Answer:
4.25m/sE. None of the option is correctExplanation:
Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.
Mathematically.
mu + MU = (m+M)v
m and M are the masses of the bullet and the block respectively
u and U are their respective velocities
v is their common velocity
from the question, the following parameters are given;
m = 20g = 0.02kg
u = 960m/s
M = 4.5kg
U =0m/s (block is at rest)
Substituting this values into the formula above to get v;
0.02(960)+4.5(0) = (0.02+4.5)v
19.2+0 = 4.52v
4.52v = 19.2
Dividing both sides by 4.52
4.52v/4.52 = 19.2/4.52
v = 4.25m/s
Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s
Unpolarized light is incident upon two polarization filters in sequence. The two filters transmission axes are not aligned. If 18% of the incident light passes through this combination of filters, what is the angle between the transmission axes of the filters
Answer:
53°
Explanation:
I/Io*2= 0.18
0.18= cos²theta
Cos^-1(0.36) = 53°
A flat loop of wire consisting of a single turn of cross-sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.60 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.70
Answer:
The induced current is [tex]I = 6.25*10^{-4} \ A[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1[/tex]
The cross-sectional area is [tex]A = 8.20 cm^2 = 8.20 * 10^{-4} \ m^2[/tex]
The initial magnetic field is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at time = 1.02 s is [tex]B_t = 2.60 \ T[/tex]
The resistance is [tex]R = 2.70\ \Omega[/tex]
The induced emf is mathematically represented as
[tex]\epsilon = - N * \frac{ d\phi }{dt}[/tex]
The negative sign tells us that the induced emf is moving opposite to the change in magnetic flux
Here [tex]d\phi[/tex] is the change in magnetic flux which is mathematically represented as
[tex]d \phi = dB * A[/tex]
Where dB is the change in magnetic field which is mathematically represented as
[tex]dB = B_t - B_i[/tex]
substituting values
[tex]dB = 2.60 - 0.500[/tex]
[tex]dB = 2.1 \ T[/tex]
Thus
[tex]d \phi = 2.1 * 8.20 *10^{-4}[/tex]
[tex]d \phi = 1.722*10^{-3} \ weber[/tex]
So
[tex]|\epsilon| = 1 * \frac{ 1.722*10^{-3}}{1.02}[/tex]
[tex]|\epsilon| = 1.69 *10^{-3} \ V[/tex]
The induced current i mathematically represented as
[tex]I = \frac{\epsilon}{ R }[/tex]
substituting values
[tex]I = \frac{1.69*10^{-3}}{ 2.70 }[/tex]
[tex]I = 6.25*10^{-4} \ A[/tex]
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.
Hahahahaha. Okay.
So basically , force is equal to mass into acceleration.
F=ma
so when F=ma , we get acceleration=6m/s/s
Force is doubled.
Mass is 1/3 times original.
2F=1/3ma
Now , we rearrange , and we get 6F=ma
So , now for 6 times the original force , we get 6 times the initial acceleration.
So new acceleration = 6*6= 36m/s/s
A motorcycle travels up one side of a hill over the top and down the other side. The crest of the hill can be considered to be a circular arc with radius of 45.0 m. Determine the maximum speed that the cycle can have while moving over the crest without losing contact with the road.
Answer:
The maximum speed of the motorcycle should be 21 m/s
Explanation:
Since the hill is considered to be a circular arc, the motorcycle will experience centripetal force that tends to flip it away from the center of the hill.
Since the motorcycle does not lose contact with the ground, it means that the weight of the motorcycle downwards just balances the centripetal force on the motorcycle.
we know that the centripetal force on the motorcycle is equal to
centripetal force = [tex]\frac{mv^{2} }{r}[/tex]
where m is the mass of the motorcycle,
v is the velocity of the motorcycle,
and r is the radius of the hill = 45.0 m
Also we now that the weight of the motorcycle is equal to
weight = mg
where m is still the mass of the motorcycle,
and g is the acceleration due to gravity = 9.81 m/s
Equating the both forces since they are equal, we'll have
[tex]\frac{mv^{2} }{r}[/tex] = mg
the mass of the motorcycle will cancel out, and we'll be left with
[tex]v^{2} = gr[/tex]
[tex]v = \sqrt{gr}[/tex]
[tex]v = \sqrt{9.81*45}[/tex]
[tex]v = \sqrt{441.45}[/tex]
[tex]v[/tex] = 21 m/s
a uniform ladder of mass 100kg leans at 60° to the horizontal against a frictionless wall, calculate the reaction on the wall.
