Answers
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
Related Questions
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field
Answers
Answer:
The electric field is [tex]E = 5.25 V/m[/tex]
Explanation:
From the question we are told that
The peak magnitude of the magnetic field is [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]
Generally the peak magnitude of the electric field is mathematically represented as
[tex]E = c * B[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]
[tex]E = 5.25 V/m[/tex]
The peak magnitude of the electric field will be "5.25 V/m".
Magnetic fieldAccording to the question,
Magnetic field's peak magnitude, B = 17.5 nT or,
= 17.5 × 10⁻⁹ T
Speed of light, c = 3.0 × 10⁸ m/s
We know the relation,
→ E = c × B
By substituting the values, we get
= 3.0 × 10⁸ × 17.5 × 10⁻⁹
= 5.25 V/m
Thus the above approach is appropriate.
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If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answers
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g
Answers
Answer:
The number of turns in the secondary coil is 4145 turns
Explanation:
Given;
the induced emf on the primary coil, [tex]E_p[/tex] = 95 V
the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V
the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns
the number of turns in the secondary coil, [tex]N_s[/tex] = ?
The number of turns in the secondary coil is calculated as;
[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]
[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]
Therefore, the number of turns in the secondary coil is 4145 turns.
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.
Answers
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answers
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ from the vertical and releases her from rest.
A: What is the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing?
B: How fast will she be moving at the bottom of the swing?
C: How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Answers
Answer
A)184.9J
B)=3.63m/s
C) Zero
Explanation:
A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is
U=Mgh
Where m=28kg
g= 9.8m/s
h= difference in height between the initial position and the bottom position
We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical
h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)
=0.674m
Her Potential Energy will now
= 28× 9.8×0.674
=184.9J
B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:
E= 0.5mv^2
where
m = 28.0 kg is the mass of the child
v is the speed of the child at the bottom position
Solving the equation for v, we find
V=√2k/m
V=√(2×184.9/28
=3.63m/s
C)we can find work done by the tension in the rope is given using expresion below
W= Tdcosx
where W= work done
T is the tension
d = displacement of the child
x= angle between the directions of T and d
In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.
Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
Answers
Answer:
4 x 10¹⁵
Explanation:
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Answers
Answer:
the two ice skater have the same momentum but the are in different directions.
Paula will have a greater speed than Ricardo after the push-off.
Explanation:
Given that:
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off
The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.
So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.
Momentum is the product of mass and velocity.
SO, from the information given:
Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]
Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]
At rest ;
their velocities will be zero, i.e
[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0
The initial momentum for this process can be represented as :
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0
after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]
The we can say their final momentum is:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
Using the law of conservation of momentum as states earlier.
Initial momentum = final momentum = 0
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Since the initial velocities are stating at rest then ; u = 0
[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Given that Ricardo weighs more than Paula
So [tex]m_{Ri} > m_{Pa}[/tex] ;
Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]
The magnitude of their momentum which is a product of mass and velocity can now be expressed as:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
The ratio is
[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]
[tex]v_{Pa} >v_{Ri}[/tex]
Therefore, Paula will have a greater speed than Ricardo after the push-off.
(A) Both the skaters have the same magnitude of momentum.
(B) Paula has greater speed after push-off.
Conservation of momentum:Given that two skaters Paula and Ricardo are initially at rest.
Ricardo weighs more than Paula.
Let us assume that the mass of Ricardo is M, and the mass of Paula is m.
Let their final velocities be V and v respectively.
(A) Initially, both are at rest.
So the initial momentum of Paula and Ricardo is zero.
According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.
Initial momentum = final momentum
0 = MV + mv
MV = -mv
So, both of them have the same magnitude of momentum, but in opposite directions.
(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:
MV = mv
M/m = v/V
Now, we know that M>m
so, M/m > 1
therefore:
v/V > 1
v > V
So, Paula has greater speed.
Learn more about conservation of momentum:
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At what temperature (degrees Fahrenheit) is the Fahrenheit scale reading equal to:_____
(a) 3 times that of the Celsius and
(b) 1/5 times that of the Celsius
Answers
Answer:
C = 26.67° and F = 80°C = -20° and F = -4°Explanation:
Find:
3 times that of the Celsius and 1/5 times that of the CelsiusComputation:
F = (9/5)C + 32
3 times that of the Celsius
If C = x
So F = 3x
So,
3x = (9/5)x + 32
15x = 9x +160
6x = 160
x = 26.67
So, C = 26.67° and F = 80°
1/5 times that of the Celsius
If C = x
So F = x/5
So,
x/5 = (9/5)x + 32
x = 9x + 160
x = -20
So, C = -20° and F = -4°
A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.
Answers
Answer:
R = 36.885 km
Explanation:
In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other
The diffraction equation for slits is
a sin θ = m λ
the first minimum occurs for m = 1
sin θ = λ a
as the diffraction experiments the angles are very small, we approximate
sin θ = θ
θ = λ / a
This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form
θ = 1.22 λ / a
In this problem they give us the frequency, let's find the wavelength with the relation
c = λ f
λ = c / f
θ = 1.22 c/ f a
since they ask us for the distance between the planes, we can use the definition of radians
θ = s / R
if we assume that the distance is large, we can approximate the arc to the horizontal distance
s = x
we substitute
x / R = 1.22 c / fa
R = x f a / 1.22c
Let's reduce the magnitudes to the SI system
f = 9000 MHz = 9 109 Hz
a = 15 m
x = 100 m
let's calculate
R = 100 10⁹ 15 / (1.22 3 108)
R = 3.6885 10⁴ m
let's reduce to km
R = 3.6885 10¹ km
R = 36.885 km
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .
