Answer:
A: leaving a copper penny in vinegar until it turns green
Explanation:
The corrosion is from oxidation -- a chemical reaction between the metal and oxygen, water, and carbon dioxide in the air. Rust is the term used to describe this process when it happens to iron instead of other metals. I also took it on E2020. If you need further explanation, please comment below.
The laboratory activity which involves a chemical change is leaving a copper penny in vinegar until it turns green.
What are chemicals?The chemicals are the artificially prepared or purified and a distinct compound or substance.
The corrosion is the result of oxidation that is a chemical reaction between the metal and oxygen, water, and carbon dioxide in the air. Rust is the term used to describe the corrosion process.
Corrosion also takes place in copper due to which the copper turns green when oxidized.
Thus, The laboratory activity which involves a chemical change is leaving a copper penny in vinegar until it turns green.
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The energy stored in a wood log is transformed when the log is burned. Which of the following explanations best describes how the
chemical energy stored in the log compares to the heat and light energy produced by burning?
O A The chemical energy is equal to the amount of heat, light, and other applicable energies.
O B The chemical energy is less than the amount of heat, light, and other applicable energies.
OCThe chemical energy is more than the amount of heat, light, and other applicable energies.
D. The chemical energy stays the same; additional energy is produced as heat, light, and other applicable energies.
Answer:
D. The chemical energy stays the same; additional energy is produced as heat, light, and other applicable energies.
Hope this helps!
The chemical energy stays the same; additional energy is produced as heat, light, and other applicable energies. Thus option D is correct.
What is chemical energy ?Energy is the ability to do work It can be movement of a body to do some physical activity.
If the energy is stored in the form of chemical bonds of a complex molecule then the energy is called as chemical energy.
The energy released in the chemical reaction and produced as as a by-product, that process is known as an exothermic reaction.
For instance, chemical energy sored in biomass, batteries, natural gas, petroleum, and coal.
Dry wood also the storage of chemical energy, as it burns the chemical energy is released and converted into light energy and thermal energy
The food we eat is also an example of chemical energy storage as it is liberated during digestion process.
Thus option D is correct.
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A projectile is launched on earth at an angle θ, relative to horizontal direction. At half of its maximum height the speed of the projectile is 1.00 m/s, and at its maximum height the speed of the projectile is 0.50 m/s. What is the angle θ ?
Answer:
the angle is about 67.79 degrees
Explanation:
We know that at its maximum height, the vertical component of the projectile's launching (initial) velocity (Vyi) is zero, so at that point it total velocity equals the horizontal component of the initial velocity (Vxi = 0.5 m/s)
We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :
[tex]half\,\,max-height = \frac{v_{yi}^2}{4\,g}[/tex]
we can use this information to find the y component of the velocity at that height via the formula:
[tex]v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}[/tex]
Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:
[tex]1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\[/tex]
Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:
[tex]tan(\theta)=\frac{v_{yi}}{v_{xi}} =\frac{\sqrt{1.5} }{0.5} \\\theta= arctan(\frac{\sqrt{1.5} }{0.5})=67.79^o[/tex]
You are riding in a car that is moving at 50 m/s north.
You pull a gun out while driving and fire a bullet north at 100 m/s. If the ground is the frame of refernce, then the bullet is traveling at
m/s relative to the ground
Answer:
the bullet is traveling at 150 m/s relative to the ground
Explanation:
The two velocities are pointing in the same direction (both pointing North) therefore, they should add.
Then, from the ground frame of reference the bullet is travelling at: 50 m/s + 100 m/s = 150 m/s
The gravitational pull of the sun on Earth keeps Earth orbiting around the sun. Which statement is correct about the force that Earth exerts on the sun?
Explanation:
The gravitational force exerted by the sun on the earth keeps the earth moving or orbiting around the sun in its orbit. This force is same with force with which the earth pulls the sun or exerts on the sun.
But the sun does not orbit around the earth because of this force since the sun is more massive then earth but the earth is less massive and it is accelerated by this massive force of the sun and moves around the sun in its orbit.
Answer:
Earth pulls the sun towards itself with a force equal to the ratio of the mass of the sun to the mass of Earth
Which three statements make up the law of gravity
A) The strength of the gravitational force between two objects decreases as the distance between their centers increases.
