Answer:
a) 0.67 rad/sec in the clockwise direction.
b) 98.8% of the kinetic energy is lost.
Explanation:
Let us take clockwise angular speed as +ve
For first cylinder
rotational inertia [tex]I[/tex] = 2.0 kg-m^2
angular speed ω = +5.0 rad/s
For second cylinder
rotational inertia [tex]I[/tex] = 1.0 kg-m^2
angular speed = -8.0 rad/s
The rotational momentum of a rotating body is given as = [tex]I[/tex]ω
where [tex]I[/tex] is the rotational inertia
ω is the angular speed
The rotational momenta of the cylinders are:
for first cylinder = [tex]I[/tex]ω = 2.0 x 5.0 = 10 kg-m^2 rad/s
for second cylinder = [tex]I[/tex]ω = 1.0 x (-8.0) = -8 kg-m^2 rad/s
The total initial angular momentum of this system cylinders before they were coupled together = 10 + (-8) = 2 kg-m^2 rad/s
When they are coupled coupled together, their total rotational inertia [tex]I_{t}[/tex] = 1.0 + 2.0 = 3 kg-m^2
Their final angular rotational momentum after coupling = [tex]I_{t}[/tex][tex]w_{f}[/tex]
where [tex]I_{t}[/tex] is their total rotational inertia
[tex]w_{f}[/tex] = their final angular speed together
Final angular momentum = 3 x [tex]w_{f}[/tex] = 3[tex]w_{f}[/tex]
According to the conservation of angular momentum, the initial rotational momentum must be equal to the final rotational momentum
this means that
2 = 3[tex]w_{f}[/tex]
[tex]w_{f}[/tex] = final total angular speed of the coupled cylinders = 2/3 = 0.67 rad/s
From the first statement, the direction is clockwise
b) Rotational kinetic energy = [tex]\frac{1}{2} Iw^{2}[/tex]
where [tex]I[/tex] is the rotational inertia
[tex]w[/tex] is the angular speed
The kinetic energy of the cylinders are:
for first cylinder = [tex]\frac{1}{2} Iw^{2}[/tex] = [tex]\frac{1}{2}*2*5^{2}[/tex] = 25 J
for second cylinder = [tex]\frac{1}{2}*1*8^{2}[/tex] = 32 J
Total initial energy of the system = 25 + 32 = 57 J
The final kinetic energy of the cylinders after coupling = [tex]\frac{1}{2}I_{t}w^{2} _{f}[/tex]
where
where [tex]I_{t}[/tex] is the total rotational inertia of the cylinders
[tex]w_{f}[/tex] is final total angular speed of the coupled cylinders
Final kinetic energy = [tex]\frac{1}{2}*3*0.67^{2}[/tex] = 0.67 J
kinetic energy lost = 57 - 0.67 = 56.33 J
percentage = 56.33/57 x 100% = 98.8%
A) The angular speed of the combination of the two cylinders is; ω₃ = 0.67 rad/s
B) The percentage of the original kinetic energy lost to friction is;
percentage energy lost = 98.82%
We are given;
Rotational Inertia for first cylinder; I₁ = 2 kg.m²
Angular speed of first cylinder; ω₁ = 5 rad/s
Angular speed of second cylinder; ω₂ = 8 rad/s
Rotational Inertia for second cylinder; I₂ = 1 kg.m²
From conservation of angular momentum, we know that;
Initial angular Momentum([tex]L_{i}[/tex]) = Final angular Momentum([tex]L_{f}[/tex])
Thus;
I₁ω₁ + I₂ω₂ = I₃ω₃
Where;
ω₃ is the angular speed when the two cylinders are combined
I₃ = I₁ + I₂
I₃ = 2 + 1
I₃ = 3 kg.m²
Since the second cylinder rotates in an anticlockwise direction, then its' angular speed will be negative. Thus;
(2 * 5) + (1 * -8) = 3ω₃
10 - 8 = 3ω₃
3ω₃ = 2
ω₃ = 2/3
ω₃ = 0.67 rad/s
B) Let us find initial kinetic energy;
E_i = ¹/₂I₁ω₁² + ¹/₂I₂ω₂²
E_i = ¹/₂((2 * 5²) + (1 * 8²)
E_i = 57 J
Final kinetic energy is;
E_f = ¹/₂I₃ω₃²
E_f = ¹/₂ * 3 * 0.67²
E_f = 0.67335 J
Energy lost = 57 - 0.67335 = 56.32665 J
percentage energy lost = (56.32665/57) * 100%
percentage energy lost = 98.82%
Read more at; https://brainly.com/question/14121636
An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase the magnitude of the induced emf? A. Straighten the wire out to be flat B. Reduce the resistance of the wire of which the loop is made C. Turning the plane of the loop to be parallel to the magnetic field D. Reducing the diameter of the loop
Answer:
changing the magnetic field more rapidly
Explanation:
According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.
Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.
In a physics lab, Asha is given a 11.5 kg uniform rectangular plate with edge lengths 62.9 cm by 46.9 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .
Answer:
6.9kgm²
Explanation:
For an axis through the center of the rectangle, I = m[(w²+L²)/12
Using the parallel axis theorem, the added value of I = mR² = m[(w²/4 + L²/4]
Adding the 2 expressions,
I = (m/3)*(w²+L²)
I =6.95 kg∙m²
A small wave pulse and a large wave pulse approach each other on a string; the large pulse is moving to the right.
Sometime after the pulses have met and passed each other, which of the following statements is correct? (More than one answer may be correct)
- the large pulse continues moving to the right
- the large pulse continues unchanged, moving to the right
- the small pulse is reflected and moves off to the right with a smaller amplitude
- the small pulse is reflected and moves off to the right with its original amplitude
- the two pulses combine into a single pulse moving to the right
Answer:
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
Explanation:
The waves fulfill the principle of superposition, which states that the value of the function at a point is the algebraic sum of the waves at a given instant.
The two waves in this exercise travel in the opposite direction, so when they are close, the resulting wave is the sum of the two waves, having a complicated shape. But when the waves follow their movement, they give in the same way as the initial a,
the large amplitude wave keeps moving to the right
the small amplitude wave continues to move to the left.
When checking the answers, the correct ones are 1, 2
A ball is thrown from the ground so that it’s initial vertical and horizontal components of velocity are 40m/s and 20m/s respectively. Find the ball’s total time of flight and distance it traverses before hitting the ground.
Answer:
8 seconds
160 meters
Explanation:
Given in the y direction:
Δy = 0 m
v₀ = 40 m/s
a = -10 m/s²
Find: t
Δy = v₀ t + ½ at²
0 m = (40 m/s) t + ½ (-10 m/s²) t²
0 = 40t − 5t²
0 = 5t (8 − t)
t = 0 or 8
Given in the x direction:
v₀ = 20 m/s
a = 0 m/s²
t = 8 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (20 m/s) (8 s) + ½ (0 m/s²) (8 s)²
Δx = 160 m
A fisherman in a stream 39 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman
Answer:
30cm
Explanation:
assume that the eyes are substantially above the water so that sin(theta) is approximately theta.
( small angle approximation).
The point at which a ray leaving the fish hits the surface of the water is x to the side of the centreline and the depth of the water is d
x/d = sin( angle of incidence)
if the apparent depth of the water is h then
x/h = sin( angle of refraction)
and applying snells law
1 sin ( theta air) = 1.33 sin( theta water)
1 * x/h = 1.33 * x/d
d/h = 1.33
or h/d = 1/1.33
h/39 = 1.33
h = 39 /1.33 so that is the apparent depth of the stream assuming:-
1. Your eyes are almost directly overhead
and
2. your eyes are a significant distance above the surface of the water.
x/d = 1.33 x/h
h/d =39/1.3
= 30cm
A certain resistor dissipates 0.5 W when connected to a 3 V potential difference. When connected to a 1 V potential difference, this resistor will dissipate:
Answer:
0.056 WExplanation:
[tex]Power = IV[/tex]
From ohms law we know that
[tex]V= IR\\\\I= \frac{V}{R} \\\\Power= \frac{V}{R}*V\\\\Power= \frac{V^2}{R}[/tex]
Given data
P1 = 0.5 Watt
P2 = ?
