Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Final answer:
To determine the heat transfer per kg in the polytropic expansion of carbon dioxide gas in a piston-cylinder system, we can use the first law of thermodynamics. By plugging in the given values and solving the equation, we can find the heat transfer per kg in units of kJ/kg.
Explanation:
In this problem, we are given the initial temperature and pressure of a carbon dioxide gas in a piston-cylinder system. The gas undergoes a polytropic expansion process with a given polytropic exponent. The final temperature is also given.
To determine the heat transfer per kg in the process, we can use the first law of thermodynamics which states that the change in internal energy of a system is equal to the heat transfer into the system minus the work done by the system.
Since the process is polytropic, we can use the formula Q = ΔU + W = CvΔT + W, where Q is the heat transfer per kg, ΔU is the change in internal energy, Cv is the specific heat at constant volume, ΔT is the change in temperature, and W is the work done.
Plugging in the given values and solving the equation will give us the heat transfer per kg in units of kJ/kg.
Finally, consider the expression (6.67×10^−11)(5.97×10^24)/(6.38×10^6)^2. Determine the values of a and k when the value of this expression is written in scientific notation. Enter a and k, separated by commas.
Answer:
[tex]x=9.78\times 10^0[/tex]
Explanation:
In this case, we need to find the value of expression :
[tex]x=\dfrac{(6.67\times 10^{-11})\times (5.97\times 10^{24})}{(6.38\times 10^6)^2}[/tex]
On solving, we get the value of given expression as :
x = 9.7827
In scientific notation, we get the value of x as :
[tex]x=a\times 10^k[/tex]
[tex]x=9.78\times 10^0[/tex]
a = 9.78
k = 0
Hence, this is the required solution.
To express the given expression in scientific notation, we find a = 9.78 and k = -1 after evaluating the expression using the laws of exponentiation and division.
Explanation:The student's question involves evaluating an expression using scientific notation and expressing the result in proper scientific notation.
To solve (6.67×10−11)(5.97×1024) / (6.38×106)2, we need to use the laws of exponentiation and multiplication. Simplify the expression within the numerator and denominator separately before dividing them.
Numerator: (6.67×10−11)(5.97×1024) = 39.8309×1013
Denominator: (6.38×106)2 = 40.7044×1012
Divide the simplified numerator by the simplified denominator:
39.8309×1013 / 40.7044×1012 = 0.978×101
To express this in proper scientific notation, rewrite 0.978×101 as 9.78×10−0.
Therefore, the values of a and k when the value of this expression is written in scientific notation are a = 9.78 and k = −0.
Given two vectors A⃗ =4.00i^+7.00j^ and B⃗ =5.00i^−2.00j^ , find the vector product A⃗ ×B⃗ (expressed in unit vectors).
Answer:
[tex]-43\hat{k}[/tex]
Explanation:
given,
[tex]\vec{A} = 4 \hat{i} + 7 \hat{j}[/tex]
[tex]\vec{B} = 5 \hat{i} - 2 \hat{j}[/tex]
vector product [tex] \vec{A} \times \vec{B} = ?[/tex]
[tex]\vec{A} \times \vec{B}[/tex] = [tex]\begin{bmatrix}i & j & k\\ 4 & 7 &0 \\ 5 & -2 & 0\end{bmatrix}[/tex]
now, expanding the vector
[tex]\vec{A} \times \vec{B}= \hat{k}(-2\times 4 - 7\times 5)[/tex]
[tex]\vec{A} \times \vec{B}= -43\hat{k}[/tex]
the vector product is equal to [tex]-43\hat{k}[/tex]
At what temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph ?
