Answer: the lowest note that would be possible to blow on this brass pipe is approximately 1700 Hz
Explanation: f = c / (2L)
where c is the speed of sound in brass, which is around 340 m/s.
Changing over the length of the pipe to meters, we have:
L = 10 cm = 0.1 m
Stopping within the values, we get:
f = 340 / (2 x 0.1) = 1700 Hz
Nuclear fusion in a star produces elements up to, but no heavier than, ________.
A. iron
B. lead
C. carbon
D. nitrogen
Answer:
A. iron.
In the process of nuclear fusion, lighter elements are fused together to form heavier elements. This process releases energy and is what powers the star. However, the fusion of elements heavier than iron requires energy, rather than releasing it. Therefore, once a star has produced iron in its core, it is no longer able to sustain nuclear fusion and will eventually undergo a supernova explosion.
A 21 g block of ice is cooled to −77 ◦C. It is added to 593 g of water in an 92 g copper calorimeter at a temperature of 28◦C. Find the final temperature. The specific heat of copper is 387 J/kg · ◦C and of ice is 2090 J/kg · ◦C . The latent heat of fusion of water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg · ◦C . Answer in units of ◦C. ( 2 significant digits pls)
Answer:m 1= 36 g=0.036 kg - the mass of ice T1= −77 ◦C. - temperature of ice m2=589 g=0.589 kg - mass of water C1=2090 J/kg · ◦C - specific heat of ice λ = 3.33 × 10^5 J/kg - latent heat of fusion of water T2=26◦C. - temperature of water C2= 4186 J/kg · ◦C . - - specific heat of water m3=74 g=0.074kg - mass of copper T3=26◦C. - temperature of copper C3=387 J/kg ·◦C - specific heat of copper is and of ice is T - ? - final temperature 1 1 ( 0 − 1 ) − ℎ 0 1 − 1 2 ( − 0 ) − ℎ 2 2 ( − 2 ) − 3 3 ( − 3 ) − m 1 C 1 (0−T 1 )−ice heating to 0 o C m1λ−ice melting m 1 C 2 (T−0)−melted water heating m 2 C 2 (T−T 2 )−water cooling m 3 C 3 (T−T 3 )−copper cooling 1 1 ( 0 − 1 ) + 1 + 1 2 ( − 0 ) + 2 2 ( − 2 ) + 3 3 ( − 3 ) = 0 m 1 C 1 (0−T 1 )+m1λ+m 1 C 2 (T−0)+m 2 C 2 (T−T 2 )+m 3 C 3 (T−T 3 )=0 0.036 ⋅ 2090 ⋅ 77 + 0.036 ⋅ 3.33 ⋅ 1 0 5 + 0.036 ⋅ 4186 ⋅ + 0.589 ⋅ 2090 ⋅ ( − 26 ) + 0.074 ⋅ 387 ⋅ ( − 26 ) = 0 0.036⋅2090⋅77+0.036⋅3.33⋅10 5 +0.036⋅4186⋅T+0.589⋅2090⋅(T−26)+0.074⋅387⋅(T−26)=0 1410.344 ⋅ = 14969.368 = 10.6 1 1410.344⋅T=14969.368 T=10.61 o C Answer: = 10.6 1 T=10.61 o C
Explanation:
Macmillan Learning
When a massive star reaches the end of its life, it is possible for a supernova to occur. This may result in the formation of a very
small, but very dense, neutron star, the density of which is about the same as a neutron. A neutron has a mass of 1.7 x 10-27 kg
and an approximate radius of 1.2 x 10-15 m. The mass of the sun is 2.0 x 1030 kg.
Okay, let's break this down step-by-step:
1) A neutron has a mass of 1.7 x 10-27 kg and an approximate radius of 1.2 x 10-15 m.
So we know the mass and radius of a single neutron.
2) The mass of the sun is 2.0 x 1030 kg.
So we know the total mass of the sun, which is much greater than a neutron.
3) When a massive star reaches the end of its life, it can explode as a supernova.
This supernova can form a neutron star.
4) A neutron star has a density about the same as a neutron.
So we can conclude that a neutron star has a density of:
Density = Mass / Volume
= (1.7 x 10-27 kg) / (4/3 * pi * (1.2 x 10-15 m)3)
= 1.6 x 1017 kg/m3
5) A neutron star forms from the core collapse of a massive star during supernova.