V1= 3 Volts
V2= 1 Volt
Thus we can solve for the power dissipated as follows
[tex]P1= \frac{V1^2}{R1}\\\\P2= \frac{V2^2}{R2}[/tex]
[tex]\frac{P1}{P2} = \frac{V1^2}{V2^2}\\\\ P2=\frac{ V2^2}{ V1^2} *P1\\\\ P2=\frac{ 1^2}{ 3^2} *0.5= 0.055= 0.056 W[/tex]
The resistor will dissipate 0.056 Watt
At the pizza party you and two friends decide to go to Mexico City from El Paso, TX where y'all live. You volunteer your car if everyone chips in for gas. Someone asks how much the gas will cost per person on a round trip. Your first step is to call your smarter brother to see if he'll figure it out for you. Naturally he's too busy to bother, but he does tell you that it is 2015 km to Mexico City, there's 11 cents to the peso, and gas costs 5.8 pesos per liter in Mexico. You know your car gets 21 miles to the gallon, but we still don't have a clue as to how much the trip is going to cost (in dollars) each person in gas ($/person).
Answer:
cost_cost = $ 96
Explanation:
In this exercise we have units in the groin system and the SI system, to avoid problems let's reduce everything to the SI system
performance = 21 miles / gallon (1,609 km / 1 mile) (1 gallon / 3,785 l)
perfomance= 8,927 km / l
now let's use a direct rule of proportions (rule of three). If a liter travels 8,927 km, how many liters are needed to travel the 2015 km
#_gasoline = 2015 km (1l / 8.927 km) = 225.72 liters
Now let's find the total cost of fuel. Ns indicates that $ 0.11 = 1 peso and the liter of fuel costs 5.8 pesos
cost_litre = 5.8 peso ($ 0.11 / 1 peso) = $ 0.638
cost_gasoline = #_gasoline cost_litro
cost_gasoline = 225.72 0.638
cost_gasoline = $ 144
This cost is for the one way trip, the total round trip cost is
cost_total = 2 cost_gasoline
cost_total = $ 288
Now let's look for the cost in the vehicle, you and two people will go, for which a total of 3 people will go, so the cost per person is
cost_person = total_cost / #_people
cost_person = 288/3
cost_cost = $ 96
A bus carrying 10 people has over turned on a remote hillside during an intense thunderstorm. What three factors could contribute to creating a delay in advanced care
Answer:
The three factors that can contribute to creating a delay in advanced care for the passengers in the overturned bus include:
1. Lack of communication: Since the accident happened on the remote hillside, there is a possibility that, there would be no communication network which could have afforded them the opportunity to call medical or technical team.
2. Steep Nature of the Hill: This is another factor which will affect the care which they could have received. Steeply area tends to be difficult for climbing in or out from.
3. Thunderstorm: This factor is another reason which could contribute to delay in receiving advance care. Thunderstorm create barriers for location f the area where the bus overturned or in other situation complicate the rescue efforts of the team sent out to rescue.
Explanation:
A student uses a spring scale attached to a textbook to compare the static and kinetic friction between the textbook and the top of a lab
table. If the scale measures 1,580 g while the student is pulling the sliding book along the table, which reading on the scale could have been
possible at the moment the student overcame the static friction? (1 point)
1,860 g
820 g
1,580 g
1,140 g
Answer:
1,860 g
Explanation:
In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
Heat and thermodynamics Numerical
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
A 5.0-µC point charge is placed at the 0.00 cm mark of a meter stick and a -4.0-µC point charge is placed at the 50 cm mark. At what point on a line joining the two charges is the electric field due to these charges equal to zero?