B) The strength of the gravitational force between two objects increases as the total mass of the objects increases
C) matter attracts all other matter in the universe
D) no matter where you are in the universe you will always have the same weight.
Answer:
the answer is A, B, and C
Explanation:
D is wrong because your weight does change...it is your mass that doesn't change
Two masses m1m1 and m2m2 exert a gravitational force of 12 N onto each other when they are 6 m apart. What will the gravitational force be if the masses are moved closer to be 3 m apart?
Answer:
48N
Explanation:
because as the distance is halfed. the force on the two objects are quadrupled.
What example represents a healthy alternative to curb nicotine cravings?
O candy bars
O potato chips
O sugar-free gum
o chewing tobacco
Answer:
C.sugar-free gum
Explanation:
Which of the following is a physical property? 1.flammability 2.acidity 3.color 4.rusting
Answer:
Color is a physical property, the rest are chemical properties.
One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 2.08 km. They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What is the magnitude of the couple’s displacement? (c) What is the direction (relative to due east) of the couple’s displacement?
Answer:
(a). The distance is 9.80 km.
(b). The magnitude of the couple’s displacement is 2.94 km.
(c). The direction of the couple’s displacement is 45°.
Explanation:
Given that,
Radius of lake = 2.08 km
(a). We need to calculate the total distance
Using formula of distance
[tex]d=\dfrac{3}{4}(2\pi R)[/tex]
Put the value into the formula
[tex]d=\dfrac{3}{4}(2\pi\times2.08)[/tex]
[tex]d=9.80\ km[/tex]
(b). We need to calculate the magnitude of the couple’s displacement
Using formula of displacement
[tex]D=\sqrt{R^2+R^2}[/tex]
[tex]D=\sqrt{2R^2}[/tex]
[tex]D=\sqrt{2\times(2.08)^2}[/tex]
[tex]D=2.94\ km[/tex]
(c), The direction of the displacement is given by
Using formula of direction
[tex]\tan\theta=\dfrac{R}{R}[/tex]
[tex]\theta=\tan^{-1}(1)[/tex]
[tex]\theta=45^{\circ}[/tex]
Hence, (a). The distance is 9.80 km.
(b). The magnitude of the couple’s displacement is 2.94 km.
(c). The direction of the couple’s displacement is 45° relative to the east.
The distance they travel is approximately 6.54644 km. The magnitude of the couple's displacement is 4.16 km. The direction of their displacement is opposite to due east.
(a) The distance they travel is:
Distance = (3/4) × 2 × 3.14159 × 2.08
Distance = 3.14159 × 2.08 km
Distance = 6.54644 km
So, the distance they travel is approximately 6.54644 km.
(b) The magnitude of the couple's displacement is:
Displacement = 2 × 2.08
Displacement = 4.16 km
So, the magnitude of the couple's displacement is 4.16 km.
(c) The direction of their displacement is opposite to due east.
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The classic Millikan oil drop experiment was the first to obtain an accurate measurement of the charge on an electron. In it, oil drops were suspended against the gravitational force by a vertical electric field. Consider an oil drop with a weight of 2.9 x 10-14N, if the drop has a single excess electron, find the magnitude (in N/C) of the electric field needed to balance its weight. Your should round your answer to an integer, indicate only the number, do not include the unit.
Answer:
[tex]E=1.81\times 10^5\ N/C[/tex]
Explanation:
In Millikan oil drop experiment, oil drops were suspended against the gravitational force by a vertical electric field such that its weight is balanced by the electron force i.e.
W = qE,
W is weight, W = mg
q is charge,
E is electric field
⇒ [tex]2.9\times 10^{-14}\ N=qE[/tex]
or
[tex]E=\dfrac{2.9\times 10^{-14}\ N}{1.6\times 10^{-19}\ C}\\\\E=181250\ N/C\\\\\text{or}\\\\E=1.81\times 10^5\ N/C[/tex]
So, the magnitude of the electric field is [tex]1.81\times 10^5\ N/C[/tex].
A lounging leopard decides to come down out of her tree and hunt for her lunch in the savanna. The graph above represents the motion of the leopard as a function of time.