Answer:
[tex]500\text{N} (490\text{N}) (490.5\text{N})[/tex]
Explanation:
The reaction force is the force that is in the perpendicular direction to the wall.
We have an angle and a hypotenuse, we need to find the adjacent angle - so we can just use cos:
[tex]cos(\theta)=\frac{\text{adj}}{\text{hyp}}\\\text{hyp}*cos(\theta)=\text{adj}\\100*cos(60)=100*0.5=50\text{kg}[/tex]
However, we would like a force and not a mass.
[tex]W=mg\\W=50g\\W=500\text{N} (490\text{N}) (490.5\text{N})[/tex]
Answer 1 if you use g as 10, answer 2 if you're studying mechanics in maths, answer 3 if you're studying mechanics in physics.
Electromagnetic waves are traveling in the vacuum of space. Calculate the wavelengths of these electromagnetic waves with the following frequencies. (Enter the first wavelength in pm and the second wavelength in cm.)
(a) 2.00 x 1019 Hz
(b) 4.50 x109 Hz
Answer:
(a) 1.5×10⁻¹¹ m.
(b) 6.7×10⁻² m
Explanation:
Note: All Electromagnetic wave travels in with the same speed, which 3×10⁸ m/s
(a) Give a frequency of 2.00×10¹⁹ Hz.
Using the equation of a wave,
V = λf................ Equation 1
Where V = Speed of electromagnetic wave, λ = wavelength, f = frequency.
make λ the subject of the equation
λ = V/f................. Equation 2
Given: f = 2.00×10¹⁹ Hz.
Constant: v = 3×10⁸ m/s.
Substitute into equation 2
λ = 3×10⁸/2.00×10¹⁹
λ = 1.5×10⁻¹¹ m.
(b) Similarly using
λ = v/f
Given: f = 4.5×10⁹ Hz, and v = 3×10⁸ m/s.
Substitute these values into equation 2 above.
λ = 3×10⁸/4.5×10⁹
λ = 6.7×10⁻² m
A sample of lead has a mass of 26.00 kg and a density of 1.130 104 kg/m3 at 0°C. (Assume the average linear expansion coefficient for lead is 2.900 10-5(°C-1).)
(a) What is the density of lead at 82.00°C? (Give your answer to four significant figures.)
____ kg/m3
(b) What is the mass of the sample of lead at 82.00°C?
_____ kg
Answer:
Explanation:
coefficient of linear expansion α = 2.9 x 10⁻⁵
coefficient of volume expansion γ = 3 x 2.9 x 10⁻⁵ = 8.7 x 10⁻⁵
[tex]d_t = d_0( 1 - \gamma t )[/tex]
[tex]d_{82} = 1.13\times 10^4( 1 - 8.7\times 10^{-5}\times82 )[/tex]
= 1.13 x 10⁴ - 806.14 x 10⁻¹
= 1.13 x 10⁴ - 0.00806 x 10⁴
= 1.1219 x 10⁴ kg / m³
b ) mass of the sample will remain the same as mass does not increase or decrease with temperature .