Answer:
Electric field is zero at point 4.73 m
Explanation:
Given:
Charge place = 50 cm = 0.50 m
change q1 = 5 µC
change q2 = 4 µC
Computation:
electric field zero calculated by:
[tex]E1 =k\frac{q1}{r^2} \\\\E2 =k\frac{q2}{R^2} \\\\[/tex]
Where electric field is zero,
First distance = x
Second distance = (x-0.50)
So,
E1 = E2
[tex]k\frac{q1}{r^2}=k\frac{q2}{R^2} \\\\[/tex]
[tex]\frac{5}{x^2}=\frac{4}{(x-50)^2} \\\\[/tex]
x = 0.263 or x = 4.73
So,
Electric field is zero at point 4.73 m
A transformer consists of a 500-turn primary coil and a 2000-turn secondary coil. If the current in the secondary is 3.0 A, what is the current in the primary
Answer:
12AExplanation:
Formula for calculating the relationship between the electromotive force (emf), current and number of turns of a coil in a transformer is expressed as shown:
[tex]\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex] where;
Vs and Vp are the emf in the secondary and primary coil respectively
Ns and Np are the number if turns in the secondary and primary coil respectively
Ip and Is are the currents in the secondary and primary coil respectively
Since the are all equal to each other, then we can equate any teo of the expression as shown;
[tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
Given parameters
Np = 500-turns
Ns = 2000-turns
Is = 3.0Amp
Required
Current in the primary coil (Ip)
Using the relationship [tex]\dfrac{N_s}{N_p} = \dfrac{I_p}{I_s}[/tex]
[tex]I_p = \dfrac{N_sI_s}{N_p}[/tex]
[tex]I_p = \dfrac{2000*3}{500} \\\\I_p = \frac{6000}{500}\\ \\I_p = 12A\\[/tex]
Hence the current in the primary coil is 12Amp
The maximum gauge pressure in a hydraulic system is 15 atm. What is the largest mass that could be lifted by this system if the diameter of the piston is 65 cm
Answer:
The maximum force that can be lifted by this system is 51,478.4 kg
Explanation:
Given;
maximum gauge pressure of the hydraulic system, Hp = 15 atm = 1.52 x 10⁶ N/m²
diameter of the piston, d = 65 cm = 0.65 m
The maximum gauge pressure of the piston is given as;
[tex]Hp = \frac{F}{A}[/tex]
Where;
F is the maximum force of the piston
A is the area of the piston
[tex]A = \pi (\frac{0.65}{2} )^2\\\\A = 0.3319 \ m^2[/tex]
F = Hp x A
F = 1.52 x 10⁶N/m² x 0.3319m²
F = 504488 N
Force is given as;
F = mg
m = F/g
m = 504488/9.8
m = 51,478.4 kg
Therefore, the maximum force that can be lifted by this system is 51,478.4 kg
A velocity selector in a mass spectrometer uses a 0.100-T magnetic field. (a) What electric field strength is needed to select a speed of 4.00 . 106 m/s
Answer:
The electric field strength needed is 4 x 10⁵ N/C
Explanation:
Given;
magnitude of magnetic field, B = 0.1 T
velocity of the charge, v = 4 x 10⁶ m/s
The velocity of the charge when there is a balance in the magnetic and electric force is given by;
[tex]v = \frac{E}{B}[/tex]
where;
v is the velocity of the charge
E is the electric field strength
B is the magnetic field strength
The electric field strength needed is calculated as;
E = vB
E = 4 x 10⁶ x 0.1
E = 4 x 10⁵ N/C
Therefore, the electric field strength needed is 4 x 10⁵ N/C
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?
Using the sample in above question how many counts per second would be observed when the detector is 10 meters away from the sample?
Answer:
At 3 meter distance, the per-second count is 222.22 and at a 10 meter distance, the per-second count is 20.
Explanation:
The number of particles (N) counts are inversely proportional to the distance between the source and the detector.
By using the below formula we can find the number of counts.
[tex]N2 = \frac{(D1)^2}{(D2)^2} \times N1 \\N1 = 2000 \\D 1 = 1 \ meter \\D2 = 3 \\[/tex]
The number of count per second, when the distance is 3 meters.
[tex]= \frac{1}{3^2} \times 2000 \\= 222.22[/tex]
Number of count per second when the distance is 10 meters.
[tex]= \frac{1}{10^2} \times 2000 \\= 20[/tex]
The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people.
Near Point(cm) Far Point(cm)
Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]
Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.
1. Berenice
2. Avishka
3. Francesca
4. Edouard
Answer:
1. Berenice = 0.67 D
2. Avishka = 1.50 D
3. Francesca = 2.00 D
4. Edouard = 2.75 D
Explanation:
The farsighted people are those with near point greater than 25 cm.
They include Avishka, Berenice, Edouard and Francisca.
A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative
From the lens formula, 1/f = 1/v 1/u; but v is negative
Therefore, 1/f = 1/u - 1/v
But, power of a lens = 1/f in meters.
Therefore, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
P = 1/ 0.25 - 1/0.4 = 1.5 D
For Berenice, u = 0.25 m, v = 0.3 m
P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v = 0.80 m
P = 1/0.25 - 1/0.80 =2.75 D
For Francesca, u = 0.25 m, v = 0.50 m
P = 1/0.25 - 1/0.5 = 2.0 D
When The farsighted people are those with a near point greater than 25cm.
Berenice is = 0.67 D
Avishka is = 1.50 D
Francesca is = 2.00 D
Edouard is = 2.75 D
What is Hyperopic Vision?
