Answer:
a) ρ = 6.25 10⁵ μg / m³, b) ρ = 1 10⁷ μg / m³
Explanation:
Let's analyze the exercise a little before starting, we must know the amount of pollutant in the box, that the one that enters less the one that degrades and with this value find the density or concentration.
Let's start by finding the volume of air that goes into the box
V = Lh x
Let's find the distance of air that enters per unit of time, as it goes at constant speed
x = v₀ t
we substitute
V₀ = Lh v₀ t
At this same time, a quantity of pollutant is distributed
Q₀ = r t
the contaminant that is entering reaches the entire box, therefore the total amount of contaminant is
Q = Qo t
we substitute
Q = r t²
the net amount of pollutant that remains is that less enters the one that degraded in the same time, as they ask for the steady state
[tex]Q_{net}[/tex]= Q - k t
the pollutant concentration is
ρ = Q_net / V
V = L L h
ρ =[tex]\frac{r \ t^2 - k \ t}{ L^2 h}[/tex]
ρ = [tex](r \frac{ L^2}{v_o^2} - k \frac{L}{v_o} ) \frac{1}{L^2 h}[/tex]
ρ = [tex]\frac{r}{ v_o h} -\frac{k}{v_o L h}[/tex]
let's reduce the magnitudes to the SI system
r = 10 kg / s
L = 100 km = 100 10³ m
h = 1 km = 1 10³ m
k = dq / dt = 0.20 1/h ( 1h/3600 s) = 5.5555 10⁻⁵ 1/s
v₀ = 4 m / s
let's calculate
The volume of the box
V = (100 100 1) 109
V = 1 10¹³ m³
ρ = [tex]\frac{10}{ 4^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 4 \ 100 \ 10^3 1 \ 10^3}[/tex]
ρ = [tex]6.25 10^{-4} - 1.389 ^{-13}[/tex]
ρ = 6.25 10⁻⁴ kg / m³
let's reduce to μg / m³
ρ = 6.25 10⁻⁻⁴ kg / m³ (10⁹ μg / 1kg)
ρ = 6.25 10⁵ μg / m³
b) in case the air speed decreases to v₀ = 1 m / s
ρ= \frac{10}{ 1^2 \ 1\ 10^3 } - \frac{5.5556 \ 10^{-5}}{ 1 \ 100 \ 10^3 1 \ 10^3}
ρ = 1 10⁻² - 5.5556 10⁻¹³
ρ = 1 10⁻² kg / m³
ρ = 1 10⁷ μg / m³
What was the average speed in km/h of a car that travels 788 km in 7.1 h?
Answer:
Well the answer is 111 (rounded) km/h
Explanation:
788/7.1 is 110.9... so if you round it it would equal 111.
A Van de Graaff generator is one of the original particle accelerators and can be used to accelerate charged particles like protons or electrons. You may have seen it used to make human hair stand on end or produce large sparks. One application of the Van de Graaff generator is to create x-rays by bombarding a hard metal target with the beam. Consider a beam of protons at 1.10 keV and a current of 4.65 mA produced by the generator.
(a) What is the speed of the protons?
(b) How many protons are produced each second?
Solution :
Given that :
The energy of the protons, K.E. = 1.10 keV
[tex]$= 1.10 \times 10^3 \ eV $[/tex]
The current produced by the generator is I = 5 mA
[tex]$= 5 \times 10^{-3} \ A$[/tex]
Now [tex]$1 \ eV = 1.6 \times 10^{-19 }\ J$[/tex]
Mass of the proton, m = [tex]$1.67 \times 10^_{-27} $[/tex] kg
Charge of the proton, [tex]$q_p = 1.6 \times 10^{-19} \ C$[/tex]
a). Therefore using the formula for K.E. we can find out the velocity of the proton.