A 0.12-kg metal rod carrying a current of current 4.1 A glides on two horizontal rails separation 6.3 m apart. If the coefficient of kinetic friction between the rod and rails is 0.18 and the kinetic friction force is 0.212 N , what vertical magnetic field is required to keep the rod moving at a constant speed of 5.1 m/s
Answer:
The magnetic field is [tex]B = 8.20 *10^{-3} \ T[/tex]
Explanation:
From the question we are told that
The mass of the metal rod is [tex]m = 0.12 \ kg[/tex]
The current on the rod is [tex]I = 4.1 \ A[/tex]
The distance of separation(equivalent to length of the rod ) is [tex]L = 6.3 \ m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.18[/tex]
The kinetic frictional force is [tex]F_k = 0.212 \ N[/tex]
The constant speed is [tex]v = 5.1 \ m/s[/tex]
Generally the magnetic force on the rod is mathematically represented as
[tex]F = B * I * L[/tex]
For the rod to move with a constant velocity the magnetic force must be equal to the kinetic frictional force so
[tex]F_ k = B* I * L[/tex]
=> [tex]B = \frac{F_k}{L * I }[/tex]
=> [tex]B = \frac{0.212}{ 6.3 * 4.1 }[/tex]
=> [tex]B = 8.20 *10^{-3} \ T[/tex]
1. A coil is formed by winding 250 turns of insulated 16-gauge copper wire, that has a diameter d = 1.3 mm, in a single layer on a cylindrical form of radius 12 cm. What is the resistance of the coil? Neglect the thickness of the insulation and the resistivity of copper is ???? = 1.69 × 10−8 Ω ∙ m.
Answer:
2.39 Ω
Explanation:
Given that
Number of winnings on the coil, = 250 turns
Radius if the copper wire, r(c) = 1.3/2 = 0.65 mm
Radius of single cylinder layer, R = 12 cm
Length of the cylinderical coil, L = 250 * 2π * 12 = 188.4 m
Resistivity of copper, ρ = 1.69*10^-8 Ωm
Area is πr(c)², which is
A = 3.142 * (0.65*10^-3)²
A = 3.142 * 4.225*10^-7
A = 1.33*10^-6 m²
The formula for resistance is given as
R = ρ.L/A, if we substitute, we have
R = (1.69*10^-8 * 188.4) / 1.33*10^-6
R = 3.18*10^-6 / 1.33*10^-6
R = 2.39 Ω.
Therefore, the resistance is 2.39 Ω
A telewision weighs 8.50 pounds. How many grams is this? (Hint: You need to
use two unit conversion fractions. 1 pound equals about 0.454 kg.)
Answer:
3859 g
Explanation:
1 pound = 0.454 kg
therefore, 8.50 ponds = 0.454*8.50 = 3.859
to covert kilograms into grams you need to multiply it by 1000
=3.859*1000
= 3859 grams
What are three ways synthetic polymers affect the environment?
Answer:
-Some synthetic polymers use materials from Earth that are nonrenewable.
-They can end up as waste products that sometimes can’t be recycled.
-They sometimes release toxins into the environment.
Explanation:
Synthetic polymers are types of polymers that are man made, that is they are polymers that are made in the labs or industries from chemical substances.
Such polymers include nylon, polyester, and polyethylene among others. These polymers are used to make many clothes and plastics that we use and see around.
Synthetic polymers have a range of disadvantages which includes. being non-biodegradable, these polymers and especially plastics may end up as waste products and pollutes the environment and end up in livers and lakes and this would be toxic for aquatic animals.
g If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be
In this circuit the resistance R1 is 3Ω, R2 is 7Ω, and R3 is 7Ω. If this combination of resistors were to be replaced by a single resistor with an equivalent resistance, what should that resistance be?
Answer:
9.1Ω
Explanation:
The circuit diagram has been attached to this response.
(i) From the diagram, resistors R1 and R2 are connected in parallel to each other. The reciprocal of their equivalent resistance, say Rₓ, is the sum of the reciprocals of the resistances of each of them. i.e
[tex]\frac{1}{R_X} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
=> [tex]R_{X} = \frac{R_1 * R_2}{R_1 + R_2}[/tex] ------------(i)
From the question;
R1 = 3Ω,
R2 = 7Ω
Substitute these values into equation (i) as follows;
[tex]R_{X} = \frac{3 * 7}{3 + 7}[/tex]
[tex]R_{X} = \frac{21}{10}[/tex]
[tex]R_{X} = 2.1[/tex]Ω
(ii) Now, since we have found the equivalent resistance (Rₓ) of R1 and R2, this resistance (Rₓ) is in series with the third resistor. i.e Rₓ and R3 are connected in series. This is shown in the second image attached to this response.