So it has a mass on the order of 1-2 times that of the sun (2 x 1030 kg),
but compressed into a sphere only about 10-20 km in radius.
So its mass would be huge, around 2 x 1030 kg, but confined to a tiny volume,
giving it an immense density, around 1.6 x 1017 kg/m3, the same as a neutron.
Does this help explain the concepts and walk through the calculations? Let me know if you have any other questions!
1. Which characteristic of a substance is constant?
a phase
b mass
Ospecific heat
d kinetic energy
A charge of −0.0004 C is a distance of 3 meters from a charge of 0.0003 C. What is the magnitude of the force between them?
The magnitude of the force is 3.6 x 10^-6 N, rounded to two significant figures.
The magnitude of the force between two point charges can be calculated using Coulomb's law:
F = k * (q1 * q2) /[tex]r^2[/tex]
where F is the magnitude of the force in Newtons (N), k is Coulomb's constant (9 x 10^9 N*m^2/C^2), q1 and q2 are the magnitudes of the charges in Coulombs (C), and r is the distance between the charges in meters (m).
Plugging in the given values, we get:
F =[tex](9 * 10^{9} N*m^{2}/C^2) * (-0.0004\ C) * (0.0003 C) / (3 m)^{2}[/tex]
Simplifying the expression, we get:
F = [tex]-3.6 * 10^{-6} N[/tex]
Note that the negative sign in the result indicates that the force is attractive. The magnitude of the force is [tex]3.6 * 10^{-6} N[/tex], rounded to two significant figures.
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11.1. Is ammeter A1 connected in series or in parallel in this diagram?
Ammeter is always connected in series with the circuit in which the current is to be measured.
An ammeter is connected, to gauge the current flowing through a part or circuit. Since current in a series connection stays constant and an ammeter's resistance is relatively low, the current being measured is unaffected. For the purpose of measuring current, an ammeter is linked in series.
A shunt running parallel to the metre carries the majority of the current at high current values, hence an ammeter can measure a wide range of current values. An ammeter is represented by a circle with a capital A inside it on circuit diagrams.
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Which of the following was one of Hubble's conclusions due to red shift?
A. There are millions of galaxies in the universe, not just ours.
B. The universe is contracting.
C. Background radiation shows the Big Bang occurring.
D. The universe is expanding.
D. The universe is expanding. Hubble's observation of red shift in the light from distant galaxies led him to conclude that these galaxies were moving away from us and that the universe was expanding.
What was Hubble's observation?Hubble's most famous observation was his discovery of the relationship between the redshift of light from distant galaxies and their distance from Earth. He observed that the light from distant galaxies was shifted toward longer, redder wavelengths, which indicated that the galaxies were moving away from us. By analyzing the degree of redshift, Hubble was able to calculate the distance of these galaxies from Earth and found that they were much farther away than previously thought. This led him to conclude that the universe was expanding and laid the foundation for the Big Bang theory.
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An RLC series circuit has a 1.00 kn resistor, a 155 mH inductor, and a 25.0 nF capacitor.
(a) Find the circuit's impedance (in ) at 485 Hz.
(b) Find the circuit's impedante (in 2) at 7.50 kHz.
(c) If the voltage source has V
rms
=
___mA (at 485 Hz)
___mA (at 7.50 kHz)
(d) What is the resonant frequency (in kHz) of the circuit?
___kHz
408 V, what is Irms (in mA) at each frequency?
(e) What is Irms (in mA) at resonance?
___mA
(a) The impedance of the circuit at 485 Hz is approximately 1253.53 Ω.
(b) The impedance of the circuit at 7.50 kHz is approximately 1256.04 Ω.
(c) The current (Irms) at 485 Hz is approximately 326 mA, and the current at 7.50 kHz is approximately 325 mA.
(d) The resonant frequency of the circuit is approximately 25.74 kHz.
(e) The current (Irms) at resonance is approximately 408 mA.
