Answer:
The speed of the water wave is [tex]v = 88 \ m/s[/tex]
Explanation:
From the question we are told that
The width of the tube is [tex]L = 55 \ m[/tex]
The fundamental frequency is [tex]f = 0.80 \ Hz[/tex]
Generally the fundamental frequency is mathematically represented as
[tex]f = \frac{v}{2 * L }[/tex]
=> [tex]v = f * 2 * L[/tex]
substituting values
[tex]v = 0.8 * 2 * 55[/tex]
[tex]v = 88 \ m/s[/tex]
The speed of the water wave will be 88 m/s.
Given information:
When you slosh the water back and forth in a tub at just the right frequency, the water alternately rises and falls at each end, remaining relatively calm at the center.
The frequency of the standing wave is [tex]f=0.8[/tex] Hz.
The width of the tub is [tex]w=55[/tex] m.
Let v be the speed of the standing wave.
The speed of the wave can be calculated as,
[tex]v=2wf\\v=2\times 55\times 0.8\\v=88\rm\; m/s[/tex]
Therefore, the speed of the water wave will be 88 m/s.
For more details, refer to the link:
https://brainly.com/question/1967686
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 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.
A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N, what is the fundamental frequency
Answer:
f₀ = 158.12 HertzExplanation:
The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.
[tex]V = \sqrt{\frac{T}{\mu} }[/tex] where T is the tension in the string and [tex]\mu[/tex] is the density of the string
Given T = 600N and [tex]\mu[/tex] = 0.015 g/cm = 0.0015kg/m
[tex]V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s[/tex]
The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.
L = 2m
Substituting the derived values into the formula f₀ = V/2L
f₀ = 632.46/2(2)
f₀ = 632.46/4
f₀ = 158.12 Hertz
Hence the fundamental frequency of the string is 158.12 Hertz
Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?
Answer:
The final temperature is 61.65 °C
Explanation:
mass of copper pot [tex]m_{c}[/tex] = 2 kg
temperature of copper pot [tex]T_{c}[/tex] = 20 °C (the pot will be in thermal equilibrium with the room)
specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C
The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J
mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg
temperature of boiling water [tex]T_{w}[/tex] = 100 °C
specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C
The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J
The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J
This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation
[tex]H_{T}[/tex] = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]
where [tex]T_{f}[/tex] is the final temperature of the water and the copper
substituting values, we have
99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x
99040 = 770[tex]T_{f}[/tex] + 836.4
99040 = 1606.4[tex]T_{f}[/tex]
[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C
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 polarized laser beam of intensity 285 W/m2 shines on an ideal polarizer. The angle between the polarization direction of the laser beam and the polarizing axis of the polarizer is 16.0 ∘. What is the intensity of the light that emerges from the polarizer?
Answer:
The intensity is [tex]I_1 = 263.35 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the beam is [tex]I = 285\ W/m^2[/tex]
The angle is [tex]\theta = 16^o[/tex]
The intensity of the light that emerges from the polarizer is mathematically represented by Malus' law as
[tex]I_1 = I * cos^2 (\theta )[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
[tex]I_1 = 263.35 \ W/m^2[/tex]
If the rods with diameters and lengths listed below are made of the same material, which will undergo the largest percentage length change given the same applied force along its length?a. d, 3L b. 3d, L c. 2d, 2L d. 4d, L
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½ )/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
A current of 5 A is flowing in a 20 mH inductor. The energy stored in the magnetic field of this inductor is:_______
a. 1J.
b. 0.50J.
c. 0.25J.
d. 0.
e. dependent upon the resistance of the inductor.
Answer:
C. 0.25J
Explanation:
Energy stored in the magnetic field of the inductor is expressed as E = 1/2LI² where;
L is the inductance
I is the current flowing in the inductor
Given parameters
L = 20mH = 20×10^-3H
I = 5A
Required
Energy stored in the magnetic field.
E = 1/2 × 20×10^-3 × 5²
E = 1/2 × 20×10^-3 × 25
E = 10×10^-3 × 25
E = 0.01 × 25
E = 0.25Joules.
Hence the energy stored in the magnetic field of this inductor is 0.25Joules
Determine the value of the current in the solenoid so that the magnetic field at the center of the loop is zero tesla. Justify your answer.
Answer:
I will explain the concept of magnetic field and how it can be calculated.
Explanation:
The formula for magnetic field at the center of a loop is given as
B = μ[tex]_{o}[/tex]I / 2R
where B is the magnetic field
R is the radius of the loop
I is the current
and μ[tex]_{o}[/tex] is the magnetic permeability of free space which is a constant 4π × [tex]10^{-7}[/tex] newtons/ampere²
If the magnetic field at the center of the loop is 0, then μ[tex]_{o}[/tex]I = 0
I = 0 which means there will be no current flow in the loop.
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.
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°
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.
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!
If this is the only water being used in your house, how fast is the water moving through your house's water supply line, which has a diameter of 0.021 m (about 3/4 of an inch)?