What is the distance traveled by the leopard during the entire 11 seconds? Please include a number and uni
A bus accelerated at 1.8 m/s2 from rest for 15 s. It then traveled at constant speed for 25 s, after which it slowed to a stop with an acceleration of 1.8 m/s2. The distance traveled by the bus was:______.
Answer:
The distance traveled by the bus was 1080 m
Explanation:
Please find the attached file for explanation
Grant sprints 50 meters to the right with an average velocity of 3.0 m/s.
How many seconds did Grant sprint?
Answer:
The Answer Is 17s
Explanation:
Displacement trianglex = 50m
Time t = ?
Average velocity v = 3.0 m/s
You can rearrange the equation v = trianglex/t to solve for time t.
v = tianglex/t
t = trianglex/v
= 50m/3.0m/s
= 17s
How is pressure related to the area
The water is wide character descriptions:__________
Answer:
Explanation:
I can't name the whole characters, I will try to name as much as I can remember, thanks.
Pat Conroy (Conrack) or (C'roy)
C'roy happens to be the male teacher on Yamacraw Island.
Dr. Henry Piedmont
Dr. Henry is the Superintendent for the school district in Beaufort, South Carolina.
Ezra Bennington
Ezra is the Deputy Superintendent of Beaufort County schools, in South Carolina
Mrs. Brown
Mrs Brown is a teacher on the Yamacraw Island. A racist teacher, if I might add.
Barbara Bolling Jones Conroy
Barbara lives with a Yamacraw Island teacher.
Ted Stone
Ted is the man with an awful lot of responsibilities on Yamacraw Island.
Lou Stone
Lou is the Yamacraw Island postmaster, in addition to being the school bus driver.
Zeke Skimberry
Zeke is the Yamacraw Island maintenance man who eventually becomes friends with one of the teachers.
Babies
This is a nickname is for the students of Yamacraw Island.
That's the lot I can mention. I hope that's enough.
A proton (mass=1.67x10^-27 kg, charge= 1.60x10^-19 C) moves from point A to point under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 m/s. At point B the speed of the proton is 80 km/s. Determine the potential difference VB-VA in volts.
Answer:
[tex]VB - VA = - 33.4[/tex]
Explanation:
Generally the workdone in moving the proton is mathematically represented as
[tex]W = KE_f - KE_i[/tex]
Where [tex]KE_i \ and \ KE_f \ are\ the\ initial \ and \ final \ kinetic \ energy [/tex]
So
[tex]KE_i = \frac{1}{2} m v_a^2[/tex]
Here [tex]v_a[/tex] is the velocity at A with value 50 m/s
So
[tex]KE_i = \frac{1}{2} (1.67*10^{-27}) * 50^2 [/tex]
[tex]KE_i = 2.09 *10^{-24} \ J [/tex]
Also
[tex]KE_f = \frac{1}{2} m v_b^2[/tex]
Here [tex]v_a[/tex] is the velocity at A with value [tex]80 km/s = 80000 m/s [/tex]
=> [tex]KE_f = \frac{1}{2} (1.67*10^{-27}) * 80000^2 [/tex]
=> [tex]KE_f = 5.34 *10^{-18} \ J[/tex]
So
[tex]W = 5.34 *10^{-18} - 2.09 *10^{-24}[/tex]
[tex]W = 5.34 *10^{-18} m/s[/tex]
Now this workdone is also mathematically represented as
[tex]W = q * V[/tex]
So
[tex] q * V = 5.34 *10^{-18} [/tex]
Here [tex]q = 1.60*10^{-19} C[/tex]
So
[tex] V = \frac{5.34 *10^{-18} }{1.60*10^{-19}}[/tex]
[tex] V = 33.4 \ V [/tex]
Generally proton movement is in the direction of the electric field it means that [tex] VA>VB [/tex]
So
[tex]VB - VA = - 33.4[/tex]
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsNfs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:
Complete Question
Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in ( The diagram is shown on the first uploaded image ) has a mass m= 10kg and is being pulled by a force F on a table with coefficient of static friction μs=0.3. Four forces act on it:
The applied force F (directed θ=30∘ above the horizontal).
The force of gravity Fg=mg (directly down, where g=9.8m/s2).
