Answer:
The centripetal acceleration of the girl is 2.468 m/s²
Explanation:
Given;
number of turns, = ¹/₄ Revolution
distance traveled by the girl, d = 25 m
time of motion, t = 5.0 s
The linear speed of the of the girl is calculated as;
[tex]v = \omega \ r\\\\v =(\frac{1}{4}rev \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1}{5 \ s} ) (25 \ m)\\\\v = (0.3142 \ \frac{rad}{s} )(25 \ m)\\\\v = 7.855 \ m/s[/tex]
The centripetal acceleration of the girl is calculated as;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(7.855)^2}{25} \\\\a_c = 2.468 \ m/s^2[/tex]
Therefore, the centripetal acceleration of the girl is 2.468 m/s²
A bartender slides a beer mug at 1.3 m/s towards a customer at the end of a frictionless bar that is 1.3 m tall. The customer makes a grab for the mug and misses, and the mug sails off the end of the bar. (a) How far away from the end of the bar does the mug hit the floor
Answer:
x = 0.67 m
Explanation:
For this problem, let's use the projectile launch equations, as the jug goes through the bar, it comes out with horizontal speed vx = 1.3 m / s, which does not decrease as there is no friction.
Let's find the time or it takes to get to the floor
y = y₀ + v_{oy} - ½ g t²
in this case I go = 0 and when I get to the floor y = 0
0 = y₀ + 0 - ½ g t²
t² = 2y₀ / g
t² = 2 1.3 / 9.8 = 0.2653
t = 0.515 s
now let's find the distance traveled in this time
x = vx t
x = 1.3 0.515
x = 0.6696 m
x = 0.67 m
The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m2 , determine the surface temperature of the spacecraft when the radiation emitted equals the solar energy absorbed.
Answer:
158.32 K = -114.83 °C
Explanation:
Since P = P' where P = power absorbed and P' = power radiated
P = αAQ where α = absorptivity = 0.3, A = area of spacecraft and Q = rate of incident solar radiation = 950 W/m²
Also, P' = εσAT⁴ where ε = emissivity of spacecraft = 0.8, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²K⁴, A = area of spacecraft and T = surface temperature of spacecraft.
So, P = P'
αAQ = εσAT⁴
T⁴ = αQ/εσ
T = ⁴√(αQ/εσ)
Substituting the values of the variables into the equation, we have
T = ⁴√(0.3 × 950 W/m²/(0.8 × 5.67 × 10⁻⁸ W/m²K⁴))
T = ⁴√(285 W/m²/(45.36 × 10⁻⁸ W/m²K⁴))
T = ⁴√(6.2831 × 10⁸ K⁴))
T = 1.5832 × 10² K
T = 158.32 K
In Celsius T(C) = 158.32 - 273.15 = -114.83 °C
In the diagram above, where is light energy converted to chemical energy?
A. Sun
B. Grass
C. Grasshopper
B. Hawk
Answer:
B. Grass
Explanation:
In the diagram above, Grass is where, light energy converted to chemical energy. Hence option B is correct.
What is Visible light ?Visible light spectrum is nothing but the range of wavelength of radiation from 4000 angstrom to 7000 angstrom(Violet to Red). light is a energy packet. Every Photon having different wavelength travels with same velocity c (velocity of light). When we focus numbers of colors from visible spectrum to a point, that point appears as a white light. hence white light is composed of numbers of Colors in it.
Light itself is a energy when it falls on the grass, grass grows by taking energy from the light and convert that energy into biomass that is a chemical energy.
Hence option B is correct.
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an object has a mass of 2000kg what is its weight on earth
Answer:
19600 N
Explanation:
weight = mass x gravity
We know that gravity = 9.8 m/s^2 and mass = 2000 kg.
w = m x g
w = 2000 kg x 9.8 m/s^2
w = 19600 N
The weight of the object is 19600 N (newtons).
Answer:
the answer i 2000kg
Explanation:
The stars, Rigel and Betelgeuse, are both found in the constellation Orion. Rigel is a blue supergiant, and Betelgeuse is a red supergiant. Which of the following correctly compares the temperatures of Rigel and Betelgeuse?
Answer:
batrix
Explanation:
someone help me with this exercise ?
1. if a body with a mass of 350kg is subjected to a fare of 90n what will be its mass
?
Mass remains mass no matter what you do to it.
Two 800 cm^3 containers hold identical amounts of a monatomic gas at 20°C. Container A is rigid. Container B has a 100 cm^2 piston with a mass of 10 kg that can slide up and down vertically without friction. Both containers are placed on identical heaters and heated for equal amounts of time.
