Answer:
b) the heavier one will have twice the kinetic energy of the lighter one.
Explanation:
The kinetic energy of object with mass, m
K.E₁ = ¹/₂mv²
where;
m is mass of the object
v is the velocity of the object
Since, the two objects are falling under same acceleration due to gravity, their velocity will be increasing at the same rate
The kinetic energy of object with mass, 2m
K.E₂ = ¹/₂(2m)v²
K.E₂ = 2(¹/₂mv²)
BUT K.E₁ = ¹/₂mv²
K.E₂ = 2(K.E₁)
Therefore, the heavier one will have twice the kinetic energy of the lighter one.
b) the heavier one will have twice the kinetic energy of the lighter one.
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
Sunlight reflected from a smooth ice surface is completely polarized. Determine the angle of incidence. (nice = 1.31.)
Answer:
Explanation:
We shall apply Brewster's law to solve the problem . Let i be the angle of reflection required. According to this law ,
tan i = μ where i is angle of incidence , μ is refractive index of the medium .
Here i = ? , μ = 1.31
tani = 1.31
i = 53° approx .
At the angle of 53° incidence , the reflected ray will be completely polarised .
You have a horizontal grindstone (a disk) that is 95 kg, has a 0.38 m radius, is turning at 87 rpm (in the positive direction), and you press a steel axe against the edge with a force of 16 N in the radial direction.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone.
(b) How many turns will the stone make before coming to rest?
Answer:
Explanation:
The moment of inertia of the disk I = 1/2 m R² where R is radius of the disc and m is its mass .
putting the values
I = .5 x 95 x .38²
= 6.86 kg m²
n = 87 rpm = 87 / 60 rps
n = 1.45 rps
angular velocity ω = 2π n , n is frequency of rotation .
= 2 x 3.14 x 1.45
= 9.106 radian /s
frictional force = 16 x .2
= 3.2 N
torque created by frictional force = 3.2 x .38
= 1.216 N.m
angular acceleration = torque / moment of inertia
= - 3.2 / 6.86
α = - 0.4665 rad /s²
b ) ω² = ω₀² + 2 α θ , where α is angular acceleration
0 = 9.106² - 2 x .4665 θ
θ = 88.87 radian
no of turns = 88.87 / 2π
= 14.15 turns
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 × 10-6 C-1.
Answer:
the new length is 17.435cm
Explanation:
the new length is 17.435cm
pls give brainliest
The new length of aluminum rod is 17.435 cm.
The linear expansion coefficient is given as,
[tex]\alpha=\frac{L_{1}-L_{0}}{L_{0}(T_{1}-T_{0})}[/tex]
Given that, An aluminum rod 17.400 cm long at 20°C is heated to 100°C.
and linear expansion coefficient is [tex]25*10^{-6}C^{-1}[/tex]
Substitute, [tex]L_{0}=17.400cm,T_{1}=100,T_{0}=20,\alpha=25*10^{-6}C^{-1}[/tex]
[tex]25*10^{-6}C^{-1} =\frac{L_{1}-17.400}{17.400(100-20)}\\\\25*10^{-6}C^{-1} = \frac{L_{1}-17.400}{1392} \\\\L_{1}=[25*10^{-6}C^{-1} *1392}]+17.400\\\\L_{1}=17.435cm[/tex]
Hence, The new length of aluminum rod is 17.435 cm.
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What is the speed of light (in m/s) in air? (Enter your answer to at least four significant figures. Assume the speed of light in a vacuum is 2.997 ✕ 108 m/s.) m/s What is the speed of light (in m/s) in polystyrene? m/s
Answer:
The speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
Explanation:
To find the speed of light in air and in polystyrene we need to use the following equation:
[tex] c_{m} = \frac{c}{n} [/tex]
Where:
[tex]c_{m}[/tex]: is the speed of light in the medium
n: is the refractive index of the medium
In air:
[tex]c_{a} = \frac{c}{n_{a}} = \frac{2.997 \cdot 10^{8} m/s}{1.0003} = 2.996 \cdot 10^{8} m/s[/tex]
In polystyrene:
[tex]c_{p} = \frac{c}{n_{p}} = \frac{2.997 \cdot 10^{8} m/s}{1.6} = 1.873 \cdot 10^{8} m/s[/tex]
Therefore, the speed of light in air is 2.996x10⁸ m/s, and polystyrene is 1.873x10⁸ m/s.
I hope it helps you!