[tex]$K.E. =\frac{1}{2}mv^2$[/tex]
[tex]$v=\sqrt{\frac{2K.E.}{m}}$[/tex]
[tex]$v=\sqrt{\frac{2\times 10^3 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}}$[/tex]
[tex]$= 4.38 \times 10^5 \ m/s$[/tex]
b). We know that the current is :
[tex]$I=\frac{\Delta Q}{\Delta t}$[/tex]
Therefore, the total charge in one second is given by :
[tex]$\Delta Q = I \times \Delta t$[/tex]
[tex]$= 5 \times 10^{-3} \times 1$[/tex]
[tex]$= 5 \times 10^{-3}\ C$[/tex]
So, the number of protons in this charge is given by :
[tex]$n = \frac{\Delta Q}{q_p}$[/tex]
[tex]$=\frac{5 \times 10^{-3} }{1.6 \times 10^{-19}}$[/tex]
[tex]$= 3.13 \times 10^{16}$[/tex] protons
A bicyclist accelerates from rest to a speed of
5.0 meters per second in 10 seconds. During the
same 10 seconds, a car accelerates from a speed
of 22 meters per second to a speed of 27 meters
per second. Compared to the acceleration of the
bicycle, the acceleration of the car is
Answer:
They have the same acceleration of 0.5m/s2 (please note m/s2 is the unit for acceleration and 2 is the power of s)
Explanation:
acceleration= velocity ÷ time
and the time is said to be 10seconds
velocity of car will be the new velocity- the initial velocity = 27-22= 5
acceleration= 5÷10
acceleration= 0.5
hope this helped
now hit that crown button :)
Help with both questions I’ll mark brainliest
Answer:
gas, liquid, solid
sound cannot travel in space
Answer:
1. gas , liquid , gas
2.sound cannot travel in space
Which is an example of positive peer pressure?
O A parent telling her son to wear his seatbelt.
O A student asking others to join the leadership club.
O a teen encouraging a friend to drink alcohol.
O A teacher telling his students to have a safe weekend.
Answer:
B. A student asking others to join the leadership club.
Explanation:
A parent isn't a peer, therefore can't be peer pressure. Drinking is negative for your body, therefore not positive peer pressure. A teacher isn't a peer, therefore the only answer left is B.
Plzz answer correctly
Answer:
B.
Explanation:
Explain how a common housecat gets “worms.”eplain(science)
Answer:
Cats most commonly contract worms after coming into contact with parasite eggs or infected feces. A cat may walk through an area with eggs or infected feces, and since cats are often such fastidious groomers, they will then ingest the eggs or fecal particles as they clean their fur and feet.
Explanation:
this is the only thing in my book hope it helps
Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of masses.
Explanation:
new non law neutron means neutral then it's important that baseball and softball features small respectable range of masses soft it means that when a ball hits anything hard it comes back by the Newton Law if the baseball is big and the small boy small and then if the contract with each other they ignore triple so when a ball hits the wall if the comeback because of the Mutants and when a big ball if we throw it to the wall it doesn't come that it comes back but in a very low way because it contains less neutrons in it if it is helpful please share with me
You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 5.00 s after it was thrown. What is the speed of the rock just before it reaches the water 22.0 m below the point where the rock left your hand? Ignore air resistance.
Answer:
Explanation:
for vertical movement , time to reach the top = time to reach the hand = 2.5 s
v = u - gt
At the top , v = 0 , time t = 2.5 s
0 = u - g x 2.5
u = 2.5 x 9.8 = 24.5 m /s
velocity of throw = 24.5 m /s
So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v
v² = u² + 2 g s
= 24.5² + 2 x 9.8 x 22
= 600.25 + 431.2
= 1031.45
v = 32.11 m /s .
Displacement vector A points due east and has a magnitude of 1.9 km. Displacement vector B points due north and has a magnitude of 2.08 km. Displacement vector C points due west and has a magnitude of 2.4 km. Displacement vector D points due south and has a magnitude of 2.8 km. Find the magnitude and direction (relative to due east) of the resultant vector A + B + C + D
Answer:
Explanation:
We shall represent all the four displacement in vector form in terms of unit vector i and j where i represents unit vector towards east , j represents unit vector towards north .
Displacement of A
D₁ = 1.9 i
Displacement of B
D₂ = 2.08 j
Displacement of C
D₃ = - 2.4 i
Displacement of D
D₄ = -2.8 j
Resultant displacement
= 1.9 i + 2.08 j - 2.4 i - 2.8 j
= - 0.5 i - 0.72 j
magnitude of resultant vector
= √ ( .5² + .72² )
=√ ( .25 + .5184 )
= √ .7684
= .876 km
Both i and j are negative of resultant displacement
hence its direction is towards south of west . Angle with west is Ф .
