Answer:
the correct result is L = 0.319 m
Explanation:
This system is a physical pendulum whose angular velocity is
w² = [tex]\frac{M \ g \ d}{I}[/tex]
where d is the distance from the center of mass to the point of rotation and I is the moment of inertia of the system
The Moment of Inertia is a scalar, therefore an additive quantity
I = I_bar + I_disk
the moment of inertia of each element with respect to the pivot point can be found with the parallel axes theorem
let's use M for the mass of the bar and m for the mass of the disk
Bar
I_bar = I_{cm} + Md²
the moment of inertia of the center of mass is
I_{cm} = [tex]\frac{1}{12}[/tex] M L²
the distance from the center of mass
d = L / 2
we substitute
I_bar = [tex]\frac{1}{12}[/tex] M L² + M ([tex]\frac{L^2}{4}[/tex])
Disk
I_disk = I_{cm} + m d²
moment of inertia of the center of mass
I_{cm} = ½ m R²
the distance d is
d = L
we substitute
I_disk = 1/2 m R² + m L²
the total moment of inertia is
I = [tex]\frac{1}{12}[/tex] M L² +[tex]\frac{1}{4}[/tex] M L² + [tex]\frac{1}{2}[/tex] m r² + m L²
I = [tex]\frac{1}{4}[/tex] M L² + m L² + ½ m r²
I = L² (m + [tex]\frac{1}{4}[/tex] M) + ½ m r²
The position of the center of mass of the system can be found with the expressions
d_{cm} = [tex]\frac{1}{M} \sum r_i m_i[/tex]
d_{cm} = [tex]\frac{1}{m+M} \ ( M \frac{L}{2} + m L)[/tex]
d_{cm} = [tex]L \frac{m + M/2}{m +M }[/tex]
now we can substitute in the expression for the angular velocity
w² = (m + M) g L [tex]\frac{m + \frac{M}{2} }{m+M}[/tex] [tex]\frac{1}{L^2 (m+ \frac{M}{4} ) + \frac{1}{2} m r^2 }[/tex]
w² = g (m + [tex]\frac{1}{2}[/tex] M) [tex]\frac{L}{ L^2 ( m +\frac{1}{4} M ) + \frac{1}{2} m r^2}[/tex]
angular velocity and period are related
w = 2π/T
sustitute
4π²/T² = g (m + [tex]\frac{1}{2}[/tex] M) [tex]\frac{L}{ L^2 ( m +\frac{1}{4} M ) + \frac{1}{2} m r^2}[/tex]
L² (m + [tex]\frac{1}{4}[/tex] M) + ½ m r² = [tex]\frac{T^2}{4 \pi ^2 } \ g ( m + \frac{1}{2} M ) \ \ L[/tex]
we substitute the values and solve the second grade equation
L² (0.1 + [tex]\frac{1}{4}[/tex] 0.3) - [[tex]\frac{1^2}{4\pi ^2}[/tex] 9.8 (0.1 + 0.3/2) ] L + ½ 0.1 0.2² = 0
L² 0.175 - 0.06206 L + 0.002 = 0
the equation remains after simplifying
L² - 0.3546 La + 0.01143 = 0
solve us
L = [tex]\frac{0.3546 \ \pm \sqrt{ 0.3546^2 - 4 \ 0.01143 }}{2}[/tex]
L = [tex]\frac{0.3546 \ \pm \ 0.28288 }{2}[/tex]
L₁ = 0.319 m
L₂ = 0.036m
the correct result must have a value greater than the radius of the disk. The correct result is L = 0.319 m
red light from a He-Ne laser is at 590.5 nm in the air. it is fired at an angle of 31.0 to horizontal at a flat transparent crystal of calcite (n= 1.34 ar this frequency) .find the wavelength and frequency of the light inside the crystal and the angle from horizontal that it travels inside the calcite crystal.
Answer:
7374.4
Explanation:
I took the test
(filler so I can post)
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
A car is moving with speed 30 m/s and acceleration 4 m/s2 at a given instant. (a) Using a second-degree Taylor polynomial, estimate how far the car moves in the next second.
Answer:
68 meters moved in the next seconds
Explanation:
Given
[tex]u= 30m/s[/tex]
[tex]a = 4m/s^2[/tex]
Required
Distance covered by the car in the next second
At a point in time t, the current distance is calculated as:
[tex]s(t) = ut + \frac{1}{2}at^2[/tex]
Substitute values for a and u in the above equation.
