each of the following conversions contains an error. In each case, explain what the error is and find the correct answer to make a true statement .
a 1000 kg mg (1kg/1000g) = 1g.
b. 50m (1cm/100m)=0.5 cm
c. "Nano" is 10^-9 , so there are 10^-9 nm in a meter.
d. micro is 10^-6, so 1kg is 10^6 ug
Answers
Answer:
a. [tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b. [tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c. "Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d. micro is 10⁻⁶, so 1kg is 10⁹ ug
Explanation:
a.
The conversion factor is written inverted. The correct statement will be:
[tex]1000\ kg(\frac{1000\ g}{1\ kg}) = 1\ x\ 10^{6}\ g[/tex]
b.
The values in the conversion factor used are wrong. The correct statement will be:
[tex]50\ m (\frac{100\ cm}{1\ m} ) = 5000\ cm[/tex]
c.
Change of units is the mistake here. The correct statement will be:
"Nano" is 10⁻⁹ , so there are 10⁻⁹ meter in a nm (OR) "Nano" is 10⁻⁹ , so there are 10⁹ nm in a meter.
d.
the conversion will be as follows:
[tex]1\ kg(\frac{1000\ g}{1\ kg})(\frac{1\ \mu g}{10^{-6}\ g}) = 10^9 \mu g[/tex]
therefore, the correct statement will be:
micro is 10⁻⁶, so 1kg is 10⁹ ug
Related Questions
A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.
Answers
This question is incomplete, the complete question is;
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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).
Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.
Answer:
the work done is -88 J
Explanation:
Given the data in the question;
we know that;
Work done = F × S
where constant force F = ( 6i + 8j - 6k )
S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )
S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )
S = ( -12I + 7j + 12k )
so
Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )
Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )
Work force = -72 + 56 - 72
Work force = -88 J
Therefore, the work done is -88 J
Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of the earth. The second satellite travels at aspeed that is less than v.At what distance from the center of the earth does the secondsatellite orbit?At a distance that is less than r.At a distance equal to r.At a distance greater than r.Now assume that a satellite of mass m is orbiting the earth at a distance r from the center of the earth with speed v_e. An identical satellite is orbiting the moon at thesame distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make onerevolution compare to the time T_e it takes the satellite orbiting the earth to make onerevolution?T_m is less than T_e.T_m is equal to T_e.T_m is greater than T_e.
Answers
Answer:
a. At a distance greater than r
b. T_m is greater than T_e.
Explanation:
a. Two identical satellites orbit the earth in stable orbits. One satellite orbits with a speed vat a distance r from the center of the earth. The second satellite travels at a speed that is less than v. At what distance from the center of the earth does the second satellite orbit?
Since the centripetal force on any satellite, F equals the gravitational force F' at r,
and F = mv²/r and F' = GMm/r² where m = mass of satellite, v = speed of satellite, G = universal gravitational constant, M = mass of earth and r = distance of satellite from center of earth.
Now, F = F'
mv²/r = GMm/r²
v² = GM/r
v = √GM/r
Since G and M are constant,
v ∝ 1/√r
So, if the speed decreases, the radius of the orbit increases.
Since the second satellite travels at a speed less than v, its radius, r increases since v ∝ 1/√r.
So, the distance the second satellite orbits is at a distance greater than r
b. An identical satellite is orbiting the moon at the same distance with a speed v_m. How does the time T_m it takes the satellite circling the moon to make one revolution compare to the time T_e it takes the satellite orbiting the earth to make one revolution?
