Answer:
ΔL = 0.66 m
Explanation:
The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:
ΔL = αLΔT
where,
ΔL = Change in Length of the bridge = ?
α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹
L = Original Length of the Bridge = 1000 m
ΔT = Change in Temperature = Final Temperature - Initial Temperature
ΔT = 40°C - (-20°C) = 60°C
Therefore,
ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)
ΔL = 0.66 m
2.) Is it possible to have negative velocity but positive acceleration? If so, what would
this mean?
Answer:
Yes, yes it would
Explanation:
an electron starts from rest from a fixed point charge with q. what total potential difference accelerates the electron to being very far away from Q
Answer:
V = (k*Q)/R
Explanation:
Total potential difference accelerates the electron to being very far away from Q is;
V = (k*Q)/R
Where,
V is the Potential Difference in Joules per Coulomb
k is the constant
Q is the charge in Coulomb
R is Electron distance in cm or m
Example
An electron starts from rest 66.1 cm from a fixed point charge with Q = -0.120 μC. What total potential difference accelerates the electron from being very far away from Q
For k = 9.0*10^9 N*m^2/C^2
V = (k*Q)/R
V = (9.0*10^9 * -0.120*10^-6)/0.661
V = -1633.9 Volt.
The answer will change to positive because V = (k*Q)/R is negative at the outset and Zero far away.
The electron (with a negative charge) has a positive energy in the beginning and that gets converted into a positive kinetic energy "far away".
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength of 880 nm. What is the potential difference though which this electron was accelerated
Answer:
3x10⁴v
Explanation:
Using
Wavelength= h/ √(2m.Ke)
880nm = 6.6E-34/√ 2.9.1E-31 x me
Ke= 6.6E-34/880nm x 18.2E -31.
5.6E-27/18.2E-31
= 3 x 10⁴ Volts
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
1. Why do you see colors when you look at reflected light from a CD or DVD disk, or when you look at a soap bubble or oil film on water?
2. What do you think causes the colors on the artwork panels on the side of HLS2 (Health Sciences building) which change with time of day and the angle from which you view them?
Explanation:
1.The light reflected from the CD/DVD or soap bubble or oil film forms an interference with the surrounding light. The inference both constructive and destructive making some color appear and some disappear.
2.As light behaves as wave it will interfere differently at different angles. At certain angle it will interfere constructively and at certain angle it will interfere destructively making some color brighter and some disappear. So, at different angles the color are different.
Interference pattern is responsible for the formation of different colour when a light reflected from CD or soap bubble.
We can see colors when we look at reflected light from a CD or DVD disk, or a soap bubble or oil film on water because of the interference pattern. The colors that we see on the CD are created due to the reflection of white light from ridges in the metal. When light passes through something with many small ridges or scratches, we often see rainbow colors and interesting patterns.
These patterns are called interference patterns. White light is made up of 7 colors i.e. red, orange, yellow, green, blue, indigo, violet. The CD converts or separates the white light into 7 colors so we can conclude that interference pattern is responsible for the formation of different colour when a light reflected from CD OR soap bubble.
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which of the following best describes pseudoscience?
Answer:
The answer is A
Explanation:
Answer:
implausible or untestable scientific claims
g A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75 C/s . Part A How many electrons are delivered to the
Answer:
Approximately 5 x 10^18 electrons are delivered to the smart phone charger.
Explanation:
The electric current in a circuit is the flow of charges through a circuit with time.
The charges through the circuit are due to the electrons that flow through the circuit.
An individual electrons has a charge of -1.60 x 10^-19 C on it.
If the current through the circuit is -0.75 C/s, then the number of electrons that are delivered is gotten by dividing the charge per second by the charge on an electron.
==> -0.75/(-1.60 x 10^-19) = 4.67 x 10^18 electrons ≅ 5 x 10^18 electrons are delivered to the smart phone charger.
A cook preparing a meal for a group of people is an example of
O kinetic energy because he has the ability to make a meal
O potential energy because he has the ability to make a meal
O kinetic energy because he is making the meal
o potential energy because he is making the meal
The frequency of light emitted from hydrogen present in the Andromeda galaxy has been found to be 0.10% higher than that from hydrogen measured on Earth.