[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]
[tex]s(t) =30t + 2t^2[/tex]
Next, we generate the second degree Taylor polynomial as follows;
Calculate velocity (s'(t))
Differentiate s(t) to get velocity
[tex]s(t) =30t + 2t^2[/tex]
[tex]s'(t) =30 + 4t[/tex]
Calculate acceleration (s"(t))
Differentiate s'(t) to get acceleration
[tex]s'(t) =30 + 4t[/tex]
[tex]s"(t) =4[/tex]
When t = 0
We have:
[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]
[tex]s'(0) =30 + 4*0 = 30[/tex]
[tex]s"(0) = 4[/tex]
So, the second degree tailor series is:
[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]
To see the distance moved in the next second, we set t to 1
So, we have:
[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]
[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]
Solving s(1), s'(1) and s"(1)
We have:
[tex]s(1) =30*1 + 2*1^2 = 32[/tex]
[tex]s'(1) =30 + 4*1 = 34[/tex]
[tex]s"(1) =4[/tex]
Hence:
[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]
[tex]T_2(1) = 32 + 34 + 2[/tex]
[tex]T_2(1) = 68[/tex]
please help, no links please! I dont understand stand this question and im going to cry
Which part of this system is in the gas phase?
Answer:
I'm pretty sure its helium
Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-sectional area of 0.92 m2 and a drag coefficient of 0.0091- very low due to some evolutionary adaptations. Such a fish can sustain a speed of 30 m/s for a few seconds. Assume seawater has a density of 1026 kg/m3. a) How much power does the fish need to put out for motion at this high speed
Answer:
the required or need power is 115960.57 Watts
Explanation:
First of all, we take down the data we can find from the question, to make it easier when substituting values into formulas.
mass of swordfish m = 650 kg
Cross - sectional Area A = 0.92 m²
drag coefficient C[tex]_D[/tex] = 0.0091
speed v = 30 m/s
density p = 1026 kg/m³
Now, we determine our Drag force F[tex]_D[/tex]
Drag force F[tex]_D[/tex] = [tex]\frac{1}{2}[/tex] × C[tex]_D[/tex] × A × p × v²
Next, we substitute the values we have taken down, into the formula.
Drag force F[tex]_D[/tex] = [tex]\frac{1}{2}[/tex] × 0.0091 × 0.92 × 1026 × (30)²
Drag force F[tex]_D[/tex] = 4.294836 × 900
Drag force F[tex]_D[/tex] = 3865.3524
Now, we determine the power needed P[tex]_w[/tex]
P[tex]_w[/tex] = F[tex]_D[/tex] × v
we substitute
P[tex]_w[/tex] = 3865.3524 × 30
P[tex]_w[/tex] = 115960.57 Watts
Therefore, the required or need power is 115960.57 Watts
2(A + B)
15. The resultant of A and B is perpendicular to A
What is the angle between A and B?
(a) cos
(b) cos
La
(c) sin
(d) sin
Answer:
θ = cos^(-1) (-A/B)
Explanation:
The image of the reauktant forces A & B are missing, so i have attached it.
Now, from the attached image, we will see that;
Angle between A and B is θ
Also;
A = Bcos(180° − θ)
Now, in trigonometry, we know that;
cos(180° − θ) = -cosθ
Thus;
A = -Bcosθ
cosθ = -A/B
Thus;
θ = cos^(-1) (-A/B)
10 POINTS!! SPACE QUESTION!
Answer:
The Gas Giants have more moons.
Explanation:
Mercury-0
Venus-0
Earth-1
Mars-2
Jupiter-66
Saturn-62
Uranus-27
Neptune-13
The density of 1 kilogram of gold is
Answer:
0.02 kg/cm³
Explanation:
17. 53 A small grinding wheel is attached to the shaft of an electric motor that has a rated speed of 3600 rpm. When the power is turned off, the unit coasts to rest in 70 s. The grinding wheel and rotor have a combined weight of 6 lb and a combined radius of gyration of 2 in. Determine the average magnitude of the couple due to kinetic friction in the bearings of the motor.
Answer:
0.337 lb-in
Explanation:
From the law of conservation of angular momentum,
L' = L" where L = initial angular momentum of system and L" = final angular momentum of system
Now L = Iω + Mt where Iω = angular momentum of shaft + wheel and Mt = impulse on system due to couple M.
L' = Iω' + (-Mt) (since the moment about the shaft is negative-anticlockwise)
L' = Iω' - Mt where Iω' = angular momentum of shaft at t' = 0 + wheel and Mt = impulse on system due to couple M in time interval t = 70 s.