When The farsighted people are those with a near point greater than 25 cm. Then, They include Avishka, Berenice, Edouard, and also Francisca.
Also, A converging lens is needed to correct farsightedness, or hyperopia, thus, When the focal length, f, is positive. Also, The image formed is virtual and also on the same side of the lens. hence the image distance is negative.
Also, From the lens formula, 1/f = 1/v 1/u; but v is negative
Thus, 1/f = 1/u - 1/v
But, when the power of a lens = 1/f in meters.
Thus, P = 1/u - 1/v
For Avishka, u = 25 cm or 0.25 m, v = 0.4 m
After that, P = 1/ 0.25 - 1/0.4 = 1.5 D
Then, For Berenice, u = 0.25 m, v = 0.3 m
Now, P = 1/0.25 - 1/0.30 =0.7 D
For Edouard, u = 0.25 m, v is = 0.80 m
Then, P = 1/0.25 - 1/0.80 is =2.75 D
Now, For Francesca, u = 0.25 m, v is = 0.50 m
Therefore, P = 1/0.25 - 1/0.5 = 2.0 D
Find more information about Hyperopic Vision here:
https://brainly.com/question/25676535
How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV
Answer:
Number of turns of wire(N) = 3,036 turns (Approx)
Explanation:
Given:
Diameter = 13 Cm
emf = 5.6 v
Note:
The given question is incomplete, unknown information is as follow.
Magnetic field increases = 0.25 T in 1.8 (Second)
Find:
Number of turns of wire(N)
Computation:
radius (r) = 13 / 2 = 6.5 cm = 0.065 m
Area = πr²
Area = (22/7)(0.065)(0.065)
Area = 0.013278 m²
So,
emf = (N)(A)(dB / dt)
5.6 = (N)(0.013278)(0.25 / 1.8)
5.6 = (N)(0.013278)(0.1389)
N = 3,036.35899
Number of turns of wire(N) = 3,036 turns (Approx)
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing Photo 2.
Answer:
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Explanation:
I cannot find any attached photo, but we can proceed anyways theoretically.
The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point
i.e
[tex]E=\frac{F}{Q}[/tex]
But the force F
[tex]F= \frac{kQ1Q2}{r^2}[/tex]
But the electric field intensity due to a point charge Q at a distance r meters away is given by
[tex]E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }[/tex]
From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases
Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided
Answer:
R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor
Explanation:
For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)
Therefore we build a circuit with 4 resistors in series, in such a way that
V = i R
let the voltage
1st resistance
V = i R
R₁ = V / i
R₁ = 14.7 / 1 10⁻³
R₁ = 14.7 10³ Ω
power is
P = V i
P = 14.7 1 10⁻³
P = 14.7 10⁻³ W = 0.0147 W
a resistance of ⅛ W is indicated
2nd resistance
R₂ = 8.18 / 1 10⁻³
R₂ = 8.18 10³ Ω
Power
P = 8.18 1 10⁻³
P = 0.00818W
a 1/8 W resistor
3rd resistance
this resistance is calculated in such a way that
V₁ + V₂ + V₃ = 24.6
V₃ = 24.6 - V₁ -V₂
V₃ = 24.6 - 14.7 - 8.18
V₃ = 1.72 V
R₃ = 1.72 / 1 10⁻³
R₃ = 1.72 10³ Ω
power
P = Vi
P = 1.72 10⁻³
P = 0.00172 W
a resistance of ⅛ W
To obtain the voltage of 24.6 we use this three resistors together
4th resistance
The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage
30 = V₁ + V₂ + V₃ + V₄
V₄ = 30 - V₁ -V₂ -V₃
V₄ = 30 -14.7 - 8.18 - 1.72
V₄ = 5.4 V
R₄ = 5.4 / 1 10⁻³
R₄ = 5.4 10³ Ω
Power
P = V i
P = 5.4 10⁻³
P = 0.0054 W
⅛ W resistance
The values of these resistance are commercially
Let's check the consumption of the circuit
R_total = R₁ + R₂ + R₃ + R₄
R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³
R_total = 30 10³
the current circulating in the circuit is
i = V / R_total
i = 30/30 10³
i = 1 10⁻³ A
therefore it is within the order requirement.
for connections see attached diagram
Air bags greatly reduces the chance og injury in a car accident.explain how they do si in terms of energy transfer
Answer:
in an accident, when the body collides with the air bags, the collision time of impact between the two bodies will increase due to the presence of air bags in the car. Larger is the impact time smaller is the transformation of energy between the body and air bag. That is why air bags greatly reduce the chance of injury in a car accident.
