Answer:
The reason for the increase or decrease of the moon is due to the angular perception of the moon.
Explanation:
Also called lunar illusion, this phenomenon is due to the position in which the moon is, it can be at the zenith or on the horizon, both distances are different from each other with respect to the position of the person.
The zenith is the highest part of the sky and the horizon the lowest.
When there are landmarks such as trees, buildings or mountains on the horizon, the illusion of closeness is given and the illusion of distance is misinterpreted.
But when looking up at the sky as there is no reference point there will be a failure in the perception of size.
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
Inside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight, horizontal piece 28.0 cm long, an elbow, and a straight vertical piece ℓ = 159 cm long. A stud and a second-story floorboard hold the ends of this section of copper pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from 18.0°C to 40.2°C. (The coefficient of linear expansion of copper is
Answer:
The magnitude and direction are
7.638×10-4m
80.01°
Explanation:
We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C
ΔT = 40.2 - 18.0 = 28.5 C°
The expansion of horizontal pipe length can be calculated as
= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8
= 0.0001325 m
The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m
horizontal displacement = 0.1325 mm
= 1.356×10^-4m
vertical displacement = 0.75223mm
=7.5223×10-4m
size of total displacement can be calculated as
√(x²+y²)
Where x and y are vertical and horizontal displacement respectively
= √(0.1325)²+(0.75223)² =
= 0.7638 mm
= 7.638×10-4m
Angle below horizontal = arctan Θ
= 0.75223/0.1325
=5.6772
= arctan (5.6772)
= 80.01°
Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°
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
An interference pattern is produced by light with a wavelength 590 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.580 mm .
Required:
a. If the slits are very narrow, what would be the angular position of the first-order, two-slit, interference maxima?
b. What would be the angular position of the second-order, two-slit, interference maxima in this case?
Answer:
a. 0.058°
b. 0.117°
Explanation:
a. The angular position of the first-order is:
[tex] d*sin(\theta) = m\lambda [/tex]
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{1* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.058 ^{\circ} [/tex]
Hence, the angular position of the first-order, two-slit, interference maxima is 0.058°.
b. The angular position of the second-order is:
[tex] \theta = arcsin(\frac{m \lambda}{d}) = arcsin(\frac{2* 590 \cdot 10^{-9} m}{0.580 \cdot 10^{-3} m}) = 0.12 ^{\circ} [/tex]
Therefore, the angular position of the second-order, two-slit, interference maxima is 0.117°.
I hope it helps you!
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
During the new moon phase, why is the Moon not visible in the sky?
Answer:
Explanation:
The moon gets the light from the sun. When the moon lies between the sun and the earth, only the back portion of the moon gets the light from the sun. So the side facing the sun does not get any light and appears to be dark or does not appear at all.
Hope this helps
plz mark as brainliest!!!!!!!
Answer:
The moon is between the sun, and Earth and reflects light back towards the sun.
Explanation:
A P E X test answer. Just took the test and this is the correct answer.
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)
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.
The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.70 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface.
A) Find the position of the image of the arrow formed by paraxial rays incident on the convex surface.
B) Find the height of the image formed by paraxial rays incident on the convex surface.
C) Is the image erect or inverted?
Answer:
A) 0.1477
B) 0.65388 mm
C) object is inverted
Explanation:
The formula for object - image relationships for spherical reflecting surface is given as;
n1/s + n2/s' = = (n2 - n1)/R
Where;
n1 & n2 are the Refractive index of both surfaces
s is the object distance from the vertex of the spherical surface
s' is the image distance from the vertex of the spherical surface
R is the radius of the spherical surface
We are given;
index of refraction of glass; n2 = 1.60
s = 24 cm = 0.24 m
R = 4 cm = 0.04 m
index of refraction of air has a standard value of 1. Thus; n1 = 1
a) So, making s' the subject from the initial equation, we have;
s' = n2/[((n2 - n1)/R) - n1/s]
Plugging in the relevant values, we have;
s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]
s' = 0.1477
b) The formula for lateral magnification of spherical reflecting surfaces is;
m = -(n1 × s')/(n2 × s) = y'/y
Where;
m is the magnification
n1, n2, s & s' remain as earlier explained
y is the height of the object
y' is the height of the image
Making y' the subject, we have;
y' = -(n1 × s' × y)/(n2 × s)
We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.