L" = Iω" where Iω" = angular momentum of shaft at t" = 70 s.
Now I = moment of inertia of system = mk² where m = mass of system = W/g where W = weight of system = 6 lb and g = acceleration due to gravity = 32 ft/s². So, m = W/g = 6lb/32 ft/s² = 0.1875 lb-s²/ft and k = radius of gyration = 2 in = 2/12 ft = 1/6 ft.
So, I = mk² = (0.1875 lb-s²/ft) × (1/6 ft)² = 0.00521 lb-ft-s², ω' = initial angular speed of system = 3600 rpm = 3600 × 2π/60 = 120π rad/s = 377 rad/s, ω" = final angular speed of system = 0 rad/s (since it stops), t' = 0 s, f" = 70 s and M = couple on system
So,
Iω' - Mt" = Iω"
Substituting the values of the variables into the equation, we have
Iω' - Mt" = Iω"
0.00521 lb-ft-s² × 377 rad/s - M × 70 s = 0.00521 lb-ft-s² × 0 rad/s"
0.00521 lb-ft-s² × 377 rad/s - 70M = 0
1.964 lb-ft-s = 70M
M = 1.964 lb-ft-s/70 s
M = 0.0281 lb-ft
M = 0.0281 lb × 12 in
M = 0.337 lb-in
When ultraviolet light with a wavelength of 400.0 nm falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV.
What is the maximum kinetic energy K0 of the photoelectrons when light of wavelength 310 nm falls on the same surface?
Use h = 6.63×10−34 J⋅s for Planck's constant and c = 3.00×108 m/s for the speed of light and express your answer in electron volts.
Answer:
Explanation:
energy of photon having wavelength of 400 nm = 1237.5/400 eV
= 3.1 eV.
Maximum kinetic energy of photoelectrons = 1.1 eV .
Threshold energy Ф = 3.1 - 1.1 = 2 eV .
energy of photons having wavelength of 310 nm = 1237.5 / 310 eV = 4 eV .
Maximum kinetic energy of photoelectrons = energy of photons - Threshold energy
= 4 - 2 = 2 eV .
Required kinetic energy K₀= 2 eV.
Which form of energy increases when a spring is compressed?
Answer:
When the spring compresses, elastic potential energy increases.
Answer:
the answer is b
Explanation:
elastic potential energy
A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it
Answer:
w = 0.319 rad / s
Explanation:
This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.
initial instant. Before the squirrel jumps
L₀ = m v r
final instant. After the trough and the squirrel are together
L_f = (I_fetter + I_ardilla) w
angular momentum is conserved
L₀ = L_f
m v r = (I_fetter + I_ardilla) w
w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]
the moment inercial ofbody is
I_thed = 2.00 kg m²
We approach the squirrel to a specific mass
I_ardilla = m r²
we substitute
w = m v r / ( I_[feefer + m r²)
let's calculate
w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )
w = 0.6426 / 2.0119
w = 0.319 rad / s
In 2-3 complete sentences, analyze how scientists know dark matter and dark energy exist.
Answer:
It doesn't interact with baryonic matter and it's completely invisible to light and other forms of electromagnetic radiation, making dark matter impossible to detect with current instruments. But scientists are confident it exists because of the gravitational effects it appears to have on galaxies and galaxy clusters.
Explanation:
Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity
Answer:
C
Explanation:
Primacy means being first or important so thats not an important facial display as the others.
A cart of mass m is moving with negligible friction along a track with known speed v1 to the right. It collides with and sticks to a cart of mass 4m moving with known speed v2 to the right. Which of the two principles, conservation of momentum and conservation of mechanical energy, must be applied to determine the final speed of the carts, and why
Answer:
conservation of linear momentum
We were told that two objects became stuck together hence we have to use the principle of conservation of momentum to obtain the final velocities of the carts.
What is conservation of momentum ?The principle of conservation of momentum lets us know that the momentum before collision is equal to the momentum after collision. As such we can write; m1u1 + m2u2 = m1v1 + m2v2.
We can use this thus principle to obtain the final speeds of the carts since the two objects that collided became stuck together.
Learn more about conservation of momentum: https://brainly.com/question/11256472
How much energy is supplied to a 9 V bulb if it is switched on for 3 minutes and takes a current of
0.2 A ?
Answer:
0.01j
Explanation:
the energy equals the work done by the bulb.