What is the impedance of the circuit?a) The impedance (Z) of an RLC series circuit can be calculated using the formula:
Z = √(R^2 + (ωL - 1/ωC)^2)
where;
R is the resistance,L is the inductance, C is the capacitance, and ω is the angular frequency in radians per second.Given:
R = 1.00 kΩ = 1000 Ω
L = 155 mH = 0.155 H
C = 25.0 nF = 25.0 × 10^(-9) F
f = 485 Hz
First, we need to convert the frequency from Hz to radians per second:
ω = 2πf
Substituting the given values into the impedance formula:
Z = √((1000)^2 + ((2π × 485) × 0.155 - 1/(2π × 485 × 25.0 × 10^(-9)))^2)
Calculating Z:
Z = √((1000)^2 + (481.663)^2) ≈ 1253.53 Ω
(b) Given:
f = 7.50 kHz = 7500 Hz
ω = 2πf
Substituting the given values into the impedance formula:
Z = √((1000)^2 + ((2π × 7500) × 0.155 - 1/(2π × 7500 × 25.0 × 10^(-9)))^2)
Calculating Z:
Z = √((1000)^2 + (568.28)^2) ≈ 1256.04 Ω
So, the impedance of the circuit at 7.50 kHz is approximately 1256.04 Ω.
(c) Given:
Vrms = 408 V
At 485 Hz:
Irms = Vrms / Z (using the impedance calculated in part (a))
Irms = 408 / 1253.53 ≈ 0.326 A = 326 mA
At 7.50 kHz:
Irms = Vrms / Z (using the impedance calculated in part (b))
Irms = 408 / 1256.04 ≈ 0.325 A = 325 mA
(d) The resonant frequency of an RLC circuit can be calculated using the formula:
f_resonant = 1 / (2π√(LC))
Given:
L = 155 mH = 0.155 H
C = 25.0 nF = 25.0 × 10^(-9) F
Substituting the given values into the resonant frequency formula:
f_resonant = 1 / (2π√(0.155 × 25.0 × 10^(-9)))
Calculating f_resonant:
f_resonant ≈ 25.74 kHz
(e) At resonance, the impedance of the inductor (ωL) and the capacitor (1/(ωC)) cancel each other out, resulting in the minimum impedance of the circuit.
Therefore, at resonance, the impedance (Z) of the circuit is equal to the resistance (R) only.
Given:
R = 1.00 kΩ = 1000 Ω
Vrms = 408 V
Using Ohm's law, we can calculate the current (Irms) at resonance:
Irms = Vrms / R
Irms = 408 / 1000 = 0.408 A = 408 mA
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(a) Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) 401.15
(b)How much work (in J) does she do if her center of mass rises 0.260 m?
(c) What is her useful power output (in W) if she does 30 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)
The force is 400.2 N
The work done is 120 J
The power is 48W
What is Force?Force is a physical concept that describes the influence that one object has on another object, causing it to accelerate or deform. Force can be defined as any influence that changes the motion of an object, such as a push or a pull.
How to solve:
under equilibrium condition
F * 1.45 m =68 kg * 9.81 m/s^2 *0.87 m
F =400.2 N
b)
work done = m*g*h =68 kg*9.81 m/s^2*0.180 m =120.0744 J =120 J
c)
power =120.0744 J *(24 /60 s) =48.02976 W = 48 W
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A magnifying glass has a converging lens of focal length of 10.3 cm. At what distance from a nickel should you hold this lens to get an image with a magnification of +1.53?
The magnifying glass should be held 16.4 cm away from the nickel to obtain an image with a magnification of +1.53.
Using the thin lens equation, 1/f = 1/o + 1/i, where f is the focal length, o is the object distance, and i is the image distance, and the magnification equation, M = -i/o, where M is the magnification, we can solve for the object distance.
First, solve for the image distance i:
M = -i/o
1.53 = -i/o
i = -1.53o
Then, substitute i into the thin lens equation:
1/0.103 = 1/o + 1/(-1.53o)
Solving for o, we get:
o = 16.4 cm
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NO USE OF AI BOTS TO answer question pls answer thanks
Answer:
19.2 centimeters or 0.192 meters. Both are the same.
Explanation:
T = 2π√(L/g)
T = 60.0 s / 14 ≈ 4.29 s
L = (gT²) / (4π²) ≈ 0.384 m
h + 0.384 m = height above the field
H = h + 0.192 m
H = h
h ≈ 0.192 m or 19.2 cm.