Thus;
y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)
y' = - 0.00065388021 m = -0.65388 mm
C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.
Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.
Thus, the object is inverted since m is negative.
A person of 70 kg standing on an un-deformable horizontal surface. She bends her knees and jumps up from rest, achieving a launching speed of 1.7 m/s. The launching process lasts 0.1 second. Calculate the average force exerted by the surface on the person during the launch.
Answer:
1190 N
Explanation:
Force: This can be defined as the product of mass and velocity. The unit of force is Newton(N).
From the question,
F = ma................. Equation 1
Where F = average force, m = mass, a = acceleration.
But,
a = (v-u)/t................ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t.............. Equation 3
Given: m = 70 kg, v = 1.7 m/s, u = 0 m/s (from rest), t = 0.1 s.
Substitute into equation 3
F = 70(1.7-0)/0.1
F = 1190 N.
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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An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic field B = 0.4 T at a constant frequency of 55 Hz. What is the maximum induced emf?
a. 625 V
b. 110 V
c. 421 V
d. 332 V
e. 200 V
Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
A stonecutter's chisel has an edge area of 0.7 cm2. If the chisel is struck with a force of 42 N, what is the pressure exerted on the stone
Answer:
The pressure is [tex]P = 583333 \ N/m^2[/tex]
Explanation:
From the question we are told that
The area of the edge is [tex]A = 0.72 cm^2 = 0.72 *10^{-4}\ m[/tex]
The force is [tex]F = 42 \ N[/tex]
The pressure is mathematically represented as
[tex]P = \frac{F}{A}[/tex]
substituting values
[tex]P = \frac{42}{0.72*10^{-4}}[/tex]
[tex]P = 583333 \ N/m^2[/tex]
Water is draining from an inverted conical tank with base radius 8 m. If the water level goes down at 0.03 m/min, how fast is the water draining when the depth of the water is 6 m
Answer:
0.03/π m/min
Explanation:
See attached file pls
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.
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!
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 object on a level surface experiences a horizontal force of 12.7 N due to kinetic friction. The coefficient of kinetic friction is 0.42.
What is the mass of the object? (Express your answer to two significant figures)kg
Answer:
The mass of the object is 3.08 kg.
Explanation:
The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.
Let the mass of the object = m.
The coefficient of kinetic friction, n = 0.42
Therefore,
Force, F = n × mg
12.7 = 0.42 × 9.8 × m
m = 3.08 kg
The mass of the object is 3.08 kg.
Mars Rover When the Mars rover was deployed on the surface of Mars in July 1997, radio signals took about 12 minmin to travel from Earth to the rover.
How far was Mars from Earth at that time?
Answer:
s = 2.16 x 10¹¹ m
Explanation:
Since, the waves travelling from Earth to the Mars rover are electromagnetic. Therefore, there speed must be equal to the speed of light. So, from the equation given below:
s = vt
where,
s = the distance between Earth and Mars = ?
v = speed of the wave = speed of light = 3 x 10⁸ m/s
t = time taken by the radio signals to reach the rover from Earth
t = (12 min)(60 s/1 min) = 720 s
Therefore,
s = (3 x 10⁸ m/s)(720 s)
s = 2.16 x 10¹¹ m
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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What is the wavelength of electromagnetic radiation which has a frequency of 3.818 x 10^14 Hz?
Answer:
7.86×10⁻⁷ m
Explanation:
Using,
v = λf.................. Equation 1
Where v = velocity of electromagnetic wave, λ = wave length, f = frequency.
make λ the subject of the equation
λ = v/f............... Equation 2
Note: All electromagnetic wave have the same speed which is 3×10⁸ m/s.