Workdone=
[tex]workdone = \frac{power}{time} [/tex]
power=voltage×current
=9×0.2
=1.8 W.
THEREFORE,
time=3×60
= 180s
workdone=1.8/180
=0.01 j
An object’s
✔ mass
will remain constant throughout the universe, but its
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
Answer:
mass, weight, strength of gravity increases, weight decreases
Explanation:
got it on edge
Answer:
An object’s
✔ mass
will remain constant throughout the universe, but its
✔ weight
can change from planet to planet.
If you increase the mass of a planet, what happens to its gravity?
✔ strength of gravity increases
If the gravity on a planet decreases, what happens to the weight of an object on that planet?
✔ weight decreases
Explanation:
right on edge 22
A boat travels west at 20km/h. The journey lasts 3hours. How far has the boat travelled? *
A)60km
B)60km[W]
C)17km[W]
D) 6.6km[W]
Answer:
B)60km[W]
Explanation:
The boat travels 20km/h. So every hour the boat goes 20 miles. So if one hour equals 20km. Then 3 hours will be 3*20km which equals 60km. The boat is also going west. So you should consider putting that in your answer as well. So the answer would be B)60km[W].
Hope that helps!
In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the kind used in smallpox vaccine). The procedure involved measuring the frequency of oscillation of a tiny sliver of silicon (just 32.0 nm long) with a laser, first without the virus and then after the virus had attached itself to the silicon. The difference in mass caused a change in the frequency. We can model such a process as a mass on a spring.
Required:
a. Find the ratio of the frequency with the virus attached ( fS+V) to the frequency without the virus (fS) in terms of mV and mS, where mV is the mass of the virus and mS is the mass of the silicon sliver.
b. In some data, the silicon sliver has a mass of 2.13×10^-16 g and a frequency of 2.04×10^15 Hz without the virus and 2.85×1014 with the virus. What is the mass of the virus in grams?
Answer:
a) m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1) , b) m_v = 1.07 10⁻¹⁴ g
Explanation:
a) The angular velocity of a simple harmonic motion is
w² = k / m
where k is the spring constant and m is the mass of the oscillator
let's apply this expression to our case,
silicon only
w₉² = [tex]\frac{K}{m_s}[/tex]
k = w₀² m_s
silicon with virus
w² = [tex]\frac{k}{m_s + m_v}[/tex]
k = w² (m_v + m_s)
in the two expressions the constant k is the same and q as the one property of the silicon bar, let us equal
w₀² m_s = w² (m_v + m_s)
m_v = ([tex]\frac{w_o}{w}[/tex])² m_s - m_s
m_v = m_s (([tex]\frac{w_o}{w}[/tex])² - 1)
b) let's calculate
m_v = 2.13 10⁻¹⁶ [([tex]\frac{20.4}{2.85}[/tex])² - 1)]
m_v = 1.07 10⁻¹⁴ g
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The block is suspended by wires from the ceiling and is initially at rest. After the bullet is embedded in the block, the block swings up to a maximum height of 0.295 cm above its initial position. What is the velocity of the bullet on leaving the gun's barrel
Answer:
The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
Explanation:
Given;
mass of the bullet, m₁ = 2.47 g = 0.00247 kg
mass of the wooden block, m₂ = 2.43 kg
initial velocity of the wooden block, u₂ = 0
height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m
let the initial velocity of the bullet on leaving the gun's barrel = v₁
let final velocity of the bullet-wooden block system after collision = v₂
Apply the principle of conservation of linear momentum;
Total initial momentum = Total final momentum
m₁v₁ + m₂u₂ = v₂(m₁ + m₂)
0.00247v₁ + 2.43 x 0 = v₂(2.43 + 0.00247)
0.00247v₁ = 2.4325v₂ -------(1)
The kinetic energy of the bullet-block system after collision;
K.E = ¹/₂(m₁ + m₂)v₂²
K.E = ¹/₂ (2.4325)v₂²
The potential energy of the bullet-block system after collision;
P.E = mgh
P.E = (2.4325)(9.8)(0.00295)
P.E = 0.07032
Apply the principle of conservation of mechanical energy;
K.E = P.E
¹/₂ (2.4325)v₂² = 0.07032
1.21625 v₂² = 0.07032
v₂² = 0.07032 / 1.21625
v₂² = 0.0578
v₂ = √0.0578
v₂ = 0.24 m/s
Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;
0.00247v₁ = 2.4325v₂
0.00247v₁ = 2.4325 (0.24)
0.00247v₁ = 0.5838
v₁ = 0.5838 / 0.00247
v₁ = 236.36 m/s
Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.
state newton first law of motion
Newton’s first law of motion states that there must be a cause—which is a net external force—for there to be any change in velocity, either a change in magnitude or direction. An object sliding across a table or floor slows down due to the net force of friction acting on the object.
got it off g lol..