A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the right when the driver turns on the motor. The boat speeds up for 6.0\,\text s6.0s6, point, 0, start text, s, end text with an acceleration of 4.0\,\dfrac{\text m}{\text s^2}4.0 s 2 m 4, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction leftward.
The question is incomplete. Here is the entire question.
A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?
Answer: Δx = - 42m
Explanation: The jetboat is moving with an acceleration during the time interval, so it is a linear motion with constant acceleration.
For this "type" of motion, displacement (Δx) can be determined by:
[tex]\Delta x = v_{i}.t + \frac{a}{2}.t^{2}[/tex]
[tex]v_{i}[/tex] is the initial velocity
a is acceleration and can be positive or negative, according to the referential.
For Referential, let's assume rightward is positive.
Calculating displacement:
[tex]\Delta x = 5(6) - \frac{4}{2}.6^{2}[/tex]
[tex]\Delta x = 30 - 2.36[/tex]
[tex]\Delta x[/tex] = - 42
Displacement of the boat for t=6.0s interval is [tex]\Delta x[/tex] = - 42m, i.e., 42 m to the left.
Describe how, using a positively-charged rod and two neutral metal spheres, we canmake one sphere positive without touching it to the rod. You might want to draw adiagram to help you.
Answer:
se the principle of induction.
place the two metallic spheres together, now we bring the positively charged bar closer to the first sphere.
The charge that was induced in the sphere is distributed as infirm as possible,
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive
Explanation:
For this exercise we will use that the electric charge is not created, it is not destroyed and charges of the same sign repel.
Let's use the principle of induction. We place the two metallic spheres together, one in front of the other, now we bring the positively charged bar closer to the first sphere.
Here the positive charge of the bar repels the positive charge of the sphere, but as this is mocil it moves as far away as possible, until the negative charge that remains neutralizes the positive charge of the bar.
The charge that was induced in the sphere is distributed as infirm as possible, most of it in the furthest sphere, since the Coulomb force decreases.
At this time I separate the spheres and move the bar away, by separating the spheres the excess positive charge in the last sphere cannot be neutralized, therefore this sphere remains with a positive charge.
A person looks horizontally at the edge of a swimming pool. If its length is 5 m, and the pool is filled to the surface, to what depth (in m) could the observer see
Answer:
The observer could see to a depth of 4.38 m
Explanation:
Please check attachment for diagram.
Mathematically, from Snell law;
n1sin theta = n2 sin theta
1 * sin 90 = n2 * sin θR
where n2 = 1.33
1/1.33 = sin θR
Sin θR = 0.7519
θR = arc sin 0.7519
θR = 48.76
Now to get the height, we use the triangle
Using trigonometric ratio;
Tan( 90- θR) = H/5
H = 5 Tan( 90 - θR)
H = 5 Tan( 90-48.76)
H = 5 Tan41.24
H = 4.38 m
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Required:
a. Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
b. Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm^2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?
Answer:
1. 6.99x 10^-6V/m
2. 18m
Explanation:
See attached file
A converging lens 7.50 cm in diameter has a focal length of 330 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of resolving power of the human eye. Part A If the resolution is diffraction limited, how far away can an object be if points on it transversely 4.10 mm apart are to be resolved (according to Rayleigh's criterion) by means of light of wavelength 600 nm
Answer:
D Is 430m
Explanation:
See attached file
A scuba diver fills her lungs to capacity (6.0 L) when 10.0 m below the surface of the water and begins to ascend to the surface. Assume the density of the water in which she is swimming is 1000 kg/m3 and use g = 10 m/s2A. Assuming the temperature of the air in her lungs is constant, to what volume must her lungs expand when she reaches the surface of the water?B. What effect would the warming of the air in her lungs have on the volume needed when she surfaces?C. Assuming the temperature of the air in her lungs is constant, what effect does her ascent have on the vrms of the air molecules in her lungs?
Answer:
Explanation:
As temperature is constant , we shall apply Boyle's law
P₁V₁ = P₂V₂
P₁ = pressure at depth of 10 m
= P + hdg , h = 10 , d = 10³ , g = 10
P is atmospheric pressure which is 10⁵ Pa
P₁ = 10⁵ + 10 x 10³ x 10
= 2 x 10⁵
applying the formula
2 x 10⁵ x 6 = 10⁵ x v
v = 2 x 6 = 12 L
volume will be doubled at the surface .
B )
warming of air at the surface will increase the volume of air in her lungs so so she will need more lung capacity .