Is this galaxy approaching or receding from the Earth, and at what speed?
Answer:
3x10^5m/s
Explanation:
See attached file
Explanation:
The speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
Doppler's Effect
According to the Doppler effect, the difference between the frequency at which light wave leave a source and reaches an observer is caused by the relative motion of the observer and the wave source.
Given that the difference in the frequency is 0.10 %. The speed of light emitted from the galaxy can be calculated by the Doppler effect.
[tex]\dfrac {\Delta f}{f} = \dfrac {v}{c}[/tex]
Where f is the frequency of the light, v is the speed of light emitted from the galaxy and c is the speed of light emitted from the earth.
[tex]\dfrac {0.10 f}{100 f} = \dfrac {v}{3\times 10^8}[/tex]
[tex]v = 3\times 10^5\;\rm m/s[/tex]
Hence we can conclude that the speed of the light emitted from the earth is approaching the galaxy at [tex]3\times 10^5\;\rm m/s[/tex].
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NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low mass sail and the energy and momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective? Why?
(b) How large a sail is necessary to propel a 10000kg
spacecraft against the gravitational force of the sun? Express your result in square kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun.
The gravitational constant is G=6.67×10−11m3⋅s−2⋅kg−1.
The mass of the sun is Ms=1.99×1030kg.
Answer:
The complete question is
NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 x 10^26 W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.
a) The sail should be reflective because, an incident electromagnetic wave, in this case, light wave, impacts twice the energy density on a reflective sail, and hence twice the force on a totally reflective sail as would be impacted on a sail that is totally absorbing.
For totally reflective, F = (2I/c)A ....1
for totally reflective, F = (I/c)A ....2
where I is the intensity of the light
c is the speed of light = 3 x 10^8 m/s
A is the area the sail
b) The intensity of the light from the sun = power/area
==> I = [tex]\frac{3.9*10^{26}}{4\pi r^{2} }[/tex]
where r is the distance from the sun and the sail
The Force from the sail from equation 1 is therefore
[tex]F[/tex] = [tex]\frac{2*3.9*10^{26}*A}{4\pi r^{2} *3*10^{8}}[/tex] = [tex]2.069*10^{17}\frac{A}{r^{2}}[/tex]
gravitational force between the sail and the sun [tex]F_{g}[/tex] = [tex]\frac{GMm}{r^{2}}[/tex]
where
G is the gravitational constant = 6.67 x 10^−11 m^3⋅s−2⋅kg−1.
m is the mass of the sail = 10000 kg
M is the mass of the sun = 1.99 x 10^30 kg.
==> [tex]F_{g}[/tex] = [tex]\frac{6.67*10^{-11}*1.99*10^{30}*10000}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
Equating the forces, we have
[tex]2.069*10^{17}\frac{A}{r^{2}}[/tex] = [tex]\frac{1.33*10^{24}}{r^{2}}[/tex]
the distance cancels out
A = (1.33 x 10^24)/(2.069 x 10^17) = 6428226.196 m^2
==> 6428.2 km^2
c) The force of the solar radiation is proportional to the intensity of the sun from the light, and the intensity is inversely proportional to the square of the distance from the source. Also, the force of gravitation is inversely proportional to the square of the distance, so they both cancel out.
If an electromagnetic wave has components Ey = E0 sin(kx - ωt) and Bz = B0 sin(kx - ωt), in what direction is it traveling?
Answer:
Its traveling in the +x direction
Explanation:
The E-field is in the +y-direction, and the B-field is in the +z-direction, so it must be moving along the +x-direction, since the E-field, B-field and the direction of moving are all at right angles to each other.
The electromagnetic wave is travelling in the +x direction.
Electromagnetic waves are waves formed as a result of vibrations between an electric field and a magnetic field.
Given that:
Ey = E0 sin(kx - ωt)
Hence the electric field is moving in the +y direction.