Because these resistors are connected in series, they can be replaced by a single resistor with an equivalent resistance R. Where R is the sum of the resistances of the two resistors: Rₓ and R3. i.e
R = Rₓ + R3
Rₓ = 2.1Ω
R3 = 7Ω
=> R = 2.1Ω + 7Ω = 9.1Ω
Therefore, the combination of the resistors R1, R2 and R3 can be replaced with a single resistor with an equivalent resistance of 9.1Ω
Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is the magnitude of the magnetic field at a point that is between the two wires and 1.0 mm from one of the two wires
Answer:
Explanation:
Magnetic field at a a point R distance away
B = μ₀ / 4π X 2I / R where I is current
Magnetic field due to one current
= 10⁻⁷ x 2 x 24 / 1 x 10⁻³
48 x 10⁻⁴ T
Magnetic field due to other current
= 10⁻⁷ x 2 x 24 / 3x 10⁻³
16 x 10⁻⁴ T
Total magnetic field , as they act in opposite direction, is
= (48 - 16 ) x 10⁻⁴
32 x 10⁻⁴ T .
For the cellar of a new house, a hole is dug in the ground, with vertical sides going down 2.10 m. A concrete foundation wall is built all the way across the 8.90 m width of the excavation. This foundation wall is 0.189 m away from the front of the cellar hole. During a rainstorm, drainage from the street fills up the space in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force that the water causes on the foundation wall. For comparison, the weight of the water is given by 2.10 m ✕ 8.90 m ✕ 0.189 m ✕ 1000 kg/m3 ✕ 9.80 m/s2 = 34.6 kN.
Answer:
The force on the foundation wall is [tex]F_f = 191394 \ N[/tex]
Explanation:
From the question we are told that
The depth of the hole's vertical side is [tex]d = 2.10 \ m[/tex]
The width of the hole is [tex]b = 8.90 \ m[/tex]
The distance of the concrete wall from the front of the cellar is [tex]c = 0.189 \ m[/tex]
Generally the area which the water from the drainage covers is mathematically represented as
[tex]A = d * b[/tex]
substituting values
[tex]A = 2.10 * 8.90[/tex]
[tex]A = 18.69 \ m^2[/tex]
Now the gauge pressure exerted on the foundation wall is mathematically evaluated as
[tex]P_g = \rho * d_{avg} * g[/tex]
Here is the average height foundation wall where the pressure of the water is felt and it is evaluated as
[tex]d_{avg} = \frac{h_1 + h_2 }{2}[/tex]
where [tex]h_1[/tex] at the height at bottom of the hole which is equal to [tex]h_1 = 0[/tex]
and [tex]h_2[/tex] is the height at the top of the hole [tex]h_2 = d = 2.10[/tex]
[tex]d_{avg} = \frac{0 + 2.10 }{2}[/tex]
[tex]d_{avg} = 1.05[/tex]
Where [tex]\rho[/tex] is the density of water with constant value [tex]\rho = 1000 \ kg/m^3[/tex]
substituting values
[tex]P_g = 1000 * 1.05 * 9.8[/tex]
[tex]P_g = 10290 \ Pa[/tex]
Then the force exerted by the water on the foundation wall mathematically represented as
[tex]F_f = P_g * A[/tex]
substituting values
[tex]F_f = 10290 * 18.69[/tex]
[tex]F_f = 191394 \ N[/tex]
I swing a ball around my head at constant speed in a circle with circumference 3 m. What is the work done on the ball by the 10 N tension force in the string during one revolution of the ball
Answer:
The work done on the ball by the tension force is 0 J.
Explanation:
The work can be calculated as follows:
[tex]W = |F|\cdot |d|cos(\theta)[/tex]
Where:
F: is the tension force = 10 N
d: is the displacement = ball's circumference = 3 m
θ: is the angle between the force and the distance = 90°
Hence, the work is:
[tex]W = |10| \cdot |3| cos(90) = 0 J[/tex]
Since the tension force and the displacement vector are orthogonal, the work done on the ball is zero.