Answer:
it state that everybody in the universe is state that" universe continues its state of rest or uniform motion in a straight path unless it is acted upon by external force."
You need friction created by your tires and the road ____
to control your speed and direction.
Answer:
surface
Explanation:
You need friction created by your tires and the road surface
to control your speed and direction.
Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________
a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.
Answer:
d. the star is a member and also a part of an eclipsing binary star system.
Explanation:
If any star happens to be brighter for an extended period of time, however, at some times, it becomes dimmer, is due to the fact that the star is being overshadowed (hiding behind another star that is known as eclipse).
The above-mentioned eclipsing binary star system is essentially what has been defined. It occurs when two stars' orbit planes are so similar that one star will obscure (the light) of the other.
Thus, option D is correct.
g An airplane is flying through a thundercloud at a height of 1500 m. (This is a very dangerous thing to do because of updrafts, turbulence, and the possibility of electric discharge.) If a charge concentration of 25.0 C is above the plane at a height of 3000 m within the cloud and a charge concentration of -40.0 C is at height 850 m, what is the electric field at the aircraft
Answer:
[tex]523269.9\ \text{N/m}[/tex]
Explanation:
q = Charge
r = Distance
[tex]q_1=25\ \text{C}[/tex]
[tex]r_1=3000\ \text{m}[/tex]
[tex]q_2=40\ \text{C}[/tex]
[tex]r_2=850\ \text{m}[/tex]
The electric field is given by
[tex]E=E_1+E_1\\\Rightarrow E=k(\dfrac{q_1}{r_1^2}+\dfrac{q_2}{r_2^2})\\\Rightarrow E=9\times 10^9\times (\dfrac{25}{3000^2}+\dfrac{40}{850^2})\\\Rightarrow E=523269.9\ \text{N/m}[/tex]
The electric field at the aircraft is [tex]523269.9\ \text{N/m}[/tex]
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller pipe that is 10.0 cm in diameter. Determine: a) The speed of the water flowing through the 10.0 cm diameter pipe. b) The mass of water that flows through the larger pipe in 1.00 minute. c) The mass of water that flows through the smaller pipe in 1.00 minute.
Answer:
a) v₂ = 30 m/s
b) m₁ = 12600 kg
c) m₂ = 12600 kg
Explanation:
a)
Using the continuity equation:
[tex]A_1v_1 = A_2v_2[/tex]
where,
A₁ = Area of inlet = π(0.15 m)² = 0.07 m²
A₂ = Area of outlet = π(0.05 m)² = 0.007 m²
v₁ = speed at inlet = 3 m/s
v₂ = speed at outlet = ?
Therefore,
[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]
v₂ = 30 m/s
b)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₁ = mass of water flowing in = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]
m₁ = 12600 kg
c)
[tex]m_1 = \rho A_1v_1t[/tex]
where,
m₂ = mass of water flowing out = ?
ρ = density of water = 1000 kg/m³
t = time = 1 min = 60 s
Therefore,
[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]
m₂ = 12600 kg
While flying at an altitude of 5.75 km, you look out the window at various objects on the ground. If your ability to distinguish two objects is limited only by diffraction, find the smallest separation between two objects on the ground that are distinguishable. Assume your pupil has a diameter of 4.0 mm and take ???? = 460 nm.
Answer:
the smallest separation between two objects is 0.8067 m
Explanation:
Given the data in the question;
Altitude h = 5.75 km = 5750 m
Diameter D = 4.0 mm = 0.004 m
λ = 460 nm = 4.6 × 10⁻⁷ m
Now, Using Rayleigh criterion for Airy disks resolution.
we know that, Minimum angular separation for resolving two points is;
θ = 1.22λ / D
so we substitute
θ = (1.22 × 4.6 × 10⁻⁷) / 0.004
θ = 5.612 × 10⁻⁷ / 0.004
θ = 1.403 × 10⁻⁴ rad
so minimum separation [tex]d_{min[/tex] = θh
so we substitute
[tex]d_{min[/tex] = (1.403 × 10⁻⁴) × 5750 m
[tex]d_{min[/tex] = 0.8067 m
Therefore, the smallest separation between two objects is 0.8067 m
A star with the same mass and diameter as the sun rotates about a central axis with a period of about 24.0 days. Suppose that the sun runs out of nuclear fuel and collapses to form a white dwarf star with a diameter equal to that of the earth. Assume the star acts like a solid sphere and that there is no loss of mass in the process. You will need some data from the inside front cover of you text. (a) What would be the new rotation period (s) of the star? (b) What is the ratio of final to initial kinetic energies (Kf /Ki)?