Required:
a. Will the final temperature of the gas in A be greater, less than, or equal to the temperature in B?
b. Show both processes on a single PV diagram.
c. What are the initial pressures in containers A and B?
d. Suppose the heaters have 25 W of power and are turned on for 15s. What is the final volume of container B?
Answer:
1) Final Temperature of the gas in A will be GREATER than the temperature in B
2) Diagram of both processes on a single PV has been uploaded below
3) The Initial pressures in containers A and B is 3039.87 J/liters
4) the final volume of container B is 923.36 cm³
Explanation:
Given that;
Temperature = 20°C = 293 K
mass of piston = 10 kg
Area = 100cm³
Volume V = 800 cm³ = 0.8 L
ideal gas constant R = 8.3 J/K·mol
1)
Final Temperature of the gas in A will b GREATER than the temperature in B
2)
Diagram of both processes on a single PV has been uploaded below,
3)
Initial pressures in containers A and B
PV = nRT
P = RT/V
we substitute
P = (8.3 × 293) / 0.8
P = 2431.9 / 0.8
P = 3039.87 J/liters
Therefore, The Initial pressures in containers A and B is 3039.87 J/liters
4)
Given that;
power = 25 W
time t = 15s
the final volume of container B = ?
we know that;
work done = power × time
work done = 25 × 15 = 375
Also work done = P( V₂ - V₁ )
so we substitute
375 = 3039.87 ( V₂ - 0.8 )
( V₂ - 0.8 ) = 375 / 3039.87
V₂ - 0.8 = 0.12336
V₂ = 0.12336 + 0.8
V₂ = 0.92336 Litres
V₂ = 923.36 cm³
Therefore, the final volume of container B is 923.36 cm³
First to answer gets brainliest
Answer:
Sodium (K)
Explanation:
The equation r(t)= (3t+9)i+(sqrt(2)t)j+(t^2)k is the position of a particle in space at time t. Find the angle between the velocity and acceleration vectors at time t=0. What is the angle?
Answer:
θ = 90º
Explanation:
The velocity is given by
v = [tex]\frac{dr}{dt}[/tex]
calculate
v = 3 i ^ + √2 j ^ + 2t k ^
acceleration is defined by
a = dv / dt
a = 2 k ^
one way to find the angle is with the dot product
v. a = | v | | a | cos θ
cos θ= v.a / | v | | a |
Let's look for the value of each term
v. a = 4 t
| v | = [tex]\sqrt{3^2 + 2 + (2t)^2 }[/tex] = [tex]\sqrt{ 11 + 4t^2}[/tex]
| a | = 2
they ask us for the angle for time t = 0
v. a = 0
| v | = √11 = 3.317
we substitute
cos θ = 0 /√11
cos θ = 0
therefore the angles must be θ = 90º
A car is traveling along a straight road at a velocity of +30.0 m/s when its engine cuts out. For the next 1.79 seconds, the car slows down, and its average acceleration is . For the next 4.03 seconds, the car slows down further, and its average acceleration is . The velocity of the car at the end of the 5.82-second period is +18.4 m/s. The ratio of the average acceleration values is = 1.53. Find the velocity of the car at the end of the initial 1.79-second interval.
Answer:
first value+2nd +3rd
Explanation:
thug life and there
The star Rho1Rho1 Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho1Rho1 Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a)the orbital speed and (b)the orbital period of the planet of Rho1Rho1 Cancri
Answer:
8.27 * 10^4 m/s ; 14.51 days
Explanation:
Mass of star, Ms = 0.85 mass of earth
Radius of star, Rs = 0.11 radius of earth
Mass of earth = 1.99 * 10^30 kg
Radius of earth round the sun = 1.5 * 10^11 m
Ms = 0.85 * 1.99 * 10^30 = 1.6915 * 10^30 kg
Rs = 0.11 * 1.5 * 10^11 = 0.165 * 10^11 m
1.) The Orbital speed, v
V = sqrt(Gm/r) ; G = 6.67 * 10^11
V = sqrt(6.67 * 10^-11 * 1.6915 * 10^30) / 0.165 * 10^11
V = sqrt(11.282 * 10^19 / 0.165 * 10^11)
V = sqrt(68.377606 * 10^8)
V = 82690.752
V = 8.27 * 10^4 m/s
2.)
T = 2πr / v
(2π * 0.165 * 10^11) /8.27 * 10^4
1.03672 * 10^11 / 8.27 * 10^4
1253598.0 s
Converting seconds to days:
1 sec = 1/ (60 * 60 * 24) = 1/86400 days
Hence,
1253598.0 * (1/86400)
= 14.509236 days
= 14.51 days
An object undergoes constant acceleration after starting from rest and then travels 5m in the first second. Determine how far it will go in the next second
The speed will be 10 m/s after the 1st and 20 m/s after the 2nd for an average of 15 m/s. So it will travel 15 m during that 2nd second
Mark as brainlist
The object, which undergoes constant acceleration after starting from rest, will go in the next second 15 m.