Since the speed of the satellite, v = √GM/r where M = mass of planet
Since the satellite is orbiting at the same distance, r is constant
So, v ∝ √M
Since mass of earth M' is greater than mass of moon, M", the speed of satellite circling moon, v_m is less than v the speed of satellite circling earth at the same distance, r
Now, period T = 2πr/v where r = radius of orbit and v = speed of satellite
Since r is constant for both orbits, T ∝ 1/v
Now, since the speed of the speed of the satellite on earth orbit v is greater than the speed of the satellite orbiting the moon, v_m, and T ∝ 1/v, it implies that the period of the satellite orbiting the earth, T_e is less than the period of the satellite orbiting the moon, T_m since there is an inverse relationship between T and v. T_e is less T_m implies T_m is greater than T_e
So, T_m is greater than T_e.
you are given a set of facts regarding a lens : object heigh, and dostance to objects. Given this jnformation, how can you tell if you're dealing with a concave or convex lens
Answers
Answer:
concave curves inward like an hourglass and convex is an outward curve like a football
Explanation:
hope this helps
A hoop rolls with constant velocity and without sliding along level ground. Its rotational kinetic energy is:______a- half its translational kinetic energyb- the same as its translational kinetic energyc- twice its translational kinetic energyd- four times its translational kinetic energy
Answers
Answer:
The same as its translational KE.
The easy way to do this is to make up numbers and use them.
So, I'll say m=2 and r=3. I will also say v=3 .
Rot. Inertia of a hoop is mr^2. So the rot KE is: 1/2 (mr^2)(w^2)
note: (1/2*I*w^2)
Translational kinetic energy is basically normal KE, so 1/2(m)(v^2)
Now, lets plug our made up values in:
Rot Ke : 1/2 (9*2)(3/3) *note w = v/r
Tran Ke: 1/2(2)(9)
Rot Ke: 9
Tran Ke: 9
9=9, same.
If the pressure of a gas is really due to the random collisions of molecules with the walls of the container, why do pressure gauges – even very sensitive ones – give perfectly steady readings? Shouldn't the gauge be continually jiggling and fluctuating? Explain.
Answers
Answer:
there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
Explanation:
The pressure measurement is carried out by calibrating the force exerted by the air on a surface of known area, suppose a small area 1 mm² = 0.01 cm²
To find out if the random movement of air molecules affects the pressure reading, let's calculate the number of molecules that reaches the pressure gauge.
In a system at atmospheric pressure and in a volume of 1 m³ (walls of 1 m each) there is one mole of air molecules, this mole is evenly distributed, so how many molecules fall on our surface
# _molecule = 6.02 10²³ 0.01 10⁻⁴ / 1
#_molecular = 6.02 10¹⁷ molecules per second
therefore the variation of the number of molecules is not very important
Consequently there is no fluctuation in the measurement because the quantity of molecule is too large and a quantity of some molecules is imperceptible.
instrument used in measurement Amount of substance
Answers
Answer:
For liquids: A measuring cylinder is used.
For solid: Over flow can is used
Answer:
i think a measuring cylinder
The latent heat of vaporization of water is roughly 10 times the latent heat of fusion of water. The amount of heat required to boil away 1 kg of water is __________ the amount of heat required to melt 1 kg of ice.
Answers
Answer:
The amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
Explanation:
let the latent heat of fusion of ice = L
then, the latent heat of vaporization of water = 10L
The heat of fusion of 1 kg of ice = 1 x L = L
The heat of vaporization 1 kg of water = 1 x 10L = 10L
Therefore, the amount of heat required to boil away 1 kg of water is 10 times the amount of heat required to melt 1 kg of ice
eididjsmsisijsjsiakaksannnahshsjeejekekekkeie
Answers
Answer:
ok
Explanation:
nskakkskdnsksmskzkksnsnxksjsjos
The image shows the right-hand rule being used for a current-carrying wire.
An illustration with a right hand with fingers curled and thumb pointed up.
Which statement describes what the hand shows?
When the current flows down the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
When the current flows down the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Answers
Answer:
The answer is (D): When the current flows up the wire, the magnetic field flows in on the left side of the wire and out on the right side of the wire.
Explanation:
if p=2i+4j+3k and q=I+5j-2k,find P×q.
