Answer:
a) 0.46875
b) 8 km
Explanation:
Smokestack height ( H ) = 50 m
speed of pollutant / wind speed = 3 m/s
Half width = 2 [tex]\sqrt{2Dt }[/tex] = 50 m ---- ( 1 )
a) If L = 2 km
value of turbulent diffusion coefficient D
back to equation 1
50 = 2 √ 2 * D * ( 2000/3 )
2500 = 4 * 2 * D * ( 2000/3 )
D = 2500 / ( 8 * ( 2000/3 ) )
= 0.46875
where : time to travel ( t ) = Distance / speed = 2000 / 3
b) when the smoke stack = 50 * 2 = 100 m
L = 800 m = 8 km
attached below is the detailed solution
A person pulls on a 9 kg crate against a 22 Newton frictional force, using a rope attached to the center of the crate. If the The crate began with a speed of 1.5 m/s and speeded up to 2.7 m/s while being pulled a horizontal distance of 2.0 meters. What is the work in J done by the force applied by the rope on the crate
Answer:
uninstell this apps nobody will give u ans its happening to me alsoi was in exam hall i thought this app will give answer but no
A ball of mass m is dropped from a height h above the ground. neglecting air resistance then determine the speed of the ball when it is at a height y above the ground and determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.
Answer:
Explanation:
kinematic equation (g will have a negative value if we assume UP is positive)
v² = u² + 2as
a) v = √(0² + 2(g)(y - h))
b) v = √(vi² + 2(g)(y - h))
How much charge is stored on two parallel-plate capacitors by the 12V battery if one is filled with air and the other is filled with a dielectric (k=3.00)
The question is incomplete, the complete question is;
how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air, and the other is filled with a dielectric for which k = 3.00; both capacitors have a plate area of 5.00×10 −3 m 2 and a plate separation of 2.00 mm.
The capacitance of the capacitor is the quantity of charge stored by the capacitor.
Given that;
C1= εo k * A/d
εo = permittivity of free space
C1 = 8.85 x 10-12 farad per meter *1 * 5.00×10 −3 m 2/2 * 10^-3
= 2.21 * 10^-11 F
C2 = 8.85 x 10-12 * 3 * 5.00×10 −3 m 2/2 * 10^-3
= 6.63 * 10^-11 F
q1 = C1V1 = 2.21 * 10^-11 C * 12 V
= 2.65 * 10^-10 C
q2 = C2V2 = 6.63 * 10^-11 F * 12 V
= 7.96 * 10^-10 C
qtotal = 2.65 * 10^-10 C + 7.96 * 10^-10 C
qtotal = 1.061 * 10^-9C
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An electron is released from rest at a distance of 9.00 cm from a fixed proton. How fast will the electron be moving when it is 3.00 cm from the proton
Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg
crate by a light rope. The light rope remains taut. Compared to the 6.00-kg crate,
the lighter 4.00-kg crate
Please explain why any of these multiple choices is correct!
Answer:
B. is subject to a smaller net force but same acceleration.
Explanation:
F = m*a
So because our force applied is constant from the women pulling on the rope which means the acceleration is the same on both the 4kg create and the 6kg create. The only thing that changes here is the mass of the creates, so there is more tension force between the women and the 6kg create then there is between the 4kg create and the 6kg. It takes less force to move the 4kg create therefore the tension force is less between the two creates.
The net force on both crates is the same and the acceleration of both crates is the same.
The given parameters;
mass of the crate, m = 6 kgmass of the second crate, = 4 kgThe force on the 4kg crate is calculated as follows;
[tex]F_{4kg } = T + F[/tex]
The force on the 6kg crate is calculated as follows;
[tex]F_{6 kg} = -T + F[/tex]
The net force on both crates is calculated as follows;
[tex]\Sigma F= -T + F - (T + F)\\\\\Sigma F= -2T[/tex]
Thus, we can conclude that the net force on both crates is the same and the acceleration of both crates is the same.
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a. The molecules of a magnet are independent...
Answer:
variable
Explanation:
A spherical, concave shaving mirror has a radius of curvature of 0.983 m. What is the magnification of a person's face when it is 0.155 m from the vertex of the mirror (answer sign and magnitude)
Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
[tex]\frac{1}{v} + \frac{1}{u} + \frac{1}{f}[/tex] .................2
put here value and we get
[tex]\frac{1}{v} = \frac{1}{0.155} - \frac{1}{4915}[/tex]
v = 0.155 mso
Magnification will be here as
m = [tex]- \frac{v}{u}[/tex]
m = [tex]\frac{0.155}{0.155}[/tex]
m = 1Answer:
The magnification is 1.5.