What is acceleration?Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).
Given parameters:
initial velocity of object: u = 0.
time = 1 second.
distance travelled: d= 5 m.
So, acceleration of the object: a = 2d/t² = (2×5)/1² m/s² = 10 m/s².
Hence, it will go in the next second = 1/2×a(2²-1²) m
= 1/2×10×3 m.
= 15 m.
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A race driver has made a pit stop to refuel. After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 7.0 m/s2; after 3.8 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 73.3 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car
Answer:
t = 13.3 s
Explanation:
The distance traveled by both cars once they are in the main speedway, assuming that the acceleration of the refueling car is constant, is given by the following kinematic equation:[tex]x = v_{o}*t + \frac{1}{2} * a * t^{2} (1)[/tex]
The refueling car (which we will call car 1) in the moment that enters to the main speedway, has achieved a speed that can be found from the definition of acceleration, rearranging terms, as follows:[tex]v_{f1} = a* t = 7.0m/s2*3.8s = 26.6 m/s (2)[/tex]
So, since vf1 = v₀ in (1), we get:[tex]x_{1} = v_{f1}*t + \frac{1}{2} * a * t^{2} (3)[/tex]
Now, for the other car (which we will call car 2), due to is moving at a constant speed, a=0, so we can write the following equation for x₂:[tex]x_{2} = v_{f2}*t = 73.3m/s*t (4)[/tex]
When the entering car catches up the other car, both distances will be equal each other, so x₁ = x₂, as follows:[tex]26.6m/s*t + \frac{1}{2} * 7.0m/s2* t^{2} = 73.3m/s*t[/tex] Rearranging, simplifying and solving for t:[tex]t =\frac{2*(73.3m/s-26.6m/s}{7.0m/s2} = 13.3 s (5)[/tex]If a motorbike accelerates from 15m/s to 25m/s in 15 seconds how far does it travel in that time
Answer:
distance = 0.2330142 miles
= 0.2330142 mi
Explanation:
not sure if its right though....
When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of the square. Suppose three identical charges of q are placed on the remaining three corners of the square. What is the magnitude of the net electric field at the center of the square
Answer:
Explanation:
Net electric field at the centre will be zero .
Since all the charges are equal and they all are symmetrically situated around the centre . So the electric field produced by each will cancel out each other and hence the resultant electric field will be zero . It happens because electric field is a vector quantity and therefore it adds up vectorially . All the four electric field will form two pairs , in each pair electric fields are acting in opposite direction . So they all cancel out to zero .
Determine the gravitational potential energy, in kJ, of 1 m3 of liquid water at an elevation of 30 m above the surface of Earth. The acceleration of gravity is constant at 9.7 m/s2 and the density of the water is uniform at 1000 kg/m3. Determine the change in gravitational potential energy if the elevation decreases by 10 m.
Answer:
Explanation:
Gravitational potential energy = mgh where m is mass , g is acceleration due to gravity and h is height from the ground .
In the first case mass m = volume x density
= 1 x 1000 = 1000 kg
height h = 30 m
potential energy = 1000 x 30 x 9.8 = 294000 J = 294 kJ .
When height decreases by 10 m , potential decreases as follows .
Decrease in potential energy
= mass x gravitational energy x decrease in height
= 1000 x 9.8 x 10
= 98000 J
= 98 kJ .
The Earth’s orbit _____.
is an ellipse
goes around the moon
is a circle
causes day and night
Ang larong Latin at Sisiw ay________________________.
larong pinoy
Explanation:
ito ay larong Pinoy
2. What makes Newton's 3rd Law an unbalanced force?
Answer:
Explanation:
a
At last year's Homecoming Pep Rally, Trudy U. Skool (attempting to generate a little excitement) slid down a 42.0 degree incline from the sports dome to the courtyard below. The coefficient of friction between Trudy's jeans and the incline was 0.650. Determine Trudy's acceleration along the incline.