TanФ = .5184 / .25 = 2.0736
Ф = 64.25° .
From east direction is = 180 + 64.25 = 244.25° .
A transformer has 150 turns in the primary coil and 350 turns in its secondary coil. If the primary coil has a voltage of 200 volts, how many volts will the secondary coil have?
242 volts
288
353
467
Answer:
467 volts
Explanation:
Vs/Vp = Ns/Np
Vs = Ns/Np × Vp
Vs = 350/150 × 200 = 7/3 × 200
Vs = 467 volts
How do you classify flammable liquid to gas and solid?
Answer:
A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.A flammable liquid enjoys the attention of at least three different federal agencies: the DOT in matters of its transportation, OSHA as it might affect workplace safety, and the EPA concerning its cradle-to-grave management.Add to that a gaggle of state, regional, and local authorities; and we feel compelled to begin this blog entry with our favorite caveat: get expert advice before deciding what to do with that rusting drum of stale gasoline out back.Explanation:
I hope it help you;)
true or false please help me now.
Calibration graphs can be used to determine unknown concentrations in electrochemical
Answer:
false
Explanation:
Two blocks collide on a frictionless surface, as shown above. They have a combined mass of 10 kg and a speed of 2.5 m/s. Before the collision, one of the blocks was at rest. This block had a mass of 8 kg. What was the speed of the second block?
Answer:
12.5 m/s
Explanation:
Excuse my scribbles!
I had to work backwards using the inelastic collision formula for this problem.
Formula: V=(M₁V₁+M₂V₂)/(M₁+M₂)V= Combined SpeedM₁= Block 1's MassV₁= Block 1's Velocity M₂= Block 2's MassV₂= Block 2's VelocityStep 1: Substitute in the values provided in the problem
Combined mass: 10kgCombined speed: 2.5m/sBlock 1's mass: 8kgBlock 1's speed: 02.5=(8*0)+(?*?)/(8+?)
Step 2: Subtract block 1's mass from the combined mass to determine block 2's mass
10-8=2 Block 2's mass is 2.
2.5=(8*0)+(2*x)/(8+2)
Now simplify.
2.5=(2*x)/(8+2)
2.5=2x/10
Step 3: Multiply both sides by the reciprocal
(5)2.5=2x/10(5)
12.5=x
Answer is checked in the attached images!
A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:
Complete question is;
A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at the same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 3 ft/s. How fast is the weight rising when the worker has walked:
A) 10 feet
B) 30 feet
Answer:
A) 0.728 ft/s
B) 1.8 ft/s
Explanation:
Let the the position of the worker in ft be denoted by s.
Since he begins to walk away at a constant rate of 3 ft/s, then;
ds/dt = 3 ft/s
Now, the rope will form a triangle, with width "s" and the height 40. Since distance from the connection point to the ceiling = 40 ft
Using pythagoras theorem, we can find the length of the rope on this side of the pulley.
Hence, the length of rope on this side of the pulley = √(s² + 40²)
Meanwhile, on the other side the length will be;
(80) - √(s² + 40²)
Also, height of the weight will be;
h = 40 - ((80) - √(s² + 80²))
h = √(s² + 80²) - 40
Differentiating this, we have;
dh/dt = (ds/dt) × (s/√(s² + 40²))
From earlier, we saw that ds/dt = 3 ft/s
Thus;
dh/dt = 3s/√(s² + 40²)
A) when he has walked 10 ft, it means that s = 10. Thus;
dh/dt = (3 × 10)/√(10² + 40²)
dh/dt = 0.728 ft/s
B) when he has walked 30 ft, it means that s = 30. Thus;
dh/dt = (30 × 3)/√(30² + 40²)
dh/dt = 1.8 ft/s
If a total 50 J of work are done on an object, it's energy...
Answer:
0.0119502868 kilocalorie
Explanation:
Answer:
increases by 50
Explanation:
Which statement is true of a glass lens that diverges light in air?