Answer:
T= 89.25 K
Explanation:
Given that
V= 590 mph
We know that
1 mph = 0.44 m/s
That is why ,V= 263.75 m/s
We know that speed of the gas molecule is given as
[tex]V=\sqrt{\dfrac{3RT}{M}}[/tex]
R= 8.314 J/mol.k
M= 32 g/mol = 0.032 kg/mol
T=Temperature in Kelvin unit
[tex]V^2=\dfrac{3RT}{M}[/tex]
[tex]T=\dfrac{V^2\times M}{3R}[/tex]
[tex]T=\dfrac{263.76^2\times 0.032}{3\times 8.314}\ K[/tex]
T= 89.25 K
Therefore the average temperature ,T = 89.25 K
Answer:
temperature does the average speed of an oxygen molecule equal that of an airplane moving at 590 mph = 89.24 K
Explanation:
Average speed of oxygen molecule is given by
[tex]v= \sqrt{\frac{3RT}{M} }[/tex]
[tex]590\times0.44704 = 263.75[/tex] m/s
R= 8.314 J/mol K = universal gas constant
M= molecular weight of oxygen = 32 g/mol =0.032 Kg/mol
now plugging these values to find T we get
[tex]263.75=\sqrt{\frac{3(8.314)(T)}{0.032}}[/tex]
solving the above equation we get
T= 89.24 K
How much stronger is the gravitational pull of the Sun on Earth, at 1 AU, than it is on Saturn at 10 AU?
The density of liquid oxygen at its boiling point is 1.14 kg/Lkg/L , and its heat of vaporization is 213 kJ/kgkJ/kg . How much energy in joules would be absorbed by 2.0 LL of liquid oxygen as it vaporized? Express your answer to two significant figures and include the appropriate units.
Answer:
heat absorbed = 4.9 × [tex]10^{5}[/tex] J
Explanation:
given data
density of liquid oxygen = 1.14 kg/L
volume = 2 L
heat of vaporization = 213 kJ/kg
solution
first we get here mass of liquid that is
mass of liquid = density × volume ......................1
mass of liquid = 1.14 × 2
mass of liquid = 2.28 kg
so here we get now heat absorbed that is
heat absorbed = mass × heat of vaporization
heat absorbed = 2.28 kg × 213 kJ/kg
heat absorbed = 485.640 kJ
heat absorbed = 4.9 × [tex]10^{5}[/tex] J
To find the energy absorbed by 2.0 L of liquid oxygen as it vaporizes, you first convert the volume to mass using the given density and then multiply by the heat of vaporization. The resulting energy is approximately 4.86 x 10^5 J or 486 kJ.
Explanation:The energy absorbed by a volume of liquid oxygen as it vaporizes can be calculated using the formula Q = mLv. In this equation, 'm' represents mass, 'Lv' represents the heat of vaporization, and 'Q' represents the total energy absorbed.
Firstly, convert the volume of liquid oxygen to mass. The density of liquid oxygen at its boiling point is 1.14 kg/L, so the mass of 2.0 L of liquid oxygen would be (1.14 kg/L) * (2.0 L) = 2.28 kg.
Then, use the heat of vaporization and the calculated mass to find the total energy. The heat of vaporization of oxygen is 213 kJ/kg, so Q = (2.28 kg) * (213 kJ/kg) = 485.64 kJ. This needs to be expressed in joules by multiplying by 10^3, resulting in 485640 J. Therefore, the energy absorbed by 2.0 L of liquid oxygen as it vaporizes is approximately 4.86 x 10^5 J.
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An exploration submarine should be able to descend 1200 m down in the ocean. If the ocean density is 1020 kg/m3, what is the maximum pressure on the submarine hull?
Answer:
11995200 N/m²
Explanation:
Pressure: The is the ratio of force to the surface area in contact. The S.I unit of pressure is N/m².
Generally pressure in fluid can be expressed as
P = ρgh......................... Equation 1
Where P = maximum pressure on the submarine hull, ρ = Density of ocean, h = depth of ocean, g = acceleration due to gravity.
Given: h = 1200 m, ρ = 1020 kg/m³
Constant: g = 9.8 m/s²
Substitute into equation 1
P = 1200(1020)(9.8)
P = 11995200 N/m²
Hence the maximum pressure on the submarine hull = 11995200 N/m²
Final answer:
The maximum pressure on a submarine hull that descends 1200 m in the ocean, with ocean density of 1020 kg/m³, is calculated using the formula P = [tex]P_{o}[/tex] + ρgh. With an atmospheric pressure of 101325 Pa, the final pressure would be 12116485 Pa at that depth.