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer?
O The nail exerts a much smaller force on the hammer in the opposite direction
O The nail exerts a much smaller force on the hammer in the same direction.
The nail exerts an equal force on the hammer in the same direction.
O The nail exerts an equal force on the hammer in the opposite direction.
Answer:
reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.
This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.
The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of the hammer is much greater, therefore the pressure is small.
When reviewing the final statements, the correct one is the quarter
The nail exerts an equal force on the hammer in the opposite direction.
Based on the passage, why is it important that different ethnic groups worked together on the strike? The groups needed to avoid speaking to one another because they wouldn’t understand. The different ethnic groups believed in being separate. The groups needed to trick the owners. They needed to be able to unite even though they spoke different languages.
Answer:D
Explanation:I got it right
Answer:
They needed to be able to unite even though they spoke different languages.
Explanation:
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistance. Find (a) the height of the tabletop above the floor; (b) the horizontal distance from the edge of the table to the point where the book strikes the floor; (c) the horizontal and vertical components of the book's velocity, and the magnitude and direction of its velocity, just before the book reaches the floor.
Answer:
(a) 0.613 m
(b) 0.385 m
(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s
v = 3.68 m/s², θ = 72.6° below the horizontal
Explanation:
(a) Take down to be positive.
Given in the y direction:
v₀ = 0 m/s
a = 10 m/s²
t = 0.350 s
Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²
Δy = 0.613 m
(b) Given in the x direction:
v₀ = 1.10 m/s
a = 0 m/s²
t = 0.350 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²
Δx = 0.385 m
(c) Find: vₓ and vᵧ
vₓ = aₓt + v₀ₓ
vₓ = (0 m/s²) (0.350 s) + 1.10 m/s
vₓ = 1.10 m/s
vᵧ = aᵧt + v₀ᵧ
vᵧ = (10 m/s²) (0.350 s) + 0 m/s
vᵧ = 3.50 m/s
The magnitude is:
v² = vₓ² + vᵧ²
v = 3.68 m/s²
The direction is:
θ = atan(vᵧ / vₓ)
θ = 72.6° below the horizontal
The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.
1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.
Answer:
1) f = 214 Hz , 2) answer is c , 3) f = 428 Hz , 4) f₂ = 428 Hz , f₃ = 643Hz
Explanation:
1) A tube with both ends open, the standing wave has a maximum amplitude and a node in its center, therefore
L = λ / 2
λ = 2L
λ = 2 0.8
λ = 1.6 m
wavelength and frequency are related to the speed of sound (v = 343 m / s)
v =λ f
f = v / λ
f = 343 / 1.6
f = 214 Hz
2) In this case the air comes out through the open hole, so we can assume that the length of the tube is reduced
λ' = 2 L ’
as L ’<L₀
λ' <λ₀
f = v / λ'
f' > fo
the correct answer is c
3) in this case the length is L = 0.40 m
λ = 2 0.4 = 0.8 m
f = 343 / 0.8
f = 428 Hz
4) the different harmonics are described by the expression
λ = 2L / n n = 1, 2, 3
λ₂ = L
f₂ = 343 / 0.8
f₂ = 428 Hz
λ₃ = 2 0.8 / 3
λ₃ = 0.533 m
f₃ = 343 / 0.533
f₃ = 643 Hz
4,1) as we have two maximums at the ends, all integer multiples are present
the answer is C
E) the length of an open pipe created that has a wavelength of lam = 1.6 m is requested
in this pipe there is a maximum in the open part and a node in the closed part, so the expression
L = λ / 4
L = 1.6 / 4
L = 0.4 m
the answer is C
F) in this type of pipe the general expression is
λ = 4L / n n = 1, 3, 5 (2n + 1)
therefore only odd values can produce standing waves
λ₃ = 4L / 3
λ₃ = 4 0.4 / 3
λ₃ = 0.533
f₃ = 343 / 0.533
f₃ = 643 Hz
A steel ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring constant is 6.00 N/m. When the ball is halfway between its equilibrium position and its maximum displacement from equilibrium, its speed is 26.1 cm/s. (a) What is the mass of the ball (in kg)? kg (b) What is the period of oscillation (in s)? s (c) What is the maximum acceleration of the ball? (Enter the magnitude in m/s2.) m/s2