The normal force N (directly up).
The force of static friction fs (directly left, opposing any potential motion).
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:
Fcosθ−μsN=0
Fsinθ+N−mg=0
In order to find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N)
Note the diagram is shown on the first uploaded image
Answer:
The expression for F is [tex]F = \frac{\mu_s * m* g}{[\mu_s * sin (\theta )] * cos (\theta)}[/tex]
Explanation:
Generally from the diagram we see that
[tex]Fcos\theta -f_s = 0[/tex]
From the question we are told that
[tex]f_s = \mu_s * N[/tex]
So
[tex]Fcos\theta - \mu_s * N = 0[/tex]
=> [tex] N = \frac{Fcos(\theta)}{\mu_s}[/tex]
Also from the diagram
[tex]Fsin(\theta )+N - F_g = 0[/tex]
Here [tex]F_g = m * g[/tex]
So
=> [tex]Fsin(\theta )+ \frac{Fcos(\theta)}{\mu_s} - m* g = 0 [/tex]
=> [tex]F = \frac{\mu_s * m* g}{[\mu_s * sin (\theta )] * cos (\theta)}[/tex]
A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 4.55 m/s. If the roof is pitched at 22.0° below the horizon and the roof edge is 2.90 m above the ground, find the time the baseball spends in the air and the horizontal distance from the roof edge to the point where the baseball lands on the ground. HINT Solve for the time of flight using vy2 = v0y2 − 2gΔy and then vy = v0y − gt. Solve for the horizontal distance using the horizontal displacement equation. Click the hint button again to remove this hint. (a) the time the baseball spends in the air (in s) 0.434 Incorrect: Your answer is incorrect. s (b) the horizontal distance from the roof edge to the point where the baseball lands on the ground (in m) m
Answer:
Explanation:
The time that the baseball spends in the air is known as time of flight and it is expressed as shown;
T = Using(theta)/g where;
U is the velocity of the baseball
g is the acceleration due to gravity.
Given parameters
U = 4.55m/s
theta = 22.0°
g = 9.81m/s²
Substituting the values in the formula;
T = 4.55sin22°/9.81
T = 4.55(0.3746)/9.81
T = 1.7045/9.81
T = 0.1738second
Hence the time flight of the baseball is 0.1738second
b) The horizontal distance covered by the ball is called the RANGE in projectile.
Range = U√2H/g
U = 4.55m/s
H is the maximum height = 2.90m
g = 9.81m/s²
Substitute the given parameters into the formula
Range = 4.55√2(2.90)/9.81
Range = 4.55√5.8/9.81
Range = 4.55√0.5912
Range = 4.55(0.7689)
Range = 3.4986m
Hence the horizontal distance from the roof edge to the point where the baseball lands on the ground is 3.4986m
A wind turbine with a rotor diameter of 40 m produces 90 kW of electrical power when the wind speed is 8 m/s. The density of air impinging on the turbine is 1.2 kg/m3. What fraction of the kinetic energy of the wind impinging on the turbine is converted to electrical energy
Answer:
0.233
Explanation:
Given that
Diameter of rotor, d = 40 m
Power of rotor, P = 90 kW
Speed of the wind, v = 8 m/s
Density of air, p = 1.2 kg/m³
It is a known fact that
KE = ½mv², where mass flow rate, m
m = p.A.v, where Area, A
A = πd²/4
A = (3.142 * 40²)/4
A = 3.142 * 1600/4
A = 3.142 * 400
A = 1256.8 m², substitute for A in the mass flow rate equation
m = p.A.v
m = 1.2 * 1256.8 * 8
m = 12065.28, substitute for m in the KE equation
KE = ½mv²
KE = ½ * 12065.28 * 8²
KE = 12065.28 * 32
KE = 386088.96 W or
KE = 386.1 kW
Fraction of kinetic energy converted to electric energy is
Fraction = Electric Power / Total KE
Fraction = 90 / 386.1
Fraction = 0.233
What does Rockstrom mean by the need to "bend the curves"?
Answer: Rockstrom meant that it's going to be a challenging decade and that people will need to work hard and also bend the curve by thinking around the curve.