Therefore, the work done on the ball by the tension force is 0 J.
I hope it helps you!
The work done on the ball by the 10 N tension force is zero ( 0 Joules).
Given that:
the circumference(displacement d) of the ball = 3 mthe tension force of the ball = 10 Nthe angle θ between the tension force and the displacement =90°∴
Using the work equation;
W = F × d cos θ
W = 10×3× cos (90)
W = 10 × 3 × 0
W = 0 Joules
Learn more about work done here:
https://brainly.com/question/13662169?referrer=searchResults
You want the output current from the secondary coil of a transformer to be 10 times the input current to the primary coil. The ratio of the number of turns N2/N1 must be:_____________.
A. 100
B. 10
C. 1
D. 0.1
Answer:
D. 0.1
Explanation:
Using transformer equation,
N2/N1 = I1/I2................... Equation 1
Where N2 = secondary coil, N1 = primary coil, I1 = input current, I2 = output current.
make I2 the subject of the equation
I2 = I1/(N2/N1)............ Equation 2
From equation 2 above, For the output current of the secondary coil to be 10 times the input current, N2/N1 = 0.1
Hence the right option is D. 0.1
(a) Find the magnitude of an earthquake that has an intensity that is 37.25 (that is, the amplitude of the seismograph reading is 37.25 cm). (Round your answer to one decimal place.)
Answer:
The magnitude of an earthquake is 5.6.
Explanation:
The magnitude of an earthquake can be found as follows:
[tex] M = log(\frac{I}{S}) [/tex]
Where:
I: is the intensity of the earthquake = 37.25 cm
S: is the intensity of a standard earthquake = 10⁻⁴ cm
Hence, the magnitude is:
[tex]M = log(\frac{I}{S}) = log(\frac{37.25}{10^{-4}}) = 5.6[/tex]
Therefore, the magnitude of an earthquake is 5.6.
I hope it helps you!
The temperature gradient between the core of Mars and its surface is approximately 0.0003 K/m. Compare this temperature gradient to that of Earth. What can you determine about the rate at which heat moves out of Mars’s core compared to Earth?
Answer:
The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.
Explanation:
Answer:
The temperature gradient between the core of Mars and its surface is lower than that on Earth. So, heat moves outward more slowly on Mars than on Earth.
Explanation:
Edmentum sample answer
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
A student is trying to decide what to wear.His bedroom is at 20.0 °C.His skin temperature is 35.0 °C.The area of his exposed skin is 1.50 m².People all over the world have skin that is dark in the infrared,with emissivity about 0.900.Find the net energy transfer from his body by radiation in 10.0 min.
Answer:
vgghcgkxcjpfiffj,ncfzfzujfzxxxoifkc xuzrusdoxTXcxgifdhh
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar.
Answer:
a) 6738.27 J
b) 61.908 J
c) [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Explanation:
The complete question is
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.
Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?
Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?
Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.
moment of inertia is given as
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex]
where m is the mass of the flywheel,
and r is the radius of the flywheel
for the flywheel with radius 1.1 m
and mass 11 kg
moment of inertia will be
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]*11*1.1^{2}[/tex] = 6.655 kg-m^2
The maximum speed of the flywheel = 35 m/s
we know that v = ωr
where v is the linear speed = 35 m/s
ω = angular speed
r = radius
therefore,
ω = v/r = 35/1.1 = 31.82 rad/s
maximum rotational energy of the flywheel will be
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]31.82^{2}[/tex] = 6738.27 J
b) second flywheel has
radius = 2.8 m
mass = 16 kg
moment of inertia is
[tex]I[/tex] = [tex]\frac{1}{2}[/tex][tex]mr^{2}[/tex] = [tex]\frac{1}{2}[/tex][tex]*16*2.8^{2}[/tex] = 62.72 kg-m^2
According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels
for the first flywheel, rotational momentum = [tex]Iw[/tex] = 6.655 x 31.82 = 211.76 kg-m^2-rad/s
for their combination, the rotational momentum is
[tex](I_{1} +I_{2} )w[/tex]
where the subscripts 1 and 2 indicates the values first and second flywheels
[tex](I_{1} +I_{2} )w[/tex] = (6.655 + 62.72)ω
where ω here is their final angular momentum together
==> 69.375ω
Equating the two rotational momenta, we have
211.76 = 69.375ω
ω = 211.76/69.375 = 3.05 rad/s
Therefore, the energy stored in the first flywheel in this situation is
E = [tex]Iw^{2}[/tex] = 6.655 x [tex]3.05^{2}[/tex] = 61.908 J
c) one third of the initial energy of the flywheel is
6738.27/3 = 2246.09 J
For the car, the kinetic energy = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
where m is the mass of the car
[tex]v_{car}[/tex] is the velocity of the car
Equating the energy
2246.09 = [tex]\frac{1}{2}mv_{car} ^{2}[/tex]
making m the subject of the formula
mass of the car m = [tex]\frac{4492.18}{v_{car} ^{2} }[/tex]
Which examples are simple machines?