Answers
Answer:
[tex]p\times q=-23i+7j+6k[/tex]
Explanation:
We are given that
p=2i+4j+3k
q=i+5j-2k
We have to find pxq
We know that
[tex]p\times q=\begin{vmatrix} i&j &k\\ 2&4 & 3\\ 1& 5 & -2\end{vmatrix}[/tex]
[tex]p\times q=i(-8-15)-j(-4-3)+k(10-4)[/tex]
[tex]p\times q=-23i+7j+6k[/tex]
Hence,[tex]p\times q=-23i+7j+6k[/tex]
A spring scale hung from the ceiling stretches by 6.1cm when a 2.0kg mass is hung from it. The 2.0kg mass is removed and replaced with a 2.8kg mass.What is the stretch of the spring?
Answers
A block of mass m is moved over a distance d. An applied force F is directed perpendicularly to the block’s displacement. How much work is done on the block by the force F?
Answers
zero
Explanation:
Work W is defined as
W = F•d = Fdcos(theta)
and it is a dot product of the force and displacement and theta is angle between F and d Since the force is perpendicular to d, angle is 90° thus cos90 = 0. Hence work is zero.
A scenario where reaction time is important is when driving on the highway. During the delay between seeing an obstacle and reacting to avoid it (or to slam on the brakes!) you are still moving at full highway speed. Calculate how much distance you cover in meters before you start to put your foot on the brakes if you are travelling 65 miles per hour.
Answers
Answer:
66.83 meters
Explanation:
After a quick online search, it seems that scientists calculate the average reaction time of individuals as 2.3 seconds between seeing an obstacle and putting their foot on the brakes. Now that we have this reaction time we need to turn the miles/hour into meters/second.
1 mile = 1609.34 meters (multiply these meters by 65)
65 miles = 104,607 meters
1 hour = 3600 seconds
Therefore the car was going 104,607 meters every 3600 seconds. Let's divide these to find the meters per second.
[tex]\frac{104,607}{3600} = \frac{29.0575 meters}{1 second}[/tex]
Now we simply multiply these meters by 2.3 seconds to find out the distance covered before the driver puts his/her foot on the brakes...
29.0575m * 2.3s = 66.83 meters
The lever of a car lift has an area of 0.2 meters squared, and the area of the lift under the car is 8
meters squared. If you push with a force of 3 newtons, how much force will be applied to the
car?
Answers
Answer:
THE ANSWER IS SOMETHING LIKE 55
What is the friction force on a box that has a mass of 15kg as it slides across the floor. The coefficient of friction of the not very clean floor is 0.25
please explain everything including formula used
Answers
Answer:
36.75 N
Explanation:
Applying
F = mgμ................. Equation 1
Where F = Friction force on the box, m = mass of the box, g = acceleration due to gravity of the box, μ = coefficient of static friction
From the question,
Given: m = 15 kg, μ = 0.25
Constant: g = 9.8 m/s²
Substitute these values into equation 1
F = 15(9.8)(0.25)
F = 36.75 N
Hence the friction force on the box is 36.75 N
A car moving in a straight line uniformly accelerated speed increased from 3 m / s to 9 m / s in 6 seconds. With what acceleration did the car move?
a.
2 m/s2
b.
1 m/s2
c.
0 m/s2
d.
3 m/s2
Answers
Answer:
b) 1 m/s
I am sure...........
A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string
Answers
Answer:
9.42 m/s
Explanation:
Applying,
V' = Aω.............. Equation 1
Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.
But,
ω = 2πf................. Equation 2
Where f = frequency, π = pie
And,
f = v/λ................ Equation 3
Where, λ = wave length, v = velocity
Also,
v = √(T/μ)................. Equation 4
Where T = Tension, μ = linear density.
From the question,
Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m
Substitute into equation 4
v = √(50/0.005)
v = √(10000)
v = 100 m/s
Also Given: λ = 2.0 m
Substitute into equation 3
f = 100/2
f = 50 Hz.
Substitute the value of f into equation 2
Where π = constant = 3.14
ω = 2(3.14)(50)
ω = 314 rad/s
Finally,
Given: A = 3.0 cm = 0.03 m
Substitute into equation 1
V' = 0.03(314)
V' = 9.42 m/s
a. A horse pulls a cart along a flat road. Consider the following four forces that arise in this situation.