Also, Bz = B0 sin(kx - ωt)
Hence the magnetic field is moving in the +z direction
Electric fields and magnetic fields (E and B) in an EM wave are perpendicular to each other and are also perpendicular to the direction of propagation of the wave.
Therefore the direction of the wave is travelling in the +x direction.
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Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields. Which of the following statements correctly describe the various applications listed above?
a) All these technologies use radio waves, including low-frequency microwaves.
b) All these technologies use radio waves, including high-frequency microwaves.
c) All these technologies use a combination of infrared waves and high-frequency microwaves.
d) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f) All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m.
g) All the technologies emit waves with a wavelength in the range 0.01 to 10.0 km.
Answer:
d) Microwave ovens emit in the same frequency band as some wireless Internet devices.
Explanation:
Microwave are radio waves of short wavelength, from about 10 centimetres to one millimetre, in the Super High Frequency and the Extremely High Frequency bands. Microwaves can penetrate into materials and deposit their energy below the surface which is why is is used in microwave heating found in microwave oven. Transmission of data sometimes involves the use of microwaves to send and receive information over a long distance. Microwaves are the mainly used in radar, used for satellite communication, and wireless networking technologies such as Wi-Fi.
A 10-cm-long thin glass rod uniformly charged to 6.00 nC and a 10-cm-long thin plastic rod uniformly charged to - 6.00 nC are placed side by side, 4.4 cm apart. What are the electric field strengths E1 to E3 at distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line connecting the midpoints of the two rods?
A. Specify the electric field strength E1
B. Specify the electric field strength E2
C. Specify the electric field strength E3
Answer:
A) E(r) = 1.3957 × 10^(5) N/C
B) E(r) = 9.8864 × 10⁴ N/C
C) E(r) = 1.13 × 10^(5) N/C
Explanation:
We are given;
q = 6 nc = 6 × 10^(-9) C
L = 10 cm = 0.1 m
d = 4.4 cm = 0.044 m
r1 = 1 cm = 0.01 m
r2 = 2 cm = 0.02 m
r3 = 3 cm = 0.03 m
Formula for the electric field strength in this question is given as;
E(r) = q/(2π(ε_o)rL) + q/(2π(ε_o)(d - r)L)
When factorized, we have;
E(r) = q/(2π(ε_o)L) × [(1/r) + (1/(d - r))]
Plugging in the relevant values for q/(2π(ε_o)L)
We know that (ε_o) has a constant value of 8.854 × 10^(−12) C²/N².m
Thus; q/(2π(ε_o)L) = (6 × 10^(-9))/(2π(8.854 × 10^(−12)0.1) = 1078.53
Thus;
E(r) = 1078.52 [1/r + 1/(d - r)]
A) E1 is at r = 1 cm = 0.01m
Thus;
E(r) = 1078.52 (1/0.01 + (1/(0.044 - 0.01))
E(r) = 1.3957 × 10^(5) N/C
B) E2 is at r = 2 cm = 0.02 m
Thus;
E(r) = 1078.52 (1/0.02 + (1/(0.044 - 0.02))
E(r) = 9.8864 × 10⁴ N/C
C) E2 is at r = 3 cm = 0.03 m
Thus;
E(r) = 1078.52 (1/0.03 + (1/(0.044 - 0.03))
E(r) = 1.13 × 10^(5) N/C
Suppose you drop paperclips into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating paper clips, explain whether the momentum and kinetic energy increase, decrease, or stay the same.
Answer:
Stay the same
Explanation:
Since, friction is negligible:
Initial Momentum = Final Momentum
Initial KE = Final KE
m1 * v1 = m2 * v2
When m increases v decreases.
The momentum and kinetic energy remain the same if you drop paper clips into an open cart rolling along a straight horizontal track with negligible friction.
What is friction?Between two surfaces that are sliding or attempting to slide over one another, there is a force called friction. For instance, friction makes it challenging to push a book down the floor. Friction always moves an object in a direction that is counter to the direction that it is traveling or attempting to move.