Given: f = 3.818×10¹⁴ Hz
Constant: v = 3×10⁸ m/s
Substitute these values into equation 2
λ = 3×10⁸/3.818×10¹⁴
λ = 7.86×10⁻⁷ m
Hence the wavelength of the electromagnetic radiation is 7.86×10⁻⁷ m
The wavelength of this electromagnetic radiation is equal to [tex]7.86 \times 10^{-7} \;meters[/tex]
Given the following data:
Frequency = [tex]3.818\times 10^{14}\;Hz[/tex]Scientific data:
Velocity of an electromagnetic radiation = [tex]3 \times 10^8\;m/s[/tex]
To determine the wavelength of this electromagnetic radiation:
Mathematically, the wavelength of an electromagnetic radiation is calculated by using the formula;
[tex]Wavelength = \frac{Speed }{frequency}[/tex]
Substituting the given parameters into the formula, we have;
[tex]Wavelength = \frac{3 \times 10^8}{3.818\times 10^{14}}[/tex]
Wavelength = [tex]7.86 \times 10^{-7} \;meters[/tex]
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where c is the speed of light and G is the universal gravitational constant. RBH gives the radius of the event horizon of a black hole with mass ????. In other words, it gives the radius to which some amount of mass ???? would need to be compressed in order to form a black hole. The mass of the Sun is about 1.99×1030 kg. What would be the radius of a black hole with this mass?
Answer:
The radius of the black hole will be 2949.6 m.
Explanation:
The radius of this black hole will be the Schwarzschild radius of the mass of the sun
[tex]r_{s}[/tex] = [tex]\frac{2GM}{c^{2} }[/tex]
where
G is the gravitational constant = 6.67 x 10^-11 m^3⋅kg^-1⋅s^-2
M is the mass of the sun = 1.99×10^30 kg
c is the speed of light = 3 x 10^8 m/s
substituting values into the equation, we have
[tex]r_{s}[/tex] = [tex]\frac{2*6.67*10^{-11}*1.99*10^{30} }{(3*10^{8} )^{2} }[/tex] = 2949.6 m
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
If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y-direction, the wave is traveling in the
Answer:
The wave is travelling in the ±z-axis direction.
Explanation:
An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the electric and magnetic field.
In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.
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.
Learn more about the Boyle's law here:
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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
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
beam of white light goes from air into water at an incident angle of 58.0°. At what angles are the red (660 nm) and blue (470 nm) parts of the light refracted? (Enter your answer to at least one decimal place.) red ° blue °
Answer:
For red light= 39.7°
Blue light 39.2°
Explanation:
Given that refractive index for red light is 1.33 and that of blue light is 1.342
So angle of refraction for red light will be
Sinစi/ (sinစ2) =( nw)r/ ni
Sin 58° x 1.000293/1.33. =( sinစ2)r
0.64= sinစ2r
Theta2r = 39.7°
For blue light
Sinစi/ (sinစ2) =( nw)b/ ni
Sin 58° x 1.000293/1.342 =( sinစ2)b
0.632= sinစ2r
Theta2b= 39.19°
a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of
Complete question:
A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).
Answer:
(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³
(b) the total energy stored in the magnetic field is 0.121 J
Explanation:
Given;
length of the solenoid, L = 98.6 cm = 0.986 m
cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²
number of turns of the solenoid, N = 1310 turns
The magnitude of the magnetic field inside the solenoid is given by;
B = μ₀nI
B = μ₀(N/L)I
Where;
μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A
[tex]B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T[/tex]
(a) Calculate the energy density of the magnetic field inside the solenoid
[tex]u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3[/tex]
(b) Find the total energy stored in the magnetic field
U = uV
U = u (AL)
U = 50.53 (24.3 x 10⁻⁴ x 0.986)
U = 0.121 J