Explanation:
To calculate the maximum pressure on a submarine hull that can descend 1200 m in the ocean, we use the formula for the pressure exerted by a fluid at a certain depth. The formula is P = [tex]P_{o}[/tex] + ρgh, where P is the pressure at depth, [tex]P_{o}[/tex] is the atmospheric pressure on the surface, ρ (rho) is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
Atmospheric pressure [tex]P_{o}[/tex] is approximately 101325 Pa (or 1 atm). The density of sea water is given as 1020 kg/m³ and the depth is 1200 m. Taking g as 9.8 m/s², the standard acceleration due to gravity, we can substitute the values into the formula:
P = 101325 Pa + (1020 kg/m³)×(9.8 m/s²)×(1200 m)
P = 101325 Pa + 12003360 Pa
P = 12116485 Pa
Therefore, the maximum pressure on the submarine hull at a depth of 1200 m would be 12116485 Pascal (Pa).
How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr?
The reaction produces -4.95 kJ of heat when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr.
The equation of the reaction is;
CO(g) + H2(g) -------> CH2O(g)
The heat of reaction is obtained from;
Enthalpy of products - Enthalpy of reactants = (-116kJ/mol) - (-110.5 kJ/mol)
= -5.5 kJ/mol
Number of moles of CO is obtained from;
PV = nRT
P = 112 kPa or 1.1 atm
T = 85°C + 273 = 358 K
n = ?
R = 0.082 atmLK-1mol-1
V = 15.0 L
n = PV/RT
= 1.1 atm × 15.0 L/ 0.082 atmLK-1mol-1 × 358 K
= 0.56 moles
Number of moles of H2
n = PV/RT
P= 744 torr or 0.98 atm
V = 14.4 L
T = 75°C + 273 = 348 K
n = 0.98 atm × 14.4 L/0.082 atmLK-1mol-1 × 348 K
n = 0.49 moles
We can see that H2 is the limiting reactant here hence 0.49 moles of formaldehyde is produced.
If 1 mole of formaldehyde produces -5.5 kJ of heat
0.49 moles of formaldehyde produces -5.5 kJ × 0.49 moles / 1 mole
= -4.95 kJ of heat
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Why do atoms absorb and reemit radiation at characteristic frequencies?
To answer this question it is necessary to use the Bohr Model as a theoretical reference. According to this model, the energy levels of an atom are divided by discrete and characteristic energies. When there is the emission or absorption of a photon with a characteristic energy there is a transition of another energy level, which is equal to the level of atomic energy. Since the energy of a photo is directly proportional to its frequency, the emitted or absorbed photons have characteristic frequencies to the difference in energy between atomic energy levels.
The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequency ωωomega of these oscillations? Use the approximation d≪ad≪a to simplify your calculation; that is, assume that d2+a2≈a2d2+a2≈a2. Express your answer in terms of given charges, dimensions, and constants.
Answer:
[tex]\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }[/tex]
Explanation:
Additional information:
The ball has charge [tex]-q_0[/tex], and the ring has positive charge [tex]+Q[/tex] distributed uniformly along its circumference.
The electric field at distance [tex]z[/tex] along the z-axis due to the charged ring is
[tex]E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.[/tex]
Therefore, the force on the ball with charge [tex]-q_0[/tex] is
[tex]F=-q_oE_z[/tex]
[tex]F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}[/tex]
and according to Newton's second law
[tex]F=ma=m\dfrac{d^2z}{dz^2}[/tex]
substituting [tex]F[/tex] we get:
[tex]- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}[/tex]
rearranging we get:
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0[/tex]
Now we use the approximation that
[tex]z^2+a^2\approx a^2[/tex] (we use this approximation instead of the original [tex]d^2+a^2\approx a^2[/tex] since [tex]z<d[/tex], our assumption still holds )
and get
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0[/tex]
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0[/tex]
Now the last equation looks like a Simple Harmonic Equation
[tex]m\dfrac{d^2z}{dz^2}+kz=0[/tex]
where
[tex]\omega=\sqrt{ \dfrac{k}{m} }[/tex]
is the frequency of oscillation. Applying this to our equation we get:
[tex]m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}[/tex]
[tex]\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}[/tex]
If the pressure inside the cylinder increases to 1.6 atm, what is the final volume, in milliliters, of the cylinder?
This is a physics problem related to Boyle's Law, which says that the volume of a gas decreases if its pressure increases, assuming constant temperature. We can't solve for the final volume specifically in this case since the initial pressure and volume aren't given, but if they were, we'd use the formula V₂ = (P₁V₁) / P₂, where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.