Select all correct answers.
a hammer
an automobile
O a pulley
an inclined plane
A planet in another solar system orbits a star with a mass of 5.0 x 1030 kg. At one point in its orbit, it is 150 x 106 km from the star and is moving at 55 km/s. What is the semimajor axis of the planet's orbit
Answer:
32
Explanation:
You have explored constructive interference from multi-layer thin films. It is also possible for interference to be destructive, a phenomenon exploited in making antireflection coatings for optical elements such as eyeglasses. In order to allow the lenses to be thinner (and thus lighter weight), eyeglass lenses can be made of a plastic that has a high index of refraction (np = 1.70). The high index causes the plastic to reflect light more effectively than does glass, so it is desirable to reduce the reflection to avoid glare and to allow more light to reach the eye. This can be done by applying a thin coating to the plastic to produce destructive interference.
a. Consider a plastic eyeglass lens with a coating of thickness d with index nc . Light with wavelength is incident perpendicular to the lens. If nc < n p , then determine an equation for d in terms of the given variables (and an integer m) in order for there to be destructive interference between the light reflected from the top of the coating and the light reflected from the coating/lens interface.
b. Repeat part a assuming that nc > n p .
c. Choose a suitable value for nc and calculate a value for d that will result in destructive interference for 500 nm light. Note that materials to use for coatings that have nc < 1.3 or nc > 2.5 are difficult to find.
d. Does the index of refraction n p of the eyeglass lens itself matter? Explain.
Answer:
a) d sin θ = m λ₀ / n
b) d sin θ = (m + ½) λ₀ / n
c) d = 2,439 10⁻⁷ m
Explanation:
For the interference these rays of light we must take as for some aspects,
* when a beam of light passes from a medium with a lower index to one with a higher index, the reflected ray has a phase change of 18º, this is equivalent to lam / 2
* when the ray penetrates the lens the donut length changes by the refractive index
λ = λ₀ / n
now let's write the destructive interference equation for these lightning bolts
d sin θ = (m´ + 1/2 + 1/2) λ / n = (m` + 1) λ₀ / n
d sin θ = m λ₀ / n
b) now nc> np
in this case there is no phase change in the reflected ray and the equation for destructive interference remains
d sin θ = (m + ½) λ₀ / n
c) select the value of nc = 2.05 of the ZnO
we calculate the thickness of the film (d)
d = m λ / (n sin 90)
in this type of interference the observation is normal, that is, the angle is 90º)
d = 1 500 10-9 / (2.05 1)
d = 2,439 10⁻⁷ m
d) the lens replacement index is very important because it depends on its relation with the film index which equation to destructively use interference
In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum
Complete Question
In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?