Answer:
a) w = 2.52 10⁷ rad / s, b) K / K₀ = 1.19 10⁴
Explanation:
a) We can solve this exercise using the conservation of angular momentum.
Initial instant. Before collapse
L₀ = I₀ w₀
Final moment. After the collapse
L_f = I w
angular momentum is conserved
L₀ = L_f
I₀ w₀ = I w (1)
The moment of inertia of a sphere is
I = 2/5 m r²
we take from the table the mass and diameter of the star
m = 1,991 10³⁰ kg
r₀ = 6.96 10⁸ m
r = 6.37 10⁶ m
to find the angular velocity let's use
w = L / T
where the length of a circle is
L = 2π r
T = 24 days (24 h / 1 day) (3600 s / 1h) = 2.0710⁶ s
we substitute
w = 2π r / T
wo = 2π 6.96 10⁸ / 2.07 10⁶
wo = 2.1126 10³ rad / s
we substitute in equation 1
w = [tex]\frac{I_o}{I}[/tex]
w = 2/5 mr₀² / 2/5 m r² w₀
w = ([tex]\frac{r_o}{r}[/tex]) ² wo
w = (6.96 10⁸ / 6.37 10⁶) ² 2.1126 10³
w = 2.52 10⁷ rad / s
b) the kinetic energy ratio
K = ½ m w²
K₀ = ½ m w₀²
K = ½ m w²
K / K₀ = (w / wo) ²
K / K₀ = 2.52 10⁷ / 2.1126 10³
K / K₀ = 1.19 10⁴
What does it mean if the reflected beam is above the incident beam? What does it mean if reflected beam is below the incident beam?
Answer:
aim at prisma and will have all colors
Explanation:
Answer:
If the vision position is above the actual image location then the light travel from the object in such a way that the angle of incidence is less than the angle of reflected ray which means that the reflected beam is above the incident beam.
Explanation:
Which of the following best describes what occurs in a fission reaction?
A.
Two low mass nuclei are joined to form one nucleus.
B.
Electrons are shared between the nuclei.
C.
A single nucleus divides into two or more nuclei and gives off energy.
D.
A chemical reaction occurs between the nuclei.
Answer:
C.A single nucleus divides into two or more nuclei and gives off energy best describes what occurs in a fission reaction.
Answer:
C.
A single nucleus divides into two or more nuclei and gives off energy.
hope it is helpful to you
An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k = 450 N/m. The object is determined to have a velocity of 0.3 m/s when passing through the equilibrium.
1. Find the amplitude of the motion
2. Find the total energy of the object at any point of its motion
Answer:
1) The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object at any point of its motion is 16.892 joules.
Explanation:
1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ([tex]U_{e}[/tex]), in joules, is equal to maximum translational kinetic energy ([tex]K[/tex]), in joules:
[tex]U_{e} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]k[/tex] - Spring constant, in newtons per meter.
[tex]A[/tex] - Amplitude, in meters.
[tex]m[/tex] - Object mass, in kilograms.
[tex]v[/tex] - Speed of the object at equilibrium, in meters per second.
If we know that [tex]k = 450\,\frac{N}{m}[/tex], [tex]m = 0.25\,kg[/tex] and [tex]v = 0.3\,\frac{m}{s}[/tex], then the amplitude of the motion is:
[tex]\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]k\cdot A^{2} = m\cdot v^{2}[/tex]
[tex]A = v\cdot \sqrt{\frac{m}{k} }[/tex]
[tex]A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }[/tex]
[tex]A \approx 0.274\,m[/tex]
The amplitude of the motion is approximately 0.274 meters.
2) The total energy of the object ([tex]E[/tex]), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ([tex]k = 450\,\frac{N}{m}[/tex], [tex]A \approx 0.274\,m[/tex])
[tex]E = U_{e}[/tex]
[tex]E = \frac{1}{2}\cdot k\cdot A^{2}[/tex]
[tex]E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}[/tex]
[tex]E = 16.892\,J[/tex]
The total energy of the object at any point of its motion is 16.892 joules.