Explanation:
radius of curvature, R = - 0.983 m
distance of object, u = - 0.155 m
Let the distance of image is v.
focal length, f = R/2 = - 0.492 m
Use the mirror equation
[tex]\frac{1}{f}=\frac{1}{v}+\frac {1}{u}\\\\\frac{-1}{0.492}=\frac{1}{v}-\frac{1}{0.155}\\\\\frac{1}{v}=\frac{1}{0.155}-\frac{1}{0.492}\\\\\frac{1}{v}=\frac{0.492-0.155}{0.155\times 0.492}\\\\\frac{1}{v}=\frac{0.337}{0.07626}\\ \\v = 0.226 m[/tex]
The magnification is given by
m = - v/u
m = 0.226/0.155
m = 1.5
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
The work done by [tex]\vec F[/tex] along the given path C from A to B is given by the line integral,
[tex]\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r[/tex]
I assume the path itself is a line segment, which can be parameterized by
[tex]\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath[/tex]
with 0 ≤ t ≤ 1. Then the work performed by F along C is
[tex]\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}[/tex]
A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s
Answer:
366 N
Explanation:
τ = Iα
FR = ½mR²α
F = ½mR(Δω/t)
F = ½(207)(1.50)(0.75)(2π) /2.00
F = 365.79919...
What is the volume of a metal block 3cm long by 2cm wide by 4cm high? What would be the volume of a block twice as long, wide, and high?
Answer:
Volume of a metal block = 24 cm^3
Volume of a block twice as long, wide and high = 192 cm^3
Explanation:
Volume of a block is measured in l*w*h and in the first block, the sides are 3, 2 and 4 and 3*2*4 = 24
Second block, just double each of the lengths to get 6*4*8 = 192
On her camping trip, Penelope was in charge of collecting firewood. The firewood she found had a mass of 120 g and a volume of 480 cm3. What is the density of the firewood? Explain the steps you took to solve this problem.
Answer:
D = .25g/cm³
Explanation:
D = m/V
D = 120g/480cm³
D = .25g/cm³
Develop a hypothesis regarding one factor you think might affect the period of a pendulum or an oscillating mass on a spring. Potential factors include the mass, the spring constant, and the length of the pendulum's string. Write down your hypothesis. 2. Design a controlled experiment to test your hypothesis. Take extreme care to keep all factors constant except the variable you are testing.
Answer:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
Explanation:
A hypothesis for the period of a pendulum is:
"The period of the pendulum varies with its length"
To test this hypothesis we can carry out a measurement of a simple pendulum keeping the angle fixed, in general the angle used is about 5º since when placing this value in radiand and the sine of this angle they differ little <5%. therefore measured the time of some oscillations, for example about 10 oscillations, changing the length of the pendulum to test the hypothesis.
If the hypothesis and the model used is correct, the relationship to be tested is
T² =(4π² /g) L
by making a graph of the period squared against the length if obtaining, os a line, the hypothesis is tested.
Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the sound intensity increase and then decrease periodically due to wave interference. How long does it take between successive instances of the sound intensity increasing
Answer:
[tex]T=1.54s[/tex]
Explanation:
From the question we are told that:
Speed of Motor 1 [tex]\omega_1=825rpm=>2 \pi 13.75[/tex]
Speed of Motor 2 [tex]\omega_1=786rpm=>2 \pi 13.1[/tex]
Therefore
Frequency of Motor 1 [tex]f_1=13.75[/tex]
Frequency of Motor 2 [tex]f_2= 13.1[/tex]
Generally the equation for Time Elapsed is mathematically given by
[tex]T=\frac{1}{df}[/tex]
Where
[tex]df=f_1-f_2[/tex]
[tex]df=13.75-13.1[/tex]
[tex]df=0.65Hz[/tex]
Therefore
[tex]T=\frac{1}{65}[/tex]
[tex]T=1.54s[/tex]
A 700N marine in basic training climbs a 10m vertical rope at constant speed in 8sec. what is power put
Answer:
875 Watts
Explanation:
P = W/t = mgh/t = 700(10)/8 = 875 Watts
a. Do the waves made by the two faucets travel faster than the waves made by just one faucet?
b. How do you know this? Describe how the two-faucet wave pattern compares with the one-faucet pattern.
c. Describe what happens to the two-faucet wave pattern as the separation of the faucets is increased.