Answer:
a
[tex]\theta = 0.3819^o[/tex]
b
[tex]y = 0.00386 \ m[/tex]
Explanation:
From the question we are told that
The slit separation is [tex]d = 150 \lambda[/tex]
The distance from the screen is [tex]D = 57.9 \ cm = 0.579 \ m[/tex]
Generally the condition for constructive interference is mathematically represented as
[tex]dsin (\theta ) = n * \lambda[/tex]
=> [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]
where n is the order of the maxima and value is 1 because we are considering the central maximum and an adjacent maximum
and [tex]\lambda[/tex] is the wavelength of the light
So
[tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]
[tex]\theta = 0.3819^o[/tex]
Generally the distance between the maxima is mathematically represented as
[tex]y = D tan (\theta )[/tex]
=> [tex]y = 0.579 tan (0.3819 )[/tex]
=> [tex]y = 0.00386 \ m[/tex]
A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round, which can be modeled as a disk with a mass of 300 kg , is spinning at 23 rpm. John runs tangent to the merry-go-round at 4.4 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg.
Required:
What is the merry-go-round's angular velocity, in rpm, after John jumps on?
Answer:
The merry-go-round's angular velocity 23.84 RPM
Explanation:
Given;
diameter of merry go round, d = 3 m
radius of the merry go round, R = 1.5 m
mass of the merry go round, m = 300 kg
angular velocity = 23 rpm
velocity of John, v = 4.4 m/s
mass of John, m = 30 kg
Apply conservation of angular momentum;
[tex]L_i = L_f[/tex]
[tex]I \omega_i + mvR = (I + mR^2)\omega _f[/tex]
where;
I is moment of inertia of disk
[tex]I = \frac{1}{2} mR^2\\\\I = \frac{1}{2} *300*1.5^2\\\\I = 337.5 \ kg.m^2[/tex]
Substitute in this value in the above equation;
[tex]337.5(2\pi \frac{23}{60} ) + (30*4.4*1.5) = (337.5 + 30*1.5^2) \omega_f\\\\812.9925 \ + \ 198 = 405 \omega _f\\\\1010.9925 = 405 \omega _f\\\\\omega _f = \frac{1010.9925}{405} \\\\\omega _f = 2.496 \ rad/s[/tex]
1 rad/s = 9.5493 rpm
2.496 rad/s = 23.84 RPM
Therefore, the merry-go-round's angular velocity 23.84 RPM
The fractional change of reacting mass to energy in a fission reactor is about 0.1%, or 1 part in a thousand. For each kilogram of uranium that is totally fissioned, how much energy is released
Answer:
9*10^13 J
Explanation:
Given that
mass of the material, m = 0.001 kg
speed of light, c = 3*10^8 m/s
To solve this, we would be adopting Einstein's mass - energy relation...
E = mc²
Where
E = Energy, in joules
m = mass, in kg
c = speed of light, in m/s
E = 0.001 * (3*10^8)²
E = 0.001 * 9*10^16
E = 9*10^13 J
Thus, the total energy released would be 9*10^13 J
A wooden artifact from a Chinese temple has a 14C activity of 41.0 counts per minute as compared with an activity of 58.2 counts per minute for a standard of zero age. You may want to reference (Pages 913 - 916) Section 21.4 while completing this problem. Part A From the half-life for 14C decay, 5715 yr, determine the age of the artifact. Express your answer using two significant figures. t
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay
[tex]N=N_0e^{-\lambda t }[/tex]
[tex].70446 =e^{-\lambda t }[/tex]
- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
What is the inductance of a coil if the coil produces an emf of 2.40 V when the current in it changes from -27.0 mA to 33.0 mA in 11.0 ms
Answer:
Inductance of a coil(L) = 0.44 H (Approx)
Explanation:
Given:
coil produces emf = 2.40 V
Old current = -27 mA
New current = 33 mA
Time taken = 11 mS
Find:
Inductance of a coil(L)
Computation:
Inductance of a coil(L) = -emf / [Δi / Δt]
Inductance of a coil(L) = -2.4 / [(-33 - 27) / 11]
Inductance of a coil(L) = -2.4 / [-5.4545]
Inductance of a coil(L) = 0.44 H (Approx)
Which scientist proved experimentally that a shadow of the circular object illuminated 18. with coherent light would have a central bright spot?
A. Young
B. Fresnel
C. Poisson
D. Arago
Answer:
Your answer is( D) - Arago