Answer:
a = - 1.8 m / s²
Explanation:
To solve this exercise let's use Newton's second law, let's start by defining a reference system with the x axis parallel to the plane, in the adjoint we can see a diagram of the forces
let's break down the weight
sin 42 = Wₓ / W
cos 42 = W_y / W
Wₓ = W sin 42
W_y = W cos 42
Y axis
N- W_y = 0
N = W_y
N = mg cos 42
X axis
fr -Wₓ = m a
a = (fr - mg sin 42) / m
the friction force has the formula
fr = μ N
fr = μ mg cos 42
we substitute
a = (μ g cos 42– g sin 42)
a = g (μ cos 42 - sin 42)
let's calculate
a = 9.8 (0.650 cos 42 - sin 42)
a = - 1.8 m / s²
The negative sign indicates that the acceleration is downward, so the boy is descending.
2. Who was the first to show ancient hieroglyphs of swimming?
Answer:
Rafi Bahalul
Explanation:
Israeli Haaretz newspaper reported Feb. 4 that Rafi Bahalul stumbled upon a 3,400-year-old Egyptian stone anchor engraved with hieroglyphs during a morning swim along Israel's northern shores last year.
In the absence of air resistance, ___ accelerates all objects at the same rate of 9.8 m/s2
Answer:
Gravity...this was proven by NASA and was examined by Leonardo da Vinci.
Answer:
Gravity
Explanation:
In the absence of air resistance, Gravity accelerates all objects at the same rate of 9.8 m/s2
A 1325 kg car and a 2050 kg pickup truck approach a curve on a highway that has a radius of 255 m. At what angle should the highway engineer bank this curve so that vehicles traveling at 75.0 mi/h can safely round it regardless of the condition of their tires
Answer:
the banking angle of the road is 24.2⁰
Explanation:
Given;
speed of the vehicles considered, v = 75 mi/h
Speed in m/s ⇒ 1 mi/h --------> 0.44704 m/s
75 mi/h --------> ?
= 75 x 0.44704 m/s = 33.528 m/s
radius of the curve, r = 255 m
The banking angle of the road is calculated as;
[tex]\theta = tan^{-1} (\frac{v^2}{rg} )\\\\\theta = tan^{-1} (\frac{33.528^2}{255\times 9.8} )\\\\\theta = tan^{-1}(0.44983)\\\\\theta =24.2^0[/tex]
Therefore, the banking angle of the road is 24.2⁰
The angle of banking is 24 degrees.
What is the angle of banking?As a driver approaches a bend two equal and opposite forces act on him which are the centripetal force and the centrifugal force. The driver will have to ben through a certain angle called the angle of banking to avoid falling off.
The angle of banking depends on the speed of the vehicle and the radius of the curve.
θ = v^2/rg
speed = 75.0 mi/h or 33.5 m/s
r = 255 m
g = 9.8 ms-1
θ = tan-1 (33.5 m/s)^2/ 255 m × 9.8 ms-1
θ = tan-1(1122.3/2499)
θ = 24 degrees
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1)A rocket expels gas at high speed for a short period of time. We are going to treat the rocket as being far away from any gravitational objects.a)Draw a momentum chart for the rocket expelling gas in space.Take the initial time before expelling gas and the final time after the rocket has finished expelling gas. The rocket has an initial constant speed.Put the rocket and the expelled gas on separate rows.b)Use your chart to explainhow the speed of the rocket changes. c)Does the rocket have to keep expelling gas to stay at a constant speed
We have that
a) We Draw a graph to follow the equation
[tex]v=\frac{m_0u}{m_0-dt}[/tex]
b) The speed of the rocket changes because Momentum is conserved because with the law of conservation in mind mass is reduced thereby causing increase in speed
c) The rocket does NOT have to keep expelling gas to stay at a constant speed because On the account of maintaining speed the Rocket need not expel gas as that will only increase speed and not maintain it
a)
Let
Mass =m
Time t
Generally, the equation for mass ejection constant is mathematically given by
[tex]\phi=\frac{d_m}{d_t}[/tex]
Therefore
[tex]m=m_0-\phi t[/tex]
where
[tex]m_0=initial\ mass[/tex]
Apply the law of conservation of momentum which states that
Conservation of momentum, states that momentum can neither be lost nor gained in an isolated system
[tex]m_o\mu=mv[/tex]
[tex]v=\frac{m_0u}{m_0-dt}[/tex]
b)
Momentum is conserved because with the law of conservation in mind mass is reduced thereby causing increase in speed
c)
On the account of maintaining speed the Rocket need not expel gas as that will only increase speed and not maintain it.Therefore the answer is NO
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True or false it is impossible to determine weather you are moving unless you can touch another object
Answer: false
Explanation:
Answer:
false
Explanation:
According to Newton's First Law of motion, an object remains in the same state of motion unless a resultant force acts on it.
John attaches a ball to a spring. The diagram below shows what happens. Which option shows the direction of the force of the ball on the spring?