Answer:
asdasd dsa dasdasd sadas dasd asdasd asd asd dsa asdd 223 aasd ada dasd sa dasd dsaa sd adsd asasd
Explanation:
The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.66c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The velocity of Enterprise 2 relative to Enterprise 1 is 0.34c. What is the velocity of Enterprise 2, as measured by the earth-based observer
Answer:
The answer is "0.82 c".
Explanation:
Given:
Spacecraft speed 1 is [tex]u = + 0.66 \ c[/tex]
Space velocity 2 relative to spacecraft 1 is [tex]v = + 0.34\ c[/tex]
The spacecraft velocity 2 measured by the Earth observation
[tex]\to u' = \frac{u +v}{1 + ( \frac{uv}{c^2})}[/tex]
[tex]= \frac{0.66 \ c +0.34\ c}{ 1+ (\frac{0.66\ c \times 0.33\ c }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (\frac{0.2178\ c^2 }{c^2})}\\\\ = \frac{1 \ c }{ 1+ (0.2178 )}\\\\ = \frac{1 \ c }{ 1.2178 }\\\\=0.82\ c[/tex]
Two train 75 km apart approach each other on parallel tracks, each moving at
15km/h. A bird flies back and forth between the trains at 20km/h until the trains pass
each other. How far does the bird fly?
Answer:
The correct solution is "37.5 km".
Explanation:
Given:
Distance between the trains,
d = 75 km
Speed of each train,
= 15 km/h
The relative speed will be:
= [tex]15 + (-15)[/tex]
= [tex]30 \ km/h[/tex]
The speed of the bird,
V = 15 km/h
Now,
The time taken to meet will be:
[tex]t=\frac{Distance}{Relative \ speed}[/tex]
[tex]=\frac{75}{30}[/tex]
[tex]=2.5 \ h[/tex]
hence,
The distance travelled by the bird in 2.5 h will be:
⇒ [tex]D = V t[/tex]
[tex]=15\times 2.5[/tex]
[tex]=37.5 \ km[/tex]
Can you please help me please?
Explanation:
CH3CH2OH
That is the answer I hope this helps
You are to connect resistors R1 andR2, with R1 >R2, to a battery, first individually, then inseries, and then in parallel. Rank those arrangements according tothe amount of current through the battery, greatest first. (Useonly the symbols > or =, for exampleseries>R1=R2>parallel.)
Answer:
The current is more in the parallel combination than in the series combination.
Explanation:
two resistances, R1 and R2 are connected to a battery of voltage V.
When they are in series,
R = R1 + R2
In series combination, the current is same in both the resistors, and it is given by Ohm's law.
V = I (R1 + R2)
[tex]I = \frac{V}{R_1 + R_2}[/tex]..... (1)
When they are connected in parallel.
the voltage is same in each resistor.
The effective resistance is R.
[tex]R = \frac{R_1R_2}{R_1 + R_2}[/tex]
So, the current is
[tex]I = \frac{V(R_1+R_2)}{R_1 R_2}[/tex]..... (2)
So, the current is more is the parallel combination.
A 0.2 oz. bullet leaves the muzzle of a rifle with a speed of 1420 ft/s. If the length of the barrel is 24 inches, what is the magnitude of the force acting on the bullet while it travels down the barrel
Answer:
196 lbf
Explanation:
v² = u² + 2as
1420² = 0² + 2a(24/12)
a = 1420²/4 = 504,100 ft/s²
F = ma = 0.2oz(1lb/16 oz)(1slug/32.2 lb)(504,100) = 195.6909...