C )
The rms value of a gas depends upon the temperature of the gas . As temperature of the gas is constant , the rms value of the gas particles will remain constant when she goes to the surface .
The lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.
What is Boyle's law?According to the law, the pressure of the gas is inversely proportional to the volume of the gas. In other words when the pressure of the gas increases the volume of the gas decreases.
We know the pressure at the 10 meters depth:
[tex]\rm P_1 = P+h\times \rho\times g[/tex]
Where P = Atmospheric pressure
h = Depth
ρ =Density of the water
We have: [tex]\rm P = 10^5 \ Pa[/tex], h = 10 meters, and [tex]\rm \rho = 1000 \ kg/m^3[/tex], and [tex]\rm g = 10 \ m/s^2[/tex]
Putting the values in the above equation, we get:
[tex]\rm P_1 = 10^5+ 10\times 1000\times 10[/tex]
[tex]\rm P_1 = 2\times 10^5[/tex]
From the Boyle's law:
[tex]\rm P_1\times V_1 = P_2\times V_2[/tex]
[tex]\rm 2\times10^5\times 6 = 10^5\times V_2[/tex]
[tex]\rm V_2 = 12 \ L[/tex]
We know that as the air at the surface warms, the volume of air in her lungs expands, requiring more lung capacity.
The temperature of the gas is constant and [tex]\rm V_{rms}[/tex] values for gas depend on the temperature of the gas, but here the temperature of the gas is constant thus, the [tex]\rm V_{rms}[/tex] will remains constant.
Thus, the lungs will expand 12 L when she reaches the surface of the water, and the warming of the air results in more lung capacity, and [tex]\rm V_{rms}[/tex] the value remains the same.
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The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter and 1.2 m long. The starter motor runs for 0.760 s until the car engine starts.Required:a. How much charge passes through the starter motor? b. How far does an electron travel along the wire while the starter motor is on?(mm)
Answer:
(a)106.4C
b)0.5676mm
Explanation:
(a)To get the charge that have passed through the starter then The current will be multiplied by the duration
I= current
t= time taken
Q= required charge
Q= I*t = 140*0.760 = 106.C
(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)
diameter of the conductor is 4.20 mm
But Radius= diameter/2= 4.20/2=
The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then
radius of the conductor is 0.0021m.
We can now calculate the area of the conductor which is
A = π*r^2
= π*(0.0021)^2 = 13.85*10^-6 m^2
We can proceed to calculate the current density below
J = 140/13.85*10^-6 = 10108303A/m
According to the listed reference:
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s
Therefore , the distance traveled is:
x = v*t = 0.7468 * 0.760 = 0.5676mm
(a) The charge passes through the starter motor is 106.4C.
(b) An electron travel along the wire while the starter motor is on 0.5676mm.
ElectronAnswer (a)
I= current
t= time taken
Q= required charge
Q= I*t
Q= 140*0.760
Q= 106.C
Answer (b)
The n electron travel along the wire while the starter motor is on:
Diameter of the conductor is 4.20 mm
Radius= diameter/2= 4.20/2
Radius =2.1mm
Radius of the conductor is 0.0021m.
A = π*r^2
A= π*(0.0021)^2
A= 13.85*10^-6 m^2
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )
Vd =0.0007468m/s
Vd =0 .7468 mm/s
The distance traveled is:
x = v*t
x= 0.7468 * 0.760
x = 0.5676mm
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A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.
Answer:
82.2 mL
Explanation:
The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;
C1V1=C2V2
Where;
C1= concentration of stock solution
V1= volume of stock solution
C2= concentration of dilute solution
V2= volume of dilute solution
V2= C1V1/C2
V2= 1.48 × 55.6/ 1.0
V2= 82.2 mL
A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy
Answer: her rotational kinetic energy increases
An electron moves on a circular orbit in a uniform magnetic field of 7.83×10-4 T. The kinetic energy of the electron is 55.3 eV. What is the diameter of the orbit?
Answer:
3.9E-8
Explanation:
We know that
Mv²/r = Bqv
So
r= mv/Bq
But E is 1/2mv² which is 5.53eV
m²v² =2m x 5.53eV
mv = √( 2 x 9.1E-31)
So
r= √( 2 x9.1E-31 x 5.53)/ 7.83x10^-4 x1.6E-19
= 3.9x10-8cm
Explanation:
Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.
Complete Question
An emf is induced in a conducting loop of wire 1.12m long as its shape is.
changed from square to circular. Find the average magnitude of the induced emf if the change in shape occurs in 0.125 ss and the local 0.504-TT magnetic field is perpendicular to the plane of the loop.