Answer:
0.273m/s
Explanation:
first find out the meaning of 0.90×10−4m3/s
literally, that is 0.9x6 = 5.4m3/s = 3•5.4m/s or 16.2 m/s
1.5 gal/min = 0.00009464 m³/s, perhaps that is what you mean?
cross-sectional area of pipe is πr² = 0.0105²π = 0.0003464 m²
so you have a a flow of 0.00009464 m³/s flowing through an area of 0.0003464 m²
they divide to 0.00009464 m³/s / 0.0003464 m² = 0.273 m/s
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]
In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in s) of the material
Answer:
2113 secondsExplanation:
The general decay equation is given as [tex]N = N_0e^{-\lambda t} \\\\[/tex], then;
[tex]\dfrac{N}{N_0} = e^{-\lambda t} \\[/tex] where;
[tex]N/N_0[/tex] is the fraction of the radioactive substance present = 1/16
[tex]\lambda[/tex] is the decay constant
t is the time taken for decay to occur = 8,450s
Before we can find the half life of the material, we need to get the decay constant first.
Substituting the given values into the formula above, we will have;
[tex]\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328[/tex]
Half life f the material is expressed as [tex]t_{1/2} = \frac{0.693}{\lambda}[/tex]
[tex]t_{1/2} = \frac{0.693}{0.000328}[/tex]
[tex]t_{1/2} = 2,112.8 secs[/tex]
Hence, the half life of the material is approximately 2113 seconds
"A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if"
Answer:
A light beam incident on a diffraction grating consists of waves with two different wavelengths. The separation of the two first order lines is great if
the dispersion is great
Which of these cannot be a resistor in a series or parallel circuit?
A)switch
B) battery
C) light bulb
D) all of these are resistors
Answer:
it is going to D. all of these are resistors
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
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
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
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
2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.
Answer:
the atom had a dense, positively charged nucleus.
Explanation:
Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.
According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.
A step-down transformer is used for recharging the batteries of portable devices. The turns ratio N2/N1 for a particular transformer used in a CD player is 2:29. When used with 120-V (rms) household service, the transformer draws an rms current of 180 mA.
Find the rms output voltage of the transformer
Answer:
8.28 V
Explanation:
Using,
N2/N1 = V2/V1.................. Equation 1
Where N2/N1 = Turn ratio of the transformer, V1 = primary/input voltage, V2 = output/secondary voltage
make V2 the subject of the equation
V2 = (N2/N1)V1............ Equation 2
Given: N2/N1 = 2:29 = 2/29, V1 = 120 V
Substitute these values into equation 2
V2 = (2/29)120
V2 = 8.28 V
Hence the rms output voltage of the transformer = 8.28 V
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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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°
An 1,820 W toaster, a 1,420 W electric frying pan, and a 55 W lamp are plugged into the same outlet in a 15 A, 120 V circuit. (The three devices are in parallel when plugged into the same socket.)
Required:
a. What current is drawn by each device?
b. Will this combination blow the 15-A fuse?
Answer:
toaster- 15.1A
electric frying pan- 11.8 A
lamp- 0.5 A
b) The combination will blow the fuse.
Explanation:
When devices are connected in parallel, the potential difference across each of the devices is the same but the current through each is different. Hence;
V= 120 V
Power= IV
For the toaster;
I= 1820/120 = 15.1 A
For the electric frying pan;
I= 1420/120 = 11.8 A
For the lamp;
55/120 = 0.5 A
Total current = 15.1 +11.8 + 0.5 = 27.4 A
The combination will blow the fuse.
Explanation:
step one:
Given data
power of toaster= 1,820 W
power of electric frying pan= 1,420 W
power of lamp= 55 W
current of the outlet= 15 A
voltage of outlet = 120 V
step two
since all three appliances are connected in parallel to the socket outlet, they will use the same voltage of 120 V and the currents will be different across each appliance,
Hence the current across the Toaster will be I₁
using P=I₁V we have
I₁= P/V
I₁= 1820/120 = 15.16 A
A. The current drawn by each device
the current across the electric frying pan will be I₂
using P=I₂V we have
I₂= P/V
I₂= 1420/120 = 11.83 A
the current across the lamp will be I₃
using P=I₃V we have
I₃= P/V
I₃= 55/120 = 0.45 A
therefore the total current drawn by all appliances will be
Total current = I₁+I₂+I₃= 15.16 +11.83+ 0.45= 27.44
B. Will this combination blow the 15-A fuse?
27.44 A > 15 A by 45% ...and this will make fuse to blow
A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?
Answer:
Explanation:
Relation between flux and inductance is as follows
φ = Li
where φ is flux associated with induction of inductance L when a current i flows through it
putting the values
3.25 x 10⁻³ x 800 = L x 2.9
L = .9 H
for induced emf in an induction , the relation is
emf induced = L di / dt
Putting the values
7.5 x 10⁻³ = .9 x di / dt
di / dt = 8.33 x 10⁻³ A / s
(a) The self inductance of the solenoid is 0.897 H.
(b) The magnitude of the rate of change of the current is 0.00836 A/s.
The given parameters;
number of turns, N = 800 turnscurrent in the solenoid, I = 2.9 flux through the solenoid, Ф = 3.25 x 10⁻³ WbThe self inductance of the solenoid is calculated as follows;
[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]
The magnitude of the rate of change of the current is calculated as follows;
[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]
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How does a negative ion differ from an uncharged atom of the same
element?
O A. The ion has a greater number of protons.
B. The ion has fewer protons.
O C. The ion has a greater number of electrons.
O D. The ion has fewer neutrons.
Answer:
C if it is a negitive ion it has more electrons because protons determine what element it is
A front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the front tires
Answer:
static friction acting opposite to the direction of travel
Explanation:
Because the Frictional force of the front wheels act to oppose the spinning, so, For the front wheels to roll without slipping, the friction must be static friction pointing in the direction of travel of the car.
Explanation:
