, but he still had to show one last person up. His brother was 15 m South.
His final position would be 5 meters north and 10 meters west.
The displacement of the Duck = √( 10² + 5²)
= 11.18 meters in NW
Thus, the final displacement of the duck would be 11.18 meters in NW.
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You and some friends visit the Florida State Fair and decide to play a game. To play the game you must slide a metal hockey-type puck up a wooden ramp so that it drops through a hole at the top of the ramp. Your prize, if you win, is a large, stuffed gorilla. You realize the secret to winning is giving the puck just enough velocity at the bottom of the ramp to make it to the hole. You estimate the distance from the bottom of the ramp to the hole at about 3 m, and the ramp appears to be inclined with an angle of 10o from the horizontal. You just got out of physics class and recall the coefficient of static friction between steel and wood is 0.1 and the coefficient of kinetic friction between steel and wood is 0.08. The mass of the puck is about 1 kg. You decide to impress your friends by sliding the puck at the precise speed on the first try so as to land it in the hole. You slide the puck at 2.0 m/sec. Do you win the stuffed poodle? Let g = 10 m/s2. Note: A slightly too hard throw will win since your aim is so good the puck would drop in the hole.
Answer:
No you didn't win the stuffed poodle
Explanation:
From the question we are told that
The distance of the ramp to the hole is d = 3 m \
The angle of inclination is [tex]\theta = 10^o[/tex]
The coefficient of static friction is [tex]\mu_s = 0.1[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.08[/tex]
The mass of the puck is m = 1 kg
The velocity of the first slide is [tex]v_1 = 2.0 m/s[/tex]
Generally the kinetic energy at the bottom of the ramp is equal the energy loss due to friction and this can be mathematically represented as
[tex]\frac{1}{2} m * v^ 2 = \mu_k * [m * g] * cos (theta ) * d[/tex]
=> [tex]\frac{1}{2} * v^ 2 = 0.08 * 9.8 * cos (10 ) * 3[/tex]
=> [tex] v= 2.17 \ m/s [/tex]
Comparing the relative velocity obtained and the velocity of your first throw we can see that you didn't win the stuffed animal
a car is moving eastward and speeding up. the momentum of the car is
what chemical diverged from trees
a resin and turpentine
b sodium
c lead
d marcotting
Answer:
a. resin and turpentine
Explanation:
The chemicals that diverged from trees are resin and turpentine.
Resin are produced by special cells in trees, most times we see them when a tree is damaged or cut. They are usually derived from pines and firs.
Turpentine is obtained by distilling resin.
Turpentine has an antiseptic property that has different uses. They are used as cleansing agents and for producing sanitary materials.
) A 1000-gallon tank currently contains 100.0 gallons of liquid toluene and a gas saturated with toluene vapor at 85°F and 1 atm. (a) What quantity of toluene (lbm) will enter the atmosphere when the tank is filled and the gas displaced? (b) Suppose that 90% of the displaced toluene is to be recovered by compressing the displaced gas to a total pressure of 5 atm and then cooling it isobarically to a temperature T(°F). Calculate T.
Answer:
A) m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm
B) T= 63.32°F
Explanation:
Given data:
1000 gallon tank currently contains 100.0 gallons of liquid toluene
and A gas saturated with toluene vapor at 85°F and 1 atm
A) Calculate quantity of toluene ( Ibm ) that will enter the atmosphere when the tank is filled
m[tex]_{T}[/tex] = [tex]n_{gas} * Y_{T} * M_{T}[/tex]
[tex]n_{gas}[/tex] (total mole of gas) = 0.3025 Ib-mole ( calculated using : [tex]\frac{PV}{RT}[/tex] )
[tex]Y_{T}[/tex] (mole fraction of toluene) = 0.0476 ( calculated using [tex]\frac{P_{T} }{P}[/tex] )
M[tex]_{T}[/tex] = 92.13 Ibm/Ib-mole
therefore: m[tex]_{T}[/tex] = 0.3025 * 0.0476 * 92.13 = 1.327 Ibm
B) using Antoine equation to solve for T
Antoine equation : [tex]log_{10} (P_{T} ) = A - \frac{B}{T+C}[/tex]
PT( partial pressure ) = 18.95 ( calculated using : [tex]y_{tb} * P[/tex] )
A = 6.95805
B = 1346.773
T = ?
C = 219.693
to calculate T make T the subject the subject of the equation
T + 219.693 = 1346.773 / 5.68044
∴ T = 17.40°C
convert T to Fahrenheit
T = 1.8 * 17.40 +32
= 63.32°F
Two vectors have magnitudes 20 m and 44 m. Which of the following cannot possibly be the magnitude of the resultant of the two vectors ?
Answer:
44M 64M
Explanation:
1pt During a phase change, as energy is added, the temperature of a substance
O A. increases
O B. is constant
O C. decreases
O D. first increases and then decreases to its original value
Answer:
D
Explanation:
Because first phase thr it will follow the law of thermodynamics as it will reach the equilibrium point then decrrases afterwards
m_Cu * sh_CuA system consists of a copper tank whose mass is 13 kilogram , 4 kilogram of liquid water, and an electrical resistor of negligible mass. The system is insulated on its outer surface. Initially, the temperature of the copper is 27 degC and the temperature of the water is 50 degC . The electrical resistor transfers 100 kilojoule to the system. Eventually the system comes to equilibrium. Determine the final equilibrium temperature, in ∘C.
Answer:
T₂ = 49.3°C
Explanation:
Applying law of conservation of energy to the system we get the following equation:
Energy Supplied by Resistor = Energy Absorbed by Tank + Energy Absorbed by Water
E = mC(T₂ - T₁) + m'C'(T'₂ - T'₁)
where,
E = Energy Supplied by Resistor = 100 KJ = 100000 J
m = mass of copper tank = 13 kg
C = Specific Heat of Copper = 385 J/kg.°C
T₂ = Final Temperature of Copper Tank
T₁ = Initial Temperature of Copper Tank = 27°C
T'₂ = Final Temperature of Water
T'₁ = Initial Temperature of Water = 50°C
m' = Mass of Water = 4 kg
C' = Specific Heat of Water = 4179.6 K/kg.°C
Since, the system will come to equilibrium finally. Therefor: T'₂ = T₂
Therefore,
(100000 J) = (13 kg)(385 J/kg.°C)(T₂ - 27°C) + (4 kg)(4179.6 J/kg.°C)(T₂ - 50°C)
100000 J = (5005 J/°C)T₂ - 135135 J + (16718.4 J/°C)T₂ - 835920 J
100000 J + 135135 J + 835920 J = (21723.4 J/°C)T₂
(1071055 J)/(21723.4 J/°C) = T₂
T₂ = 49.3°C
Blood is 92% water. Blood is
drawn using a capillary tube.
Write 1 sentence explaining how it
illustrates this characteristic of
water. Use the terms adhesion
and capillary action.
Answer:
Plasma, which constitutes 55% of blood fluid, is mostly water (92% by volume), and contains proteins, glucose, mineral ions, hormones, carbon dioxide (plasma being the main medium for excretory product transportation), and blood cells themselves.
Explanation:
Capillary motion is crucial for circulating water. Your body's cells would not hydrate without this flow, and crucial communication between your brain and body would slack off, hence blood contains water 92% water helps in capillary action.
What is Capillary action?Capillary action, also known as capillary effect or motion, is the process by which liquid moves through constricted areas without the aid of external forces like gravity but rather with the help of intermolecular forces that exist between the liquid and solid surface (s).
Adhesion: The attraction of two different molecules, such as the hydrogen and oxygen molecules found in water and plastic drinking straws.
Proteins, glucose, mineral ions, hormones, carbon dioxide (plasma is the principal medium for excretory product transfer), and blood cells themselves are contained in plasma, which makes up 55% of blood fluid. Plasma is mostly water (92% by volume) and contains water as well as several other substances.
Therefore, the water contains in the blood helps in capillary action.
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Consider the force field and circle defined below. F(x, y) = x2 i + xy j x2 + y2 = 121 (a) Find the work done by the force field on a particle that moves once around the circle oriented in the clockwise direction.
Answer: the work done by the force is 0
Explanation:
F (x², xy)
121 = 11²
so R = x² + y² = 11²
p = x². Q = xy
Δp/Δy = 0, ΔQ/Δx
using Green's theorem
woek = c_∫F.Δr = R_∫∫ ΔQ/Δx - Δp/Δy) ΔA
= (x² + y² = 121)_∫∫ yΔA
now let x = rcosФ, y = rsinФ
ΔA = rΔrΔФ
so r from 0 to 11
and Ф from 0 to 2π
= 0_∫^2π 0_∫^11 rsinФ × rΔrΔФ
= 0_∫^2π SinФΔФ 0_∫^11 r²Δr
= [ -cosФ]^2π_0 [r³/3]₀¹¹ = ( -cos2π + cos0) (11³/3) = 0
therefore the work done by the force is 0
Activity Problem 8.3 This problem asks you to compare the finite current element approximation to the infinite straight wire approximation for the isolated segment of wire carrying current of 2A to the right. The wire is 8cm long. Note, the finite current approximation will be a fairly poor approximation to the actual field which would be found by integration; the infinite wire field is a better approximation in this case. (a)Compute the magnetic field at point C a distance 2cm for the center of the wire using the infinite straight wire approximation. (b)Compute the magnetic field at point C using the finite current element approximation. (c)Compute the magnetic field at point E a distance 2cm from the end using the finite current element approximation.
Answer: (a) B = 2 x 10⁻⁵T
(b) B = 1.94 x 10⁻⁵T
(c) B = 1.8 x 10⁻⁴T
Explanation: A magnetic field due to a current passing through a straight wire is calculated using the Biot-Savart Law:
[tex]dB=\frac{\mu_{0}}{4.\pi} \frac{IdLXR}{r^{3}}[/tex]
where
dL is current length element
[tex]\mu_{0}[/tex] is permeability of free space ([tex]4.\pi.10^{-7}[/tex]T.m/A)
(a) For a infinite straight wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}}[/tex]
B = 2x10⁻⁵T
For an infinite, long and straight wire, magnetic field is 2x10⁻⁵T.
(b) For a finite wire:
[tex]B=\frac{\mu_{0}I}{2.\pi.R}\frac{L}{\sqrt{L^{2}+R^{2}} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{2.\pi.2.10^{-2}} \frac{8.10^{-2}}{\sqrt{(8.10^{-2})^{2}+(2.10^{-2})^{2}} }[/tex]
B = 1.94x10⁻⁵T
The magnetic field for a finite wire in the same conditionsas infinite wire is 1.94x10⁻⁵T.
(c) For a finite wire at a point distant from the end of the wire:
[tex]B=\frac{\mu_{0}I}{4.\pi.L\sqrt{2} }[/tex]
[tex]B=\frac{4.\pi.10^{-7}.2}{4.\pi.8.10^{-2}\sqrt{2} }[/tex]
B = 0.18x10⁻⁵T
At a point at the end, magnetic field is 1.8x10⁻⁴T.
A plane flying horizontally at a speed of 40.0 m/s and at an elevation of 160 m drops a package. Two seconds later it drops a second package. How far apart will the two packages land on the ground?
Answer:
Package 1 will land at 228.0 m, package 2 will land at 308.0 m, and the distance between them is 80.0 m.
Explanation:
To find the distance at which the first package will land we need to calculate the time:
[tex] Y_{f} = Y_{0} + V_{0y}t - \frac{1}{2}gt^{2} [/tex]
Where:
Y(f) is the final position = 0
Y(0) is the initial position = 160 m
V(0y) is initial speed in "y" direction = 0
g is the gravity = 9.81 m/s²
t is the time=?
[tex] 0 = 160 m + 0t - \frac{1}{2}9.81 m/s^{2}t^{2} [/tex]
[tex] t = \sqrt{\frac{2*160 m}{9.81 m/s^{2}}} = 5.7 s [/tex]
Now we can find the distance of the first package:
[tex] X_{1} = V_{0x}*t = 40.0 m/s*5.7 s = 228.0 m [/tex]
Then, after 2 seconds the distance traveled by plane is (from the initial position):
[tex] X_{p} = V_{0x}*t = 40.0 m/s*2 s = 80.0 m [/tex]
Now, the distance of the second package is:
[tex] X _{2} = X_{1} + X_{p} = 228.0 m + 80.0 m = 308.0 m [/tex]
The distance between the packages is:
[tex] X = X_{2} - X_{1} = 308.0 - 228.0 m = 80.0 m [/tex]
Therefore, package 1 will land at 228.0 m, package 2 will land at 308.0 m and the distance between them is 80.0 m.
I hope it helps you!
In the absence of a gravitational field, you could determine the mass of an object (of unknown composition) by:
A) applying a known force and measuring it's acceleration.
B) measuring the volume.
C) weighing it.
Answer:
A) By applying a known force, and measuring it's acceleration.
Explanation:
This is actually something that astronauts do in space as a mathmatical exercise when calculating the mass of an object since F = m × a.
Once the force, and acceleration are applied, the only unknown is the mass which can be solved by dividing force over acceleration. This is because inertial mass is equal to gravitational mass.
Which formula describes Boyle's law?
ОА.
OB.
D = m/v
V1T2 = V2T
P1V1 = P2V2
P1T2 = P2T1
Ос.
OD
Answer: P1V1 = P2V2
Explanation:
Increase in potential energy of spring
Colored lights are called additive colors. Why do you think this is so?