This question is related to the concept of Boyle's Law in physics, which states that the pressure and volume of a gas have an inverse relationship, when temperature is held constant. If the pressure of the gas inside a cylinder is increased, the volume decreases, assuming the amount of gas and the temperature remain constant.
In this case without knowing the initial pressure and volume of the cylinder, it's impossible to calculate the exact final volume after the pressure has increased to 1.6 atm. However, the base formula you'd use to find out your final volume, given you had initial values for volume (V₁) and pressure (P₁), is: P₁V₁ = P₂V₂.
In this formula, P₁ and V₁ refer to the initial pressure and volume, while P₂ and V₂ refer to the final pressure and volume. To solve for the final volume (V₂), you would rearrange the formula to be V₂ = (P₁V₁) / P₂. So, if you were given the initial pressure and volume, you could substitute those values into the formula, along with the final pressure of 1.6 atm, to find the final volume.
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In this case, the final volume is 31.1 mL.
The final volume of the helium in milliliters can be calculated using Boyle's Law, which states that the product of pressure and volume is constant for a given amount of gas at constant temperature. Initially, the pressure is 447 torr, and the volume is 86.4 mL. When the pressure increases to 1,240 torr, the final volume can be found using the formula:
P1V1 = P2V2
Solving for the final volume, V2, we get:
V2 = (P1V1) / P2
Substitute the given values:
V2 = (447 torr× 86.4 mL) / 1,240 torr
= 31.1 mL
Two construction cranes are each able to lift a maximum load of 20000 N to a height of 250 m. However, one crane can lift that load in 1 6 the time it takes the other. How much more power does the faster crane have?
Answer:
Explanation:
Given
Load [tex]W=20000\ N[/tex]
height to which load is raised [tex]h=250\ m[/tex]
Another crane take [tex]\frac{1}{6}[/tex] th time to lift the load
Energy required required to lift the Weight
[tex]E=W\times h[/tex]
[tex]E=20000\times 250[/tex]
[tex]E=5,000,000\ J[/tex]
Suppose [tex]P_1[/tex] and [tex]P_2[/tex] is the Power required to lift the weight in t and [tex]\frac{t}{6}[/tex] time
[tex]E=P_1\times t[/tex]
[tex]E=P_2\times \frac{t}{6}[/tex]
[tex]P_1\times t=P_2\times \frac{t}{6}[/tex]
thus
[tex]P_2=6P_1[/tex]
Second Crane requires 6 times more power than the slow crane
The power of the faster crane is 6 times greater than the power of the slower crane.
Explanation:To calculate the power of the cranes, we need to use the formula:
Power = Work/Time
The work done by each crane is equal to the maximum load lifted multiplied by the height lifted, so:
Work = Load x Height
Let's assume the time taken by the slower crane is t. Therefore, the time taken by the faster crane is t/6.
Now, let's calculate the power of each crane:
Power of slower crane = Work/Time = (20000 N x 250 m) / t = 5000000 Nm/t
Power of faster crane = Work/Time = (20000 N x 250 m) / (t/6) = 30000000 Nm/t
The power of the faster crane is 6 times greater than the power of the slower crane.
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The resistivity of a certain semi-metal is 10-3 Ohm-cm. Suppose we would like to prepare a silicon wafer with the same resistivity. Assuming we will use n-type dopants only, what dopant density would we choose
Complete Question
The complete question is shown on the first uploaded image
Answer:
The dopant density is ND ≈ 8.135*10¹² cm⁻³
Explanation:
The explanation is shown on the second , third ,fourth and fifth uploaded image
The guy wires AB and AC are attached to the top of the transmission tower. The tension in cable AB is 9.1 kN. Determine the required tension T in cable AC such that the net effect of the two cables is a downward force at point A. Determine the magnitude R of this downward force.
In order for the net effect at point A to be a downward force, the tension in cable AC (T) should be equal to the tension in cable AB (9.1 kN). The magnitude of the resulting downward force (R) would be the sum of the tensions in both cables, thus 2 * 9.1 kN = 18.2 kN.
Explanation:To understand this scenario, it is essential to apply the principles of equilibrium and vector sum in Physics. The tension in the wires can be considered as forces experienced by point A. According to the question, the net effect of these forces should be a downward force, implying that they should negate the opposite upward force.