1. the force of the horse pulling on the cart
2. the force of the cart pulling on the horse
3. the force of the horse pushing on the road
4. the force of the road pushing on the horse
b. Suppose that the horse and cart have started from rest; and as time goes on, their speed increases in the same direction. Which one of the following conclusions is correct concerning the magnitudes of the forces mentioned above?
1. Force 1 exceeds Force 2.
2. Force 2 is less than Force 3.
3. Force 2 exceeds Force 4.
4. Force 3 exceeds Force 4.
5. Forces 1 and 2 cannot have equal magnitudes.
Answers
Answer:
a) F₁ = F₂, F₃ = F₄, b) the correct answer is 3
Explanation:
a) In this exercise we have several action and reaction forces, which are characterized by having the same magnitude, but different direction and being applied to different bodies
Forces 1 and 2 are action and reaction forces F₁ = F₂
Forces 3 and 4 are action and reaction forces F₃ = F₄
as it indicates that the
b) how the car increases if speed implies that force 1> force3
F₁ > F₃
therefore the correct answer is 3
If 1.02 ✕ 1020 electrons move through a pocket calculator during a full day's operation, how many coulombs of charge moved through it?
Answers
Answer:
Explanation:
one electron has [tex]1.60217662*10^{-19}~coulombs~then\\\\1.02*10^{20}~electrons------->1.02*10^{20}*1.60217662*10^{-19}~coulombs= 16.3422~coulombs[/tex]
Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous
Answers
Answer:
francium
Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.
According to the model, when was the universe at its most dense?
A) During the Dark Ages where matter increased in mass.
B) Just before the Big Bang where all matter existed in a singularity.
C) During the nuclear fusion events, as the atoms become more massive.
D) Current day, as the number of galaxies, solar systems, and planets have increased.
Answers
Answer:
The Answer is D
Explanation:
Hope this helps!!!!
A swimmer heading directly across a river that is 200 m wide reaches the opposite bank in 6 min 40 s. During this swim, she is swept downstream 480 m. How fast can she swim in still water
Answers
Answer:
The speed of the swimmer in stil water is 0.5 m/s
Explanation:
Given;
total time taken to swim across = 6 mins 40 s = (6 x 60s) + 40 s = 400 s
width of the river, = 200 m
Please find the image attached for explanation.
The cycle is a process that returns to its beginning, but it does not repeat
itself.
True
False
Answers
An empty 12,954 kg railroad car, traveling at a speed of 28 m/s strikes a partially filled 17,616 kg railroad car moving in the same direction at a speed of 5 m/s. What is the total momentum of the two railroad cars AFTER the collision?
Answers
Answer:
450792 kgm/s
Explanation:
by conservation of momentum,
total momentum AFTER collision = total momentum BEFORE collision
=mv+m'v'
=12954×28+17616×5
=450792 kgm/s
Sunsets are a deep red because A) tiny particles in the air are more efficient at scattering short wavelength light than they are at scattering long wavelength light. Hence, long wavelength light ends up coming directly towards you. B) most polluting gases and dust particles in the air are reddish in color and lend their color to that of the sky. C) air molecules absorb red light more efficiently than they do blue light because of their electron orbitals. D) air molecules absorb blue light more efficiently than they do red light because of their electron orbitals.
Answers
Answer:
i think its A
Answer:tiny particles in the air are more efficient at scattering short wavelength light than they are at scattering long
Explanation:
What are the messing forces that would make the object be in equilibrium?
Answers
Answer:
A) 20 N, B) 20 N, & C) 8 N
Explanation:
For the object to be in equilibrium, the upward forces must be equal to the downward forces and the forward forces must be equal to the backward forces.
1. Determination of A and B.
Forward forces = Backward forces
A + 10 + B = 25 + 25
A + 10 + B = 50
Collect like terms
A + B = 50 – 10
A + B = 40
Assume A and B to be equal. Thus, A is 20 N and B is 20 N.
2. Determination of C
Upward forces = Downward forces
C + 112 = 20 + 100
C + 112 = 120
Collect like terms
C = 120 – 112
C = 8 N
Thus, for the object to be in equilibrium, A must be 20 N, B must be 20 N and C must be 8N.