Given:
The paperclips into an open cart rolling along a straight horizontal track with negligible friction,
Calculate the momentum, Since friction is negligible,
Initial Momentum = Final Momentum
Initial Kinetic Energy = Final Kinetic Energy
m₁ × v₁ = m₁ × v₂
When m increases, v decreases,
Thus, momentum will remain the same.
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The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the current in the filament
Answer:
energy carried by the current is given by the pointyng vector
Explanation:
The current is defined by
i = dQ / dt
this is the number of charges per unit area over time.
The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.
But the energy carried by the current is given by the pointyng vector of the electromagnetic wave
S = 1 / μ₀ EX B
It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement
A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 36.4 m/s , blows a horn whose frequency is 123 Hz .what is its speed?
Answer:
51. 7m/s
Explanation:
Take speed of sound in air = 340 m/s
fp = fs (V + Vp)/(V + Vs)
128 = 123 (340 + Vp)/(340 + 36.4)
Vp = 51.7m/s
Explanation:
Consider two parallel plate capacitors. The plates on Capacitor B have half the area as the plates on Capacitor A, and the plates in Capacitor B are separated by twice the separation of the plates of Capacitor A. If Capacitor A has a capacitance of CA-17.8nF, what is the capacitance of Capacitor? .
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
CB = 4.45 x 10⁻⁹ F = 4.45 nF
At a rock concert, a dB meter registered 131 dB when placed 2.6 m in front of a loudspeaker on the stage. The intensity of the reference level required to determine the sound level is 1.0×10−12W/m2.
a) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?
b) How far away would the sound level be 86 dB?
Answer:
Explanation:
A) 131 dB = 10*log(I / 1e-12W/m²)
where I is the intensity at 2.6 m away.
13.1 = log(I / 1e-12W/m²
1.25e13= I / 1e-12W/m²
I = 1.25 x10^1W/m²
power = intensity * area
P = I * A = 12.5W/m² * 4π(2.6m)² =1061 W ◄
B) 86 dB = 10*log(I / 1e-12W/m²)
8.6 = log(I / 1e-12W/m²)
3.98e8 = I / 1e-12W/m²
I = 3.98e-4 W/m²
area A = P / I = 1061W / 3.98e-4W/m² = 2.66e6 m²
A = 4πr²
2.66e6 m² = 4πr²
r = 14.5m ◄
By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 dB and 20 dB respectively
Answer:
The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
Explanation:
The intensity of sound is given by;
[tex]I(dB) = 10Log(\frac{I}{I_o} )[/tex]
where;
I is the intensity of the sound
I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²
The intensity of sound at a rock concert
[tex]120 = 10Log(\frac{I}{1*10^{-12}} )\\\\12 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{12}\\\\I = 1*10^{-12} *10^{12}\\\\I = 1*10^0\\\\I =1 \ W/m^2[/tex]
The intensity of sound of a whisper
[tex]20 = 10Log(\frac{I}{1*10^{-12}} )\\\\2 = Log(\frac{I}{1*10^{-12}} )\\\\\frac{I}{1*10^{-12}} = 10^{2}\\\\I = 1*10^{-12} *10^{2}\\\\I = 1*10^{-10}\\\\I =10^{-10} \ W/m^2[/tex]
Thus, the intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
A bicycle has wheels that are 60 cm in diameter. What is the angular speed of these wheels when it is moving at 4.0 m/s
Answer:
13.33 rad/s
Explanation:
Applying,
v = ωr......................... Equation 1
Where v = linear speed, ω = angular speed and r = radius.
Note that,
r = d/2................. Equation 2
Where d = diameter of the wheel.
Substitute equation 2 into equation 1
v = ωd/2............... Equation 3
make ω the subject of the equation
ω = 2v/d................ Equation 4
Given: v = 4 m/s, d = 60 cm = 0.6 m
Substitute these values into equation 4
ω = 2(4)/0.6
ω = 13.33 rad/s
Suppose that a sound source is emitting waves uniformly in all directions. If you move to a point twice as far away from the source, the frequency of the sound will be:________.
a. one-fourth as great.
b. half as great.
c. twice as great.
d. unchanged.