In an illustration, lines of force are drawn around a magnet to illustrate the magnetic
O field.
O induction.
O poles.
O force.
Need help now
Answer: A. Field
Explanation: Magnetic Field is the correct answer.
Answer:
A
Explanation:
A 0.75 m3 rigid tank initially contains air whose density is 1.18 kg/m3. The tank is connected to a high-pressure supply link through a valve. The valve is opened, and air is allowed to enter the tank until the density in the tank rises to 4.95 kg/m3. Determine, in kg, the mass of air that has entered the tank..
Answer:
2.83kg
Explanation:
Answer:
2.83kg
Explanation:
Given
initial density = 1.18 kg/m3
Final density in the tank = 4.95 kg/m3.
Let us write the mass balance first.
Change in the mass of the system=mass of the air entering the system - Mass of air out the system
Mass that entered= M2 - M1
But DENSITY= MASS/ VOLUME
Mass= volume × Density
We can expressed the mass in terms of density since density is given in the question.
Mass that entered= (volume × density)2 - ( volume × density)1
= (V ρ)2 - (V ρ)1
But V1= V2 the volume remains the same
= ( ρ2 - ρ1)v
= (4.95 kg/m3 - 1.18 kg/m3) 0.75 m3
= 3.77× 0.75
= 2.8275kg
Mass that entered= 2.83kg
therefore, mass of air that has entered the tank= 2.83kg
Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)
Answer:
III) [tex]d_{1}+ d_{2}=d_{t}[/tex]
Explanation:
I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5 with the the y being any other value.
II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)
III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]
(i) using coordinate grid notation to represent d₁, d₁ = 5y
(ii) using coordinate grid notation to represent d₂, d₂ = 5x + 5y
(ii) The sum of d₁ and d₂ is written as 5x + 10y
In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.
Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.
(i) Using coordinate grid to represent d₁ = (0, 5)
[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]
(ii) Using coordinate grid to represent d₂ = (5, 5)
[tex]d_2 = 5x + 5y[/tex]
(iii) The total vector is written as;
[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]
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Describe the life cycle of a star before it collapses into a black hole.
Describe the life cycle of a star before it becomes a black dwarf.
What is the likely outcome of our sun? *
The sun will supernova and become a black hole.
The sun will swell, encompassing the inner planets and collapses into a dwarf star.
The sun will become a pulsar.
How Do You Know?
P.S. the how do you know is only for the last question
answer:
uhm this makes no sense like at all so chile anyways
explanation:
An object from a certain height falls freely. which of the following happens PE and KE when the object is half on its way down
Answer:
A. Loses PE and gains KE
Explanation:
Statement is incomplete. Complete statement of problem is:
1. An object from a certain height falls freely. Which of the following happens to PE and KE when the object is half on its way down?
A. Loses PE and gains KE
B. Gains PE and loses KE
C. Loses both PE and KE
D. Gains both PE and KE
If we neglect the effects of any conservative force, the application of the Principle of Energy Conservation is reduced to a sum of gravitational potential ([tex]U_{g}[/tex]) and translational kinetic energies, measured in joules. That is:
[tex]U_{g,1}+K_{1} =U_{g,2}+K_{2}[/tex] (Eq. 1)
Let assume that an object falls from a height [tex]h[/tex] with a speed of zero. By definitions of gravitational potential and translational kinetic energies the previous is expanded. If final height is the half of initial value, then:
[tex]m\cdot g\cdot h = 0.5\cdot m\cdot g\cdot h +K_{2}[/tex] (Eq. 1b)
[tex]K_{2} = 0.5\cdot m\cdot g\cdot h[/tex]
[tex]K_{2} = 0.5\cdot U_{g,1}[/tex]
In a nutshell, translational kinetic energy is increased at the expense of diminishing gravitational potential energy. The correct answer is A.
It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Answer:
102 m
Explanation:
Given that It takes a minimum distance of 48.96 m to stop a car moving at 12.0 m/s by applying the brakes (without locking the wheels). Assume that the same frictional forces apply and find the minimum stopping distance when the car is moving at 25.0 m/s.
Let the stopping distance be equal to S.
According to the definition of speed,
Speed = distance / time.
make time the subject of the formula
Time = distance / speed
then, the equivalent time is:
48.96 / 12 = S / 25
Cross multiply
12S = 48.96 x 25
12S = 1224
S = 1224 / 12
S = 102 m
Therefore, the stopping distance is 102 m
Complete this problem so that it can be solved using the concept listed Angela is playing basketball
A. Acceleration
B. Speed