A.
It is thick near the center and thin at the edges.
B.
It is thin near the center and thick at the edges.
C.
It is uniformly thick.
D. It is uniformly thin.

Answer: it is thin near the center and thick at the edges
Explanation: took the test on Plato :)
A battery has an emf of ε = 15.0 V. THe terminal voltage of the battery is Vt = 11.6 V when it is delivering P = 20.0 W of power to an external load resistor R. (a) What is the value of R? (b) What is the internal resistance r of the battery?
AnswerHM???
Explanation:
I dONT KNOW
An uncharged parallel-plate capacitor is connected through an open switch to a battery of voltage VV. The switch is closed and the capacitor is allowed to charge. As the capacitor is charged, energy is transferred to it from the battery. When the capacitor is fully charged, the energy stored in the capacitor is U1U1 . The energy stored in the capacitor when the stored charge is q02q02 is
Answer:
(1/2)U₁
Explanation:
An uncharged parallel-plate capacitor is connected through an open switch to a battery of voltage VV. The switch is closed and the capacitor is allowed to charge. As the capacitor is charged, energy is transferred to it from the battery. When the capacitor is fully charged, the energy stored in the capacitor is U1U1 . The energy stored in the capacitor when the stored charge is q₁/2 is
Solution:
A capacitor is an electrical device used to store electrical energy in an electric field. The energy stored in a capacitor is given by:
U = (1/2)QV; where U is the energy stored, Q is the charge and V is the voltage applied.
The energy stored in a fully charged capacitor with a charge q₁ and battery of voltage V is given as:
U₁ = (1/2)q₁V
If the stored charge is q₁/2, the energy stored (U₂) becomes:
U₂ = (1/2)(q₁ / 2)V
U₂ = (1/2)* (1/2)q₁V
U₂ = (1/2)U₁
If a car's speed triples, how does the momentum and kinetic energy of the
car change? Answer in form (momentum change, kinetic energy change)
Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.
Explanation:
The momentum of a car (an object) is
p= mv
where
m is =the mass of the object( in this case car)
v is its= velocity
While the kinetic energy is is given by the formulae
K=1/2mv²
To determine how momentum and kinetic energy of the car changes when the speed of the object triples, We have that the new velocity,
v¹= 3v
So that the momentum change becomes
p¹=mv¹=m (3v)= 3mv
mv=p
therefore p¹= 3p
we can see that the momentum also triples.
And the kinetic energy change becomes
K¹=1/2m(v¹)²= 1/2m (3v)²
= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²
1/2mv²=K
K¹= Kinetic energy = 9k
but Kinetic energy increases 9 times
A parallel-plate capacitor is connected to a battery until it is fully charged. Then, the capacitor is disconnected from the battery and connected to two uncharged parallel plates that make up a capacitor. The potential between the plates of the initial capacitor will
Answer:
The potential between the plates will decrease.
Explanation:
An insulator is usually placed between the parallel plates and is also called a dielectric because it makes the amount of charge a capacitor can accommodate to increase at a particular potential difference.
Furthermore, the dielectric effect will make the electric field of the charged capacitor which is not connected to a source of supply to decrease.
Now, when the battery is removed, the charge Q remains constant and Capacity C will increase.
Formula for the potential difference is here;
V = Q/C
Since the numerator Q is constant and the denominator C increases, it means the potential difference V will decrease
what is borh's postulates for the hydrogen atom
Answer:
An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. ... Each orbit corresponds, to a certain energy level.
Explanation:
Hope it is helpful....
a surfer talks about riding a 20-foot wave. Which measurement of waves is the surfer describing?
frequency
amplitude
wavelength
speed
60 POINTS!!
Answer:
C. Amplitude
Explanation: Amplitude is the maximum displacement from the equilibrium of a wave. Basically the height.
two faer coin and unbayers
dice are thrown together list the
Sample space
determine the probabilities that
A head and even number
A prime number and atleast a tail
who is bill cypher and what is his origin?