A block of mass 0.260 kg is placed on top of a light, vertical spring of force constant 5 200 N/m and pushed downward so that the spring is compressed by 0.090 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise
After being released, the restoring force exerted by the spring performs
1/2 (5200 N/m) (0.090 m)² = 12.06 J
of work on the block. At the same time, the block's weight performs
- (0.260 kg) g (0.090 m) ≈ -0.229 J
of work. Then the total work done on the block is about
W ≈ 11.83 J
The block accelerates to a speed v such that, by the work-energy theorem,
W = ∆K ==> 11.83 J = 1/2 (0.260 kg) v ² ==> v ≈ 9.54 m/s
Past the equilibrium point, the spring no longer exerts a force on the block, and the only force acting on it is due to its weight, hence it has a downward acceleration of magnitude g. At its highest point, the block has zero velocity, so that
0² - v ² = -2gy
where y is the maximum height. Solving for y gives
y = v ²/(2g) ≈ 4.64 m
what are the limitation of clinical thermometer
Answer:
Their main disadvantage is that they are fairly easy to break and if they do, it results in small splinters of glass and the release of mercury which is quite toxic if absorbed into the body.
We do not use water instead of mercury in a barometer
Answer:
Because
1)Water is relatively less dense. Mercury is 13.6 times more dense than water. SO, If you use water, you have to have the length of barometer of length (or height) 13.534 times the length of mercury barometer, which may be more than 11 meter in length.
2)Also mercury as compared to water, has comparatively less specific heat and good conductor of heat, could come to the same temperature of the atmosphere more quickly.
Answer:
The atmospheric pressure at sea level 76 cm of Hg=1.013× 10⁵ pascal .
Explanation:
That is if water is used in barometer tube instead of mercury the length of the tube must be greater than 10.326 cm.so we cannot replace mercury by water in the barometer.
A bus starts from rest and accelerates at 1.5m/s squared until it reaches a velocity of 9m/s .the bus continues at this velocity and then deccelerate at -2m/s squared until it comes to stop 400m from it's starting point. how much time did the bus takes to cover the 400m?
Answer:
23s
Explanation:
s=ut+1/2at^2
the distance (s) is 400, initial velocity (u) is 0, acceleration (a) is 1.5 therefore
400=0t+1/2(1.5)t^2
400/0.75=0.75t^2/0.75
t^2=√533.33
t=23s
I hope this helps and sorry if it's wrong
In part B of the lab, when the current flows through the orange part of the wire from right to left, the wire deflects (or moves) ____. This is in accordance with the right-hand-rule.
This seems to be incomplete, as we do not have any information about the magnetic field surrounding the wire, but we can answer in a general way.
We know that for a wire of length L, with a current I, and in a magnetic field B, the force can be written as:
F = L*(IxB)
if we define the right as the positive x-axis, and knowing that the current flows to the right, we can write:
I = i*(1, 0, 0)
And the field will be some random vector that can't be parallel to the current because in that case, we do not have any force.
To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.
Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.
For example, if the magnetic field is in the positive z-axis, we will point upwards.
Now the palm of our hand tells us in which direction the force is applied.
This is the right-hand rule.
For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.
coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second later. assume g is 10m/s. how far below the top of the tower des coin 1 pass coin 2
The height below the tower at which coin 1 pass coin 2 is 89.04 m.
The given parameters:
height of the tower, h = 100 m
initial velocity of coin 1, v = 15 m/s
time spent in air by coin 1 before coin 2 was dropped = 2s
To find:
the height below the tower at which coin 1 passes coin 2Find the maximum height attained by coin 1 before falling to the ground:
[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]
Find the time taken for coin 1 to fall to the ground:
Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m
[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]
But the time taken for the coin 1 to reach 11.25 m above the tower:
[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]
Total time spent by coin 1 before reaching ground with respect to coin 2:
time = (1.5 s + 4.72 s) - 2 s
time = 4.22 s
Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.
Find the total time taken for coin 2 to fall to the ground:
Height of coin 2 above the ground = 100 m
Total time taken by coin 2 before falling to the ground is calculated as:
[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]
The time at which coin 1 will pass coin 2 is 4.22 s.
Find the height below the tower when the time is 4.22 s.
[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]
Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.
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basic source of magnetism is a) charged particles alone b)Movement of charged particles c) Magnetic dipoles d)magnetic domains
Answer:
C . Magnetic dipoles is the correct
Answer:
b). movement of charged particles.
Explanation:
These charges create the nagnetic dipoles.
Calculate the kinetic energy of a mass 2kg moving with a velocity of 0.1m/s
ANSWER-:
1/2 mv²
K.E = 1/2 mv²
K.E = 0.01 J.
Hence, the kinetic energy of a body is 0.01 Joule.
!! HOPE ITS HELP U !!
what are three effects of gravity
Answer:
effect on motation.effect on direction