To find the tension T in cable AC, it's logical to assume that the force due to this tension needs to be equal and opposite to the force exerted by the tension in wire AB, which is 9.1 kN. Therefore, T should also be 9.1 kN for the net effect at point A to be a downward force.
The magnitude R of the downward force can be determined by considering the combined effect of tensions in cables AB and AC. Since point A is in equilibrium, R will be the result of the total upward forces. Hence, R is equal to 2 times the tension in any one cable (as they are equal), which gives us R = 2 * 9.1 kN = 18.2 kN.
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For the following statements, choose the word or words inside the parentheses that serve to make a correct statement. Each statement has at least one and may have more than one correct answer. a. For a sample of an ideal gas, the product pV remains constant as long as the (temperature, pressure, volume, internal energy) is held constant. b. The internal energy of an ideal gas is a function of only the (volume, pressure, temperature). c. The Second Law of Thermodynamics states that the entropy of an isolated system always (increases, remains constant, decreases) during a spontaneous process. d. When a sample of liquid is converted reversibly to its vapor at its normal boiling point, ( q, w, p, V, T, U, H, S, G, none of these) is equal to zero for the system. e. If the liquid is permitted to vaporize isothermally and completely into a previously evacuated chamber that is just large enough to hold the vapor at 1 bar pressure, then ( q, w, U, H, S, G ) will be smaller in magnitude than for the reversible vaporizatio
Answer:
a) Temperatura, b) Temperature, c) Constant , d) None of these , e) Gibbs enthalpy and free energy (G)
Explanation:
a) the expression for ideal gases is PV = nRT
Temperature
b) The internal energy is E = K T
Temperature
c) S = ΔQ/T
In an isolated system ΔQ is zero, entropy is constant
Constant
d) all parameters change when changing status
None of these
e) Gibbs enthalpy and free energy
Consider a small frictionless puck perched at the top of a xed sphere of radius R. If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere?
Answer:
Explanation:
Let the vertical height by which it descends be h . Let it acquire velocity of v .
1/2 mv² = mgh
v² = 2gh
As it leaves the surface of sphere , reaction force of surface R = 0 , so
centripetal force = mg cosθ where θ is the angular displacement from the vertex .
mv² / r = mg cosθ
(m/r )x 2gh = mg cosθ
2h / r = cosθ
cosθ = (r-h) / r
2h / r = r-h / r
2h = r-h
3h = r
h = r / 3
Final answer:
Through conservation of energy and dynamics principles, the puck descends through a height of R/2 from the base of the sphere before losing contact, due to the gravitational force no longer providing sufficient centripetal force.
Explanation:
The question asks through what vertical height a small frictionless puck will descend before it leaves the surface of a fixed sphere of radius R, when given a tiny nudge down the sphere. Using the principles of energy conservation and dynamics, it can be determined that the puck will lose contact with the sphere when the centripetal force is no longer sufficient to provide the necessary force for circular motion, which happens at a height of R/2 from the base of the sphere. This happens because, at this point, the gravitational component acting towards the center of the sphere is exactly equal to the required centripetal force for circular motion. As a result, any further descent would mean this balance is disturbed, causing the puck to leave the surface of the sphere.
A small, charged, spherical object at the origin of a Cartesian coordinate system contains 2.60 × 10 4 more electrons than protons. What is the magnitude of the electric field it produces at the position (2.00 mm, 1.00 mm)?
Answer:
E = 7.77 N/C
Explanation:
The charge of a single electron is 1.6 x 10^{-19} C. The net charge of the object is therefore the multiplication of the number of excess electrons and the charge of a single electron:
[tex]Q = (2.6\times 10^4) \times 1.6\times 10^{-19} = 4.16 \times 10^{-15}~C[/tex]
The electric field can be found by the following formula
[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]
where 'r' can be calculated as
[tex]r = \sqrt{(2\times 10^{-3})^2 + (1\times 10^{-3})^2} = 0.0022~m\\r^2 = 4.84\times 10^{-6}[/tex]
Finally, the electric field at the position (2.00 mm, 1.00 mm) is
[tex]E = \frac{1}{4\pi(8.8\times 10^{-12})}\frac{4.16\times 10^{-15}}{4.84\times 10^{-6}} = 7.77~N/C[/tex]
The magnitude of the electric field it produces at the position is 7.5 N/C.