Answers
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?
Answers
Answer:
Fluoroscopy
Explanation:
A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?
Answers
Answer:
3x10^5m/s
Explanation:
See attached file
Explanation:
The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
Doppler's Effect
According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.
Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.
[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]
Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.
[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]
[tex]v = 3\times 10^5\;\rm m/s[/tex]
Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
To know more about the doppler effect, follow the link given below.
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What is the separation in meters between two slits for which 594 nm orange light has its first maximum at an angle of 32.8°?
Answers
Answer:
1.1micro meter
Explanation:
Given that
Constructive interference is
ma = alpha x sin theta
Alpha = 1 x 594 x10^ -9/ sin 32.8°
= 1.1 x 10^ -6m
Explanation:
CAN SOMEONE HELP ME PLEASE ITS INTEGRATED SCIENCE AND I AM STUCK
Answers
Answer:
[tex]\huge \boxed{\mathrm{Option \ D}}[/tex]
Explanation:
Two forces are acting on the object.
Subtracting 2 N from both forces.
2 N → Object ← 5 N
- 2 N - 2N
0 N → Object ← 3 N
The force 3 N is pushing the object to the left side.
The mass of the object is 10 kg.
Applying formula for acceleration (Newton’s Second Law of Motion).
a = F/m
a = 3/10
a = 0.3
A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?
Answers
Answer:
51. 7m/s
Explanation:
Take speed of sound in air = 340 m/s
fp = fs (V + Vp)/(V + Vs)
128 = 123 (340 + Vp)/(340 + 36.4)
Vp = 51.7m/s
Explanation:
Calculate the density of the following material.
1 kg helium with a volume of 5.587 m³
700 kg/m³
5.587 kg/m³
0.179 kg/m³
Answers
Answer:
[tex]density \: = \frac{mass}{volume} [/tex]
1 / 5.587 is equal to 0.179 kg/m³
Hope it helps:)
Answer:
The answer is
0.179 kg/m³Explanation:
Density of a substance is given by
[tex]Density \: = \frac{mass}{volume} [/tex]
From the
mass = 1 kg
volume = 5.583 m³
Substitute the values into the above formula
We have
[tex]Density \: = \frac{1 \: kg}{5.583 \: {m}^{3} } [/tex]
We have the final answer as
Density = 0.179 kg/m³Hope this helps you
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
Answers
Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
pls give brainliest
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
[tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]
Substitute, [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]
[tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]
Hence, The new length of aluminum rod is 17.435 cm.
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The roller coaster car reaches point A of the loop with speed of 20 m/s, which is increasing at the rate of 5 m/s2. Determine the magnitude of the acceleration at A if pA
Answers
Answer and Explanation:
Data provided as per the question is as follows
Speed at point A = 20 m/s
Acceleration at point C = [tex]5 m/s^2[/tex]
[tex]r_A = 25 m[/tex]
The calculation of the magnitude of the acceleration at A is shown below:-
Centripetal acceleration is
[tex]a_c = \frac{v^2}{r}[/tex]
now we will put the values into the above formula
= [tex]\frac{20^2}{25}[/tex]
After solving the above equation we will get
[tex]= 16 m/s^2[/tex]
Tangential acceleration is
[tex]= \sqrt{ac^2 + at^2} \\\\ = \sqrt{16^2 + 5^2}\\\\ = 16.703 m/s^2[/tex]
Expectant mothers many times see their unborn child for the first time during an ultrasonic examination. In ultrasonic imaging, the blood flow and heartbeat of the child can be measured using an echolocation technique similar to that used by bats. For the purposes of these questions, please use 1500 m/s as the speed of sound in tissue. I need help with part B and C
To clearly see an image, the wavelength used must be at most 1/4 of the size of the object that is to be imaged. What frequency is needed to image a fetus at 8 weeks of gestation that is 1.6 cm long?
A. 380 kHz
B. 3.8 kHz
C. 85 kHz
D. 3.8 MHz
Answers
Answer:
380 kHz
Explanation:
The speed of sound is taken as 1500 m/s
The length of the fetus is 1.6 cm long
The condition is that the wavelength used must be at most 1/4 of the size of the object that is to be imaged.
For this 1.6 cm baby, the wavelength must not exceed
λ = [tex]\frac{1}{4}[/tex] of 1.6 cm = [tex]\frac{1}{4}[/tex] x 1.6 cm = 0.4 cm =
0.4 cm = 0.004 m this is the wavelength of the required ultrasonic sound.
we know that
v = λf
where v is the speed of a wave
λ is the wavelength of the wave
f is the frequency of the wave
f = v/λ
substituting values, we have
f = 1500/0.004 = 375000 Hz
==> 375000/1000 = 375 kHz ≅ 380 kHz
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answers
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answers
Answer:
They both have the same acceleration
Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answers
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid
Answers
Answer:
1.37 ×10^-3 T
Explanation:
From;
B= μnI
μ = 4π x 10-7 N/A2
n= number of turns /length of wire = 1700/0.75 = 2266.67
I= 0.48 A
Hence;
B= 4π x 10^-7 × 2266.67 ×0.48
B= 1.37 ×10^-3 T
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answers
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answers
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answers
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Learn more about the average density here:
https://brainly.com/question/1371999
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated
Answers
Answer:
3x10⁴v
Explanation:
Using
Wavelength= h/ √(2m.Ke)
880nm = 6.6E-34/√ 2.9.1E-31 x me
Ke= 6.6E-34/880nm x 18.2E -31.
5.6E-27/18.2E-31
= 3 x 10⁴ Volts
48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?
Answers
Answer:
the interation blood veins
Explanation:
A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answers
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