Explanation:
Some years ago, Johan Rockström, who was the executive director of Stockholm Environment Institute, and a professor at Stockholm University, was the head of an international team that was assembled in defining planetary boundaries.
The that was gathered to speak regarding issues that pertained to the Earth ane how it can be protected from being infiltrated ad protecting it from failing. During the conversation, Rockstrom said that it's going to be a challenging decade and that people will need to work hard and also bend the curve by thinking around the curve.
A runner is jogging in a straight line at a
steady vr= 5.9 km/hr. When the runner is
L= 6.6 km from the finish line, a bird begins
flying straight from the runner to the finish
line at vb= 29.5 km/hr (5 times as fast as
the runner). When the bird reaches the finish
line, it turns around and flies directly back to
the runner.
What cumulative distance does the bird
travel? Even though the bird is a dodo, assume that it occupies only one point in space
(a "zero" length bird), travels in a straight
line, and that it can turn without loss of
speed.
Answer in units of km.
Answer:
11.88 km
Explanation:
Given that the bird begins flying straight from the runner to the finish
line at vb= 29.5 km/hr (5 times as fast as the runner). When the bird reaches the finish line, it turns around and flies directly back to the runner.
Then the first distance covered by the bird is 6.6 km.
Since the speed of the bird is five times the speed of the man, the man must have covered the distance one - fifth of the 6.6 km. That is,
1/5 × 6.6 = 1.32
Take 1.32 away from 6.6 you will get
6.6 - 1.32 = 5.28
The cumulative distance the bird
travel will be:
Cumulative distance = 6.6 + 5.28 = 11.88km
Therefore, the cumulative distance the bird travelled is 11.88 km
A cylindrical tank of methanol has a mass of 60 kg and a volume of 75 L. Determine the methanol’s weight, den-sity, and specific gravity. Take the gravitational acceleration to be 9.81 m/s2. Also, estimate how much force is needed to accelerate this tank linearly at 0.25 m/s2.
Answer:
Weight = 588.6 N
Density = 800 kg/m³
Specific Gravity = 0.8
Force = 147.15
Explanation:
W = mg
W = 60 * 9.81
W = 588.6 N
Volume from L to m³ would be
75 L = 75 * 1/1000
75 L = 0.075 m³
Recall, density = mass/volume, so then, density of methanol is.
Density = 60/0.075
Density = 800 kg/m³
Specific gravity, SG is equal to
SG = density of methanol/density of water
SG = 800/1000
SG = 0.8
the force needed to accelerate the tank, F = ma
F = 588.6 * 0.25
F = 147.15 N
Therefore, the needed variables are
Weight = 588.6 N
Density = 800 kg/m³
Specific Gravity = 0.8
Force = 147.15
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.471.47 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.2810.281 seconds to pass the length of the meter stick, how high h0h0 above the ground was the acorn before it fell, assuming that the acorn did not run into any branches or leaves on the way down?
Answer:
The value is [tex]h_a = 1.712 \ m [/tex]
Explanation:
From the question we are told that
The height of the top meter stick above the ground is [tex]h_1 = 1.47 \ m[/tex]
The time taken for the acorn to pass the length of the stick is [tex]t = 0.281 \ s[/tex]
Generally the height of the acorn at the point it is the same height with the metered stick is mathematically represented as
[tex]h = h_m = u_a * t + \frac{1}{g} t^2[/tex]
Here [tex]h_m[/tex] is height of the meter stick and the value is 1 m (This because we are told in the question that the stick is 1 meter in length ( a meter stick))
So
[tex]1 = u_a * 0.281 + \frac{1}{9.8} (0.281)^2[/tex]
=> [tex]u_a = -2.2 \ m/s[/tex]
Generally the velocity of the acorn just before passing the top of the meter stick is mathematically represented by a kinematic equation as
[tex]u^2_a = u^2 + 2gs[/tex]
here u is zero since the acorn started from rest
So
[tex] (-2.2) = 0 + 2 * 9.8 * s[/tex]
[tex]s = 0.242 \ m[/tex]
Generally the height of the acorn is
[tex]h_a = h_1 + s[/tex]
[tex]h_a = 0.242 + 1.47[/tex]\
[tex]h_a = 1.712 \ m [/tex]
The resultant velocity with respect to the ground of a fighter
plane flying at 100 km/hr air speed will be
encounters a 45 km/hr tailwind.