Option C shows the direction of the force of the ball on the spring. The direction of the force of the ball on the spring will be downwards.
What is force?Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Force is defined as the product of mass and acceleration. Its unit is Newton.
The spring is extended downward because the weight is always act downwards. The direction of the force of the ball on the spring will be downwards.
Hence, option C shows the direction of the force of the ball on the spring
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Which three statements are true of all matter?
A.
It is filled with air.
B.
It takes up space.
C.
It contains aluminum.
D.
It has mass.
E.
It is made up of atoms
Answer:
B, D and E, not all matter can be filled with air
Consider a 150-W incandescent lamp. The filament of the lamp is 5 cm long and has a diameter of 0.5 mm. The diameter of the glass bulb of the lamp is 8 cm. Determine the heat flux in W/m2 on: (a) on the surface of the filament (b) on the surface of the glass bulb (use the assumption that the surface of the bulb can be approximated as a sphere with a diameter of 8 cm), and (c) calculate how much it will cost per year to keep that lamp on for eight hours a day every day if the unit cost of electricity is $0.08/kW
Answer:
The answer is below
Explanation:
The heat flux or heat intensity is the flow of energy per unit area per unit time. measured in watts per meter² (W/m²).
a) The surface area of filament = πDL
D = diameter = 0.5 mm = 5 * 10⁻⁴ m, L = length = 5 cm = 5*10⁻² m
surface area of filament = π* 5 * 10⁻⁴ m * 5*10⁻² m = 25π * 10⁻⁶ m²
The heat flux on the surface of filament = 150 W / (25π * 10⁻⁶ m²) = 1.91 * 10⁶ W/m²
b) Since the glass tube is a sphere with diameter = 8 cm;
radius (r) = diameter / 2 = 4 cm = 4* 10⁻² m
Surface area of glass tube = 4πr² = 4π * (4* 10⁻²)² = 64π* 10⁻⁴ m²
The heat flux on the surface of filament = 150 W / (64π* 10⁻⁴ m²) = 7.46 * 10³ W/m²
c) The total energy consumed in a year = 150 W * 365 days * 8 hours/day = 438000 Wh = 438 kWh
Cost of energy = unit cost * energy consumed = $0.08/kW * 438 kWh = $35.04
Suppose Group A runs their experiment 3 times and calculates that the best estimate of the distance slid by red blocks is 15 + 3 feet. Group B runs a similar experiment 3 times and calculates that the best estimate of the distance slid by green blocks is 25 + 4 feet. Using what you learned in the above video, find the t' parameter for the comparison of the results of Groups A and B.
In comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
The given parameters;
in group A, distance traveled by the red block, s = 15 ± 3 feet
in group B, distance traveled by the green block, s = 25 ± 4 feet
To find:
the t' parameter for comparison of the results of Groups A and B.Since both groups performed the experiment at equal times, we assume the time for both motion = t
Also, assume the initial velocity of both blocks = 0
For group A, we set-up the equation of motion as follows;
[tex]s = v_0t + \frac{1}{2} at^2 \\\\15+ 3 = 0 + 0.5\times a_1t^2\\\\18 = 0.5a_1t^2\\\\t^2 = \frac{18}{0.5a_1} \\\\t^2 = \frac{36}{a_1}[/tex]
For group B, we set-up the equation of motion as follows;
[tex]25 + 4 = v_0t + \frac{1}{2} a_2t^2\\\\29 = 0.5\times a_2t^2\\\\t^2 = \frac{29}{0.5a_2} = \frac{58}{a_2}[/tex]
Solve the first equation and the second equation together;
[tex]\frac{36}{a_1} = \frac{58}{a_2} \\\\\frac{a_2}{a_1} = \frac{58}{36} \\\\\frac{a_2}{a_1} = 1.61[/tex]
The ratio of error margin of both experiments;
[tex]\frac{4}{3} = 1.33[/tex]
The resulting parameter for comparison;
[tex]parameter, t' = 1.33 \times 1.61 = 2.14 \approx 2.0[/tex]
Thus, in comparing of result of experiment A and B, the result of A is by 2 factors differ from experiment B.
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13. Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right
This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
[tex]_{-\theta }[/tex] (-) ---------------->(+) [tex]_{+\theta }[/tex]
d
so let distance between the dipoles be d
∴ P = d[tex]\Theta[/tex]
Let [tex]\Theta_{1}[/tex] = 1 nC
so
P = d[tex]\Theta[/tex]
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let [tex]\Theta_{2}[/tex] = 2 nC
so
P = d[tex]\Theta[/tex]
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let [tex]\Theta_{3}[/tex] = 3 nC
so
P = d[tex]\Theta[/tex]
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m