Answer:
The induced emf is [tex]\epsilon = 0.0863 \ V[/tex]
Explanation:
From the question we are told that
The time taken is [tex]\Delta t = 0.125 \ s[/tex]
The magnitude of the magnetic field is B = 0.504 T
The length of the loop wire is [tex]l = 1.12 \ m[/tex]
Generally the circumference of the wire when in circular form is
[tex]C = 2 \pi r[/tex]
=> [tex]l = 2 \pi r[/tex]
=> [tex]r =[/tex][tex]\frac{l}{2 \pi}[/tex]
=> [tex]r =[/tex][tex]\frac{1.12}{2 * 3.142}[/tex]
=> [tex]r =[/tex][tex]0.1782 \ m[/tex]
Now the area of the wire as a circle is
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.1782)^2[/tex]
=> [tex]A = 0.0998 \ m^2[/tex]
The length of one side of the square is
[tex]b = \frac{l}{4}[/tex]
[tex]b = \frac{1.12}{4}[/tex]
[tex]b = 0.28 \ m[/tex]
Now the area of the wire as a square is
[tex]A_s = b^2[/tex]
=> [tex]A_s =(0.28 )^2[/tex]
[tex]A_s = 0.0784 \ m^2[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = \frac{B * [A - A_s ]}{\Delta t }[/tex]
=> [tex]\epsilon = \frac{0.504 * [0.0998 - 0.0784 ]}{0.125 }[/tex]
=> [tex]\epsilon = 0.0863 \ V[/tex]
When a ray of light passes from glass to water it is?
Answer:
[tex]\huge\boxed{Refracted}[/tex]
Explanation:
When a ray of light passes from glass to water, it
1) is Slightly refracted (bending of light)
2) moves away from the normal.
Whenever a light ray travels from a denser medium to a rarer medium, it bends away from the normal.
Answer:
refraction
Explanation:
A diver shines an underwater searchlight at the surface of a pond (n = 1.33). At what angle (relative to the surface) will the light be totally reflected?
Answer:
41.2°
Explanation:
Total internal reflection is the reflection of the incident ray at the interface between two media in which one of the media has a lower refractive index than the other. It occurs when the angle of incidence in the denser medium exceeds the critical angle.
The critical angle is the angle of incidence in the denser medium when the angle of incidence in the less dense medium is 90°.
Since
n= 1/sin C
C= sin^-(1/n)
C= sin^-(1/1.33)
C= 48.8°
Hence angle of incidence= 90-48.8 = 41.2°
Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?
Answer:
L = 0.8 m
Explanation:
Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:
Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m
but,
Δx = λL/d
λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m
d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m
L = Distance between slits and screen = ?
Therefore, using the values, we get:
2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)
L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)
L = 0.8 m
An electron in a vacuum chamber is fired with a speed of 7400 km/s toward a large, uniformly charged plate 75 cm away. The electron reaches a closest distance of 15 cm before being repelled.
What is the plate's surface charge density?
Answer:
2.29e-9C/m²
Explanation:
Using E = σ/ε₀ means the force on the electron is F = eE = eσ/ε₀.
The work done on the electron is W = Fd = deσ/ε₀. This equals the kinetic energy lost, ½mv².
½mv² = deσ/ε₀
d = 75cm – 15cm = 60cm = 0.6m
σ = mv²ε₀/(2de)
. .= 9.11e-31 * (7.4e6)² * 8.85e-12 / (2 * 0.6 * 1.6e-19)
. .= 2.29e-9 C/m² (i.e. 2.29x10^-9 C/m²)
within which type of system is the total mass conserved but not the total energy
In a closed system the mass is conserved, but the energy is not conserved.
To find the answer, we have to study about different systems in thermodynamics.
What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.
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What is the smallest value of n for which the wavelength of a Balmer series line is less than 400 nm
Answer:
The smallest value is n= 2
Explanation:
The balmer equation is given below
1/λ = R(1/4 - 1/n₂²).
R= 1.0973731568508 × 10^7 m^-1
λ= 400*10^-9 m
(400*10^-9)= 1.0973731568508 × 10^7 (1/4-1/n²)
(400*10^-9)/1.0973731568508 × 10^7
= 1/4 - 1/n²
364.51 *10^-16= 1/4 - 1/n²
1/n²= 1/4 -364.51 *10^-16
1/n² = 0.25-3.6451*10^-14
1/0.25= n²
4= n²
√4= n
2= n
The smallest value is N= 2
When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The braking force on the magnet is nearly equal to its weight, so it falls very slowly. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the magnet is pushed, forcefully, toward the block, the rate of change of flux is much higher than this. When the magnet is moving much more quickly than it will fall unaided, what is the direction of the net force on the magnet?