Answer:145 km/hr
Explanation: v1+v2 100+45 =145 km/hr
Two forces P and Q act on an object of mass 9.00 kg with Q being the larger of the two forces. When both forces are directed to the left, the magnitude of the acceleration of the object is 1.20 m/s2. However, when the force P is directed to the left and the force Q is directed to the right, the object has an acceleration of 0.600 m/s2 to the right. Find the magnitudes of the two forces P and Q in newtons.
Answer:
P = 2.7 N
Q = 8.1 N
Explanation:
Q > P
Both the forces are acting towards the left
Net force = Q + P
Applying newton's 2nd law
net force = mass x acceleration
Q + P = 9 x 1.2 = 10.8 ----------------- ( 1 )
In the second case Q is acting towards the left and P acts towards the right
so they act in opposite direction
Hence net force = Q - P
Applying newton's 2nd law
net force = mass x acceleration
Q - P = 9 x .6 = 5.4 -------------------- ( 2 )
Adding ( 1 ) and ( 2 )
2 Q = 16.2
Q = 8.1 N
From ( 1 )
8.1 + P = 10.8 N .
P = 2.7 N
how many key reasons are there for time getting away from you?
Answer:
4
Explanation: the problem with this one is 75 the problem the problem with
g A very early, simple satellite consisted of an inflated spherical aluminum balloon 38 m in diameter and of mass 22 kg. Suppose a meteor having a mass of 21 kg passes within 8.0 m of the surface of the satellite. What is the magnitude of the gravitational force on the meteor from the satellite at the closest approach?
Answer:
6.37*10^-13 N
Explanation:
Given that
We r = 38/2 = 14 m
m1 = 22kg
m2 = 21 kg
so we can say that distance between center of the satellite and meteor,
d = r + h
= 14 + 8
= 22m
So the gravitational force on the meteor, is
F = G x m1 x m2/d²
= 6.67*10^-11 x 21 *22/22²
= 6.37*10^-13 N
is weight included in free body diagram
Answer:
Yes
Explanation:
I took physics last year that that was a requirement.
An air filter can remove dust particles from air but will reach capacity (saturation) at 50.0 mg. Of air containing 225 µg dust particles per m3 air is passed through the filter at 400 ft3 /min and air leaving the filter has 15.0 µg dust/m3 air, how much time, in days, is required for the filter to reach saturation
Answer:
Time to Reach Saturation = 0.0146 day
Explanation:
In order to solve this problem, we first need to calculate the dust filtered by the filter per cubic meter of air:
Filtered Dust per m³ = Dust Particles Entering per m³ - Dust Particles Leaving per m³
Filtered Dust per m³ = 225 μg/m³ - 15 μg/m³
Filtered Dust per m³ = 210 μg/m³ = 210 x 10⁻³ mg/m³
Now, we find volume flow rate of air through filter:
Volume Flow Rate of Air = (400 ft³/min)(0.3048 m/1 ft)³
Volume Flow Rate of Air = 11.33 m³/min
Now, we calculate rate of dust filtered:
Rate of Dust Filtered = (Filtered Dust per m³)(Volume Flow Rate of Dust)
Rate of Dust Filtered = (210 x 10⁻³ mg/m³)(11.33 m³/min)
Rate of Dust Filtered = 2.38 mg/min
Now, for the time required to reach saturation:
Time to Reach Saturation = (Saturation Capacity)/(Rate of Dust Filtered)
Time to Reach Saturation = (50 mg)/(2.38 mg/min)
Time to Reach Saturation = (21.02 min)(1 day/24 h)(1 h/60 min)
Time to Reach Saturation = 0.0146 day
What was the velocity of a dog that ran 50 meters in 15 seconds?
Answer:
5m/s
Explanation:
there are 3,600 seconds per hour so 5•3600 18,000 hours and 18 km/h
Answer:
velocity = 3.33 m/sec
Explanation:
velocity = distance / time
where
distance = 50 meters i
time = 15 seconds
plugin values into the formula
velocity = 50 m
15 sec
velocity = 3.33 m/sec