Answer:
d. unchanged.
Explanation:
The frequency of a wave is dependent on the speed of the wave and the wavelength of the wave. The frequency is characteristic for a wave, and does not change with distance. This is unlike the amplitude which determines the intensity, which decreases with distance.
In a wave, the velocity of propagation of a wave is the product of its wavelength and its frequency. The speed of sound does not change with distance, except when entering from one medium to another, and we can see from
v = fλ
that the frequency is tied to the wave, and does not change throughout the waveform.
where v is the speed of the sound wave
f is the frequency
λ is the wavelength of the sound wave.
Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.
Answer:
81 N
7.1 m/s²
Explanation:
Draw a free body diagram of the sphere. There are two forces:
Weight force mg pulling straight down,
and tension force T pulling up along the rope.
Sum of forces in the centripetal direction:
∑F = ma
T − mg sin 45° = m v² / r
T = m (g sin 45° + v² / r)
T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)
T = 81 N
Sum of forces in the tangential direction:
mg cos 45° = ma
a = g cos 45°
a = (10 m/s²) cos 45°
a = 7.1 m/s²
Which notation is better to use? (Choose between 4,000,000,000,000,000 m and 4.0 × 1015 m)
Answer:
4 x 10¹⁵
Explanation:
A circuit contains a single 220 pF capacitor hooked across a battery. It is desired to store three times as much energy in a combination of two capacitors by adding a single capacitor to this one.
How would you hook it up?
The capacitor is connected in series to the original capacitor
or
The capacitor is connected in parallel to the original capacitor
I believe its parallel
but now What would its value be?
Answer
The capacitor should be connected in parallel as parallel connection gives the arithmetic sum of capacitance which will give a corresponding sum of energy while capacitors in series gives the sum of the reciprocal if the individual capacitance
An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m long. The speed of sound in the room is 330 m/s. Which of the following sets of frequencies consists of frequencies which can be produced by both pipes?
a. 110Hz,220Hz, 330 Hz
b. 220Hz 440Hz 66 Hz
c. 110Hz, 330Hz, 550Hz
d. 330 Hz, 550Hz, 440Hz
e. 660Hz, 1100Hz, 220Hz
Answer:
A. 110Hz,220Hz, 330 Hz
Explanation:
for organ open at open both ends;
the length of the organ for the fundamental frequency, L = A---->N + N----->A
A---->N = λ /4 and N----->A = λ /4
L = λ /4 + λ /4 = λ /2
[tex]L = \frac{\lambda}{2} \\\\\lambda = 2L[/tex]
λ = 2 x 1.5m = 3.0 m
Wave equation is given by;
V = Fλ
Where;
V is the speed of sound
F is the frequency of the wave
F = V/ λ
F₀ = V / 2L
Where;
F₀ is the fundamental frequency
F₀ = 330 / 2(1.5)
F₀ = 330 / 3
F₀ = 110 Hz
the length of the organ for the first overtone, L = A---->N + N----->A + A----->N + N----->A
L = 4λ /4
L = λ
λ = 1.5 m
F₁ = 330 / 1.5
F₁ = 220 Hz
Thus, F₁ = 2F₀
For open organ at one end
the length of the organ for the fundamental frequency, L = N------A
L = λ /4
λ = 4L
F₀ = V/4L
F₀ = 330 / (4 x 0.75)
F₀ = 110 Hz
the length of the organ for the first overtone, L = N-----N + N-----A
L = λ/2 + λ / 4
L = 3λ /4
F₁ = 3F₀
F₁ = 3 x 110
F₁ = 330 Hz
Thus the fundamental frequency for both organs is 110 Hz,
The first overtone for the organ open at both ends is 220 Hz
The first overtone for the organ open at one end is 330 Hz
The correct option is "A. 110Hz,220Hz, 330 Hz"
The correct option is option (A)
the frequencies produced by the pipes are (A) 110Hz,220Hz, 330 Hz
Frequencies and overtones:(I) For an organ pipe open at open both ends the frequency of different modes is given by:
F = nv/2L
where
F is the frequency
L is the length of the organ pipe
v is the speed of the wave
and, n is the mode of frequency
the fundamental frequency corresponds to n = 1, given by:
F₀ = v/2L
F₀ = 330 / 2(1.5)
F₀ = 330 / 3
F₀ = 110 Hz
The first overtone corresponds to n = 2, the second overtone corresponds to n = 3, and so on...