Answer:
The net force is directed downwards.
Explanation:
Since the magnet is falling much more faster than it would unaided, then there is a net force that is accelerating the magnet downwards. We know that acceleration is due to a force acting on a mass, and in this case, the magnet is the mass. Also, the acceleration is always in the direction of the force producing it, which means that the net force on the magnet is vertically downwards.
A cylinder with rotational inertia I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω1=5.0rad/s. A second cylinder with rotational inertia I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω2=8.0rad/s. If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?
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%
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Question 5 of 10
Heat is being transferred through currents within a liquid. When will this heat
transfer mostly end?
O A. When the substance changes state and becomes a gas
O B. When the entire liquid is a single temperature
O C. When the substance is very hot on top and cold beneath
O D. When the particles stop bumping into each other
SUBMIT
Answer:
When the entire liquid is a single temperature
Explanation:
When a liquid is heated, a convection current is set up. Convection is the movement of
fluid particles in response to a temperature gradient.
When you start heating a liquid, the particles near the base of the heating vessel increase in temperature, become less dense and rise upwards while the denser particles move downwards. This convection current will continue until an equilibrium temperature is obtained throughout the liquid.
Calculate the range of wavelengths (in m) for AM radio given its frequency range is 540 to 1,600 kHz. smaller value m larger value m (b) Do the same for the ultraviolet frequency range of 760 to 30,000 THz. smaller value m larger value m
Answer:
a) λ = 555.5 m, λ = 187.5 nm
Explanation:
The velocity of a wave is given by the relation
c = λ f
λ = c /f
a) length of the radii AM
λ = 3 10⁸/540 10⁻⁷
λ= 5.555 102 m
λ = 555.5 m
f = 1600 kHz
λ = 3 108/1600 103
λ = 1.875 102 m
lam = 187.5 nm
b) light = 760 Thz = 760 10-12 Hz
λ= c/f
λ = 3 108/760 10-12
λ = 3.947 10-9
λ= 30000 Thz
λ= c/f
λ = 3 10⁸/ 30000 10-12
λ = 1 m
Two automobiles are equipped with the same singlefrequency horn. When one is at rest and the other is moving toward the first at 20 m/s , the driver at rest hears a beat frequency of 9.0 Hz.
Requried:
What is the frequency the horns emit?
Answer: f ≈ 8.5Hz
Explanation: The phenomenon known as Doppler Shift is characterized as a change in frequency when one observer is stationary and the source emitting the frequency is moving or when both observer and source are moving.
For a source moving and a stationary observer, to determine the frequency:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
where:
[tex]f_{0}[/tex] is frequency of observer;
[tex]f_{s}[/tex] is frequency of source;
c is the constant speed of sound c = 340m/s;
[tex]v_{s}[/tex] is velocity of source;
Rearraging for frequency of source:
[tex]f_{0} = f_{s}.\frac{c}{c-v_{s}}[/tex]
[tex]f_{s} = f_{0}.\frac{c-v_{s}}{c}[/tex]
Replacing and calculating:
[tex]f_{s} = 9.(\frac{340-20}{340})[/tex]
[tex]f_{s} = 9.(0.9412)[/tex]
[tex]f_{s} =[/tex] 8.5
Frequency the horns emit is 8.5Hz.
"How many wavelengths wide must a single slit be if, at a point 8o from the central maximum, there is a 72 rad phase difference between the top and bottom rays?"
Answer
82.3 wavelengths
What is the maximum speed (in units of m/s) with which a car can round a
flat horizontal curve of radius (r=60 m), if the coefficient of static friction between tires and
road is (0.4)?
A) 5
B) 15.5
€) 240
D) 25.1
Answer:
B) 15.5 m/s
Explanation:
r = 60m
μs = 0.4
using the formula max V = √r*g*μs (flat roadway)
v = sqrt(60 * 10 * 0.4)
v = 15.5 m/s
A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?
Answer:
Explanation:
According to Equations of Projectile motion :
[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]
vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec
(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec
[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]
(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m
[tex]Horizontal Range = vcosx * t[/tex]
(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m