Answer:
a) m = 0.626 kg , b) T = 2.09 s , c) a = 1.0544 m / s²
Explanation:
In a spring mass system the equation of motion is
x = A cos (wt + Ф)
with w = √(k / m)
a) velocity is defined by
v = dx / dt
v = - A w sin (wt + Ф) (1)
give us that the speed is
v = 26.1 m / s
for the point
x = a / 2
the range of motion is a = 11.0 cm
x = 11.0 / 2
x = 5.5 cm
Let's find the time it takes to get to this distance
wt + Ф = cos⁻¹ (x / A)
wt + Ф = cos 0.5
wt + Ф = 0.877
In the exercise they do not indicate that the body started its movement with any speed, therefore we assume that for the maximum elongation the body was released, therefore the phase is zero f
Ф = 0
wt = 0.877
t = 0.877 / w
we substitute in equation 1
26.1 = -11.0 w sin (w 0.877 / w)
w = 26.1 / (11 sin 0.877))
w = 3.096 rad / s
from the angular velocity equation
w² = k / m
m = k / w²
m = 6 / 3,096²
m = 0.626 kg
b) angular velocity and frequency are related
w = 2π f
frequency and period are related
f = 1 / T
we substitute
w = 2π / T
T = 2π / w
T = 2π / 3,096
T = 2.09 s
c) maximum acceleration
the acceleration of defined by
a = dv / dt
a = - Aw² cos (wt)
the acceleration is maximum when the cosine is ±1
a = A w²
a = 11 3,096²
a = 105.44 cm / s²
we reduce to m / s
a = 1.0544 m / s²
What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly
Answer:
λ = 667.85 nm
Explanation:
Let f be the frequency detected by the observer
Let v be the speed at which the observer is moving.
Now, when the direction at which the observer is moving is away from the source, we have the frequency as;
f = f_o√((1 - β)/(1 + β))
From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;
λ = λ_o√((1 + β)/(1 - β))
Where, β = Hr/c
H is hubbles constant which has a value of 0.0218 m/s • ly
c is speed of light = 3 × 10^(8) m/s
r is given as 2.40 x 10^(8) ly
Thus,
β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))
β = 0.01744
Since we are given λ_o = 656.3 nm
Then;
λ = 656.3√((1 + 0.01744)/(1 - 0.01744))
λ = 667.85 nm
As you finish listening to your favorite compact disc (CD), the CD in the player slows down to a stop. Assume that the CD spins down with a constant angular acceleration. If the CD rotates clockwise (let's take clockwise rotation as positive) at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, the angular acceleration of the CD, as it spins to a stop at -20.1 rad/s 2. How many revolutions does the CD make as it spins to a stop?
Answer:
10.8rev
Explanation:
Using
Wf²-wf = 2 alpha x theta
0²- 56.36x56.36/ 2(-20.13) x theta
Theta = 68.09 rad
But 68.09/2π
>= 10.8 revolutions
Explanation:
a toy propeller fan with a moment of inertia of .034 kg x m^2 has a net torque of .11Nxm applied to it. what angular acceleration does it experience
Answer:
The angular acceleration is [tex]\alpha = 3.235 \ rad/s ^2[/tex]
Explanation:
From the question we are told that
The moment of inertia is [tex]I = 0.034\ kg \cdot m^2[/tex]
The net torque is [tex]\tau = 0.11\ N \cdot m[/tex]
Generally the net torque is mathematically represented as
[tex]\tau = I * \alpha[/tex]
Where [tex]\alpha[/tex] is the angular acceleration so
[tex]\alpha = \frac{\tau }{I}[/tex]
substituting values
[tex]\alpha = \frac{0.1 1}{ 0.034}[/tex]
[tex]\alpha = 3.235 \ rad/s ^2[/tex]
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.
Answer:
The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
Explanation:
Given;
work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J
Apply Einstein Photo electric effect;
E = K.E + Ф
Where;
E is the energy of the incident light
K.E is the kinetic of electron
Ф is the work function of silver surface
For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.
E = Ф
hf = Ф
[tex]h\frac{c}{\lambda} = \phi[/tex]
where;
c is speed of light = 3 x 10⁸ m/s
h is Planck's constant, = 6.626 x 10⁻³⁴ J/s
λ is the wavelength of the incident light
[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]
Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