F₁ =2v/2L
F₁ = 330 / 1.5
F₁ = 220 Hz
Thus, F₁ = 2F₀
The difference between successive overtones is F₀
(II) For an organ pipe open at one end the frequency of different modes is given by:
F = nv/4L
where
F is the frequency
L is the length of the organ pipe
v is the speed of the wave
and, n is the mode of frequency
the fundamental frequency corresponds to n = 1, given by:
F₀ = V/4L
F₀ = 330 / (4 x 0.75)
F₀ = 110 Hz
For an organ pipe open at one end, only those overtones are present which correspond to odd n, that is n = 3,5,...so:
F₁ = 3F₀
F₁ = 3 x 110
F₁ = 330 Hz
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A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of the magnetic field inside the solenoid
Answer:
1.37 ×10^-3 T
Explanation:
From;
B= μnI
μ = 4π x 10-7 N/A2
n= number of turns /length of wire = 1700/0.75 = 2266.67
I= 0.48 A
Hence;
B= 4π x 10^-7 × 2266.67 ×0.48
B= 1.37 ×10^-3 T
The concentration of sodium and potassium ions in the blood and body fluid is regulated by :
Answer:
Kidney
Explanation:
One of the main function of the kidney is to maintain the homeostasis of sodium and potassium ions in the blood and body.
Aldosterone is a key steroid hormone that balances sodium and potassium ions in the blood and body fluid. Potassium and sodium ions generate electric impulse in the body which helps to perform different activities such as muscles flexing.
Kidney function for reabsorption and secretion, in which reabsorption of Na is done nd balances the sodium and potassium in the blood and body.
A disk between vertebrae in the spine is subjected to a shearing force of 375 N. Find its shear deformation, taking it to have a shear modulus of 1.60×109 N/m2. The disk is equivalent to a solid cylinder 0.750 cm high and 6.50 cm in diameter.
Answer:
5.29×10^-7
Explanation:
shear stress τ = F/ A
shear deformation δ = (VL)/ (AG)
= (τL)/ G
V=shear force
L=height of disk=6.50×10^-2
A=cross sectional area
G= shear modulus= (1.60x10^9N/m^2)
A=πd^2/4
Then substitute the values we have
4×(375N)(0.00750m)
________________ = δ
(π*0.00650^2)(1.60x10^9N/m^2)
= 5.29×10^-7
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
most likely final temperature of the mixture?
O A. 80°C
OB. 10-C
OC. 20°C
O D. 60°C
Answer:
Option (c) : 20°C
Explanation:
[tex]t(final) = \frac{w1 \times t1 + w2 \times t2}{w1 + w2} [/tex]
T(final) = 500* 10 + 100*70/600 = 20°C
A 17.0 g bullet traveling horizontally at 785 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s.
What is the maximum temperature increase that the water could have as a result of this event? (in degrees)
Answer:
The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]
The speed is [tex]v_1 = 785 \ m/s[/tex]
The mass of the water is [tex]m_w = 13.5 \ kg[/tex]
The velocity it emerged with is [tex]v_2 = 534 \ m/s[/tex]
Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then
the change in kinetic energy of the bullet = the heat gained by the water
So
The change in kinetic energy of the water is
[tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]
substituting values
[tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]
[tex]\Delta KE = 2814.1 \ J[/tex]
Now the heat gained by the water is
[tex]Q = m_w* c_w * \Delta T[/tex]
Here [tex]c_w[/tex] is the specific heat of water which has a value [tex]c_w = 4190 J/kg \cdot K[/tex]
So since [tex]\Delta KE = Q[/tex]
we have that
[tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]
[tex]\Delta T = 0.0497 \ ^oC[/tex]