1. Convert the following length into meters
a. 123.50mm
b. 560cm
c. 100dm
d. 125.89km
Answers
Using only astronomical data from the Appendix E in the textbook, calculate the speed of the planet Venus in its essentially circular orbit around the sun.
Venus = 4.87x10^24
Answers
Answer:
[tex]v=3.49\times 10^4\ m/s[/tex]
Explanation:
Given that,
Mass of Venus, [tex]M_V=4.87\times 10^{24}\ kg[/tex]
We know that,
Mass of Sun, [tex]M_s=1.98\times 10^{30}\ kg[/tex]
The distance between the center of Sun and the center of Venus is [tex]1.08\times 10^{11}\ m[/tex]
We need to find the peed of the planet Venus in its essentially circular orbit around the sun. using the formula,
[tex]v=\sqrt{\dfrac{GM_s}{r}}[/tex]
Put all the values,
[tex]v=\sqrt{\dfrac{6.67\times 10^{-11}\times 1.98\times 10^{30}}{1.08\times 10^{11}}}\\\\v=3.49\times 10^4\ m/s[/tex]
So, the speed of the planet venus is [tex]3.49\times 10^4\ m/s[/tex].
A boy throws a ball straight up with a speed of 21.5 m/s. The ball has a mass of 0.19 kg. How much gravitational potential energy will the ball have at the top of its flight? (Assume there is no air resistance.) A. 43.9 J B. 37.5 J C. 48.5 J D. 41.2 J
Answers
Answer:
Explanation:
The equation fo potential energy is PE = mgh, where m is the mass of the ball, g is the pull of gravity (constant at 9.8), and h is the max height of the ball. What we do not have here is that height. We need to first solve for it using one-dimensional equations. What we have to know above all else, is that the final velocity of an object at its max height is always 0. That allows us to use the equation
[tex]v_f=v_0+at[/tex] where vf is the final velocity and v0 is the initial velocity. We will find out how long it takes for the object to reach that max height first and then use that time to find out what that max height is. Baby steps here...
0 = 21.5 + (-9.8)t and
-21.5 = -9.8t so
t = 2.19 seconds (Keep in mind that if I used the rules correctly for sig fig's, the answer you SHOULD get is not one shown, so I had to adjust the sig fig's and break the rules. But you know what they say about rules...)
Now we will use that time to find out the max height of the object in the equation
Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:
Δx = [tex]21.5(2.19)+\frac{1}{2}(-9.8)(2.19)^2[/tex] which simplifies down a bit to
Δx = 47.1 - 23.5 so
Δx = 23.6 meters.
Now we can plug that in to the PE equation to find the PE of the object:
PE = (.19)(9.8)(23.6) so
PE = 43.9 J
Two thin conducting plates, each 56.0 cm on a side, are situated parallel to one another and 7.0 mm apart. If 10^-10 electrons are moved from one plate to the other, what is the electric field between the plates?
Answers
Answer:
[tex]E=576.5V/m[/tex]
Explanation:
From the question we are told that:
Length [tex]l=56.0cm=0.56m[/tex]
Distance apart [tex]d=7.0mm=0.007m[/tex]
Electron Transferred [tex]n=10^{-10}[/tex]
Therefore
Total Charge
Since Charge on each electron is
[tex]e=1.602*10^{-19}[/tex]
Therefore
[tex]T=1.602*10^{-19} *10^{10}[/tex]
[tex]T=1.602*10^{-9}[/tex]
Generally the equation for Charge density is mathematically given by
[tex]\rho=T/A[/tex]
Where
Area
[tex]A=0.56*0.56[/tex]
[tex]A=0.3136[/tex]
Therefore
[tex]\rho=1.602*10^{-9}/0.3136[/tex]
[tex]\rho=5.10*10^{-9}[/tex]
Generally the equation for Electric Field in the capacitor is mathematically given by
[tex]E=\frac{\rho}{e_0}[/tex]
[tex]E=\frac{5.10*10^{-9}}{8.85x10{-12}}[/tex]
[tex]E=576.5V/m[/tex]