Answer:
Bill Cipher is the true main antagonist of Gravity Falls. He is a Dream-Demon with mysterious motives and seems to have a vendetta against the Pines family, especially his old rival Stanford Pines
Explanation:
If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. the floor deforms very slightly under your weight only if you are heavy enough does the floor deform at all under your weight the floor does not deform at all under your weight
Answer:
the floor deforms very slightly under your weight
Explanation:
A trampoline is made up of a large piece of strong cloth held by springs on which you jump up and down as a sport. So, If you stand on a trampoline, it depresses under your weight. However, the floor does not deform under your weight as it is too stiff.
Therefore,
when you stand on a hard stone floor, the floor deforms very slightly under your weight.
Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay aboard the space station to be a full second youngerthan people on the ground? Please show and explain how you would set-up the problem,before you actually try to solve it. If you cannot solve it exactly, please try to offer an estimate.(5 pts)
Answer:
#_time = 7.5 10⁴ s
Explanation:
In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.
t = [tex]\frac{t_p}{ \sqrt{1- (v/c)^2} }[/tex]
where t_p is the person's own time in an immobile reference frame,
[tex]t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }[/tex]
let's calculate
we assume that the speed of the space station is constant
[tex]t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }[/tex]
[tex]t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}[/tex]
t_ = 0.99998666657 s
therefore the time change is
Δt = t - t_p
Δt = 1 - 0.9998666657
Δt = 1.3333 10⁻⁵ s
this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s
#_time = 1 / Δt
#_time =[tex]\frac{1}{1.3333 \ 10^{-5}}[/tex]
#_time = 7.5 10⁴ s
g Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of magnitude 46.0 N is exerted and the magnitude of the astronaut's acceleration is measured to be 0.834 m/s2. Calculate her mass.
a car acceleration from rest to 90km/h in 10 seconds. what is its acceleration in meter per second square?
Answer:
2.5 m/s^2
Explanation:
First, convert 90 km/hr into m/s:
90/3.6 = 25 m/s
vf = final velocity = 25 m/s
vi = initial velocity = 0 m/s
t = time = 10 seconds
a = acceleration, unknown
Then, find a using the following equation:
(vf - vi)/t = a
(25 m/s)/10 s = 2.5 m/s^2
a = 2.5 m/s^2
Hope this helps!! :)
Let’s look at a radio-controlled model car. Suppose that at time t1=2.0st1=2.0s the car has components of velocity vx=1.0m/svx=1.0m/s and vy=3.0m/svy=3.0m/s and that at time t2=2.5st2=2.5s the components are vx=4.0m/svx=4.0m/s and vy=3.0m/svy=3.0m/s . Find (a) the components of average acceleration and (b) the magnitude and direction of the average acceleration during this interval.
Answer:
[tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]
Magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex]
Explanation:
[tex]t_1=2\ \text{s}[/tex]
[tex]v_x=1\ \text{m/s}[/tex]
[tex]v_y=3\ \text{m/s}[/tex]
[tex]t_2=2.5\ \text{s}[/tex]
[tex]v_x=4\ \text{m/s}[/tex]
[tex]v_y=3\ \text{m/s}[/tex]
Average acceleration in the different axes
[tex]a_x=\dfrac{\Delta v_x}{\Delta t}\\\Rightarrow a_x=\dfrac{4-1}{2.5-2}\\\Rightarrow a_x=6\ \text{m/s}^2[/tex]
[tex]a_y=\dfrac{\Delta v_y}{\Delta t}\\\Rightarrow a_y=\dfrac{3-3}{2.5-2}\\\Rightarrow a_y=0\ \text{m/s}^2[/tex]
The components of the acceleration is [tex]a_x=6\ \text{m/s}^2[/tex] and [tex]a_y=0\ \text{m/s}^2[/tex]
The magnitude of acceleration
[tex]a=\sqrt{a_x^2+a_y^2}\\\Rightarrow a=\sqrt{6^2+0^2}\\\Rightarrow a=6\ \text{m/s}^2[/tex]
Direction
[tex]\theta=\tan^{-1}\dfrac{a_y}{a_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{0}{6}\\\Rightarrow \theta=0^{\circ}[/tex]
The magnitude of accleration is [tex]6\ \text{m/s}^2[/tex] and the direction is [tex]0^{\circ}[/tex].