The given parameters:
Number of excess electron, n = 2.6 x 10⁴Position of the excess electron, x = (2.00 mm, 1.00 mm)The position of the charged object is calculated as follows;
[tex]r^2 = (2.0 \times 10^{-3})^2 + (1.0 \times 10^{-3})^2\\\\r^2 = 5\times 10^{-6} \ m^2[/tex]
The charge of the electron is calculated as follows;
[tex]Q = nq\\\\Q = 2.6 \times 10^4 \times 1.6\times 10^{-19}\\\\Q =4.16 \times 10^{-15} \ C[/tex]
The magnitude of the electric field it produces at the position is calculated as follows;
[tex]E = \frac{F}{Q}= \frac{kQ}{r^2} = \frac{9\times 10^9 \times 4.16 \times 10^{-15}}{5\times 10^{-6}} \\\\E = 7.5 \ N/C[/tex]
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George determines the mass of his evaporating dish to be 3.375 g. He adds a solid sample to the evaporating dish, and the mass of them combined is 26.719 g. What must be the mass of his solid sample
Explanation:
The given data is as follows.
Mass of evaporating dish = 3.375 g
Total mass = Mass of solid sample + evaporating dish
That is, Mass of solid sample + evaporating dish = 26.719 g
Therefore, we will calculate the mass of solid sample as follows.
Mass of solid sample = (Mass of solid sample + evaporating dish) - mass of evaporating dish
= 26.719 g – 3.375 g
= 23.344 g
Thus, we can conclude that mass of his solid sample must be 23.344 g.
The mass of the solid sample is 23.344 g.
Explanation:In order to find the mass of the solid sample, we need to subtract the mass of the evaporating dish from the combined mass of the dish and the sample. The mass of the solid sample can be calculated by subtracting 3.375 g (mass of the evaporating dish) from 26.719 g (combined mass of dish and sample).
Mass of solid sample = 26.719 g - 3.375 g = 23.344 g.
Therefore, the mass of the solid sample is 23.344 g.
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A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibrationa. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelengthe. The length of the string is equal to four times the wavelength
Answer:
d. The length of the string is equal to one-half of a wavelength
Explanation:
A stretched string of length L, fixed at both ends, is vibrating in its third harmonic. How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration
a. The length of the sting is equal to one-quarter of a wavelength.b. The length of the string is equal to the wavelength.c. The length of the string is equal to twice the wavelength.d. The length of the string is equal to one-half of a wavelength
e. The length of the string is equal to four times the wavelength
A stretched string of length L fixed at both ends is vibrating in its third harmonic H
How far from the end of the string can the blade of a screwdriver be placed against the string without disturbing the amplitude of the vibration
d. The length of the string is equal to one-half of a wavelength
There are two points during vibration , the node and the antinode
the node is the point where the amplitude is zero.
from the third harmonics, there are two nodes. The first node is half of the wavelength which is the closest to the fixed point.
for third harmonics=3/2lamda
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their getaway car, but aren't entirely sure if they can make the jump or need to head through the building. a) If one of them drops a pebble off the edge of the roof and it hits the ground two seconds later, how fast will they hit the ground if they jump? Give answers in terms of meters per second. b) How high up are they? Give answers in terms of meters. c) Is this a safe jump?
Answer:
a) They will hit the ground with a speed of 19.6 m/s.
b) They are at a height of 20 m.
c) It is not a safe jump.
Explanation:
Hi there!
a) The equations of height and velocity in function of time of a free falling body are the following:
h = h0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where:
h = height of the object at time t.
h0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).
v = velocity of the object at time t.
Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:
v = v0 + g · t (v0 = 0)
v = g · t
v = -9.8 m/s² · 2.0 s
v = -19.6 m/s
They will hit the ground with a speed of 19.6 m/s.
b)Now, we have to use the equation of height:
h = h0 + v0 · t + 1/2 · g · t²
If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m
0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²
h0 = 1/2 · 9.8 m/s² · (2.0 s)²
h0 = 20 m
They are at a height of 20 m.
c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.
If you drop a feather and a steel hammer at the same moment, they should hit the ground at the same instant. Why doesn’t this work on Earth, and why does it work on the Moon?
Answer:
Air Resistance
Explanation:
If you were to drop both items on a plant without an atmosphere, they would both hit the ground at the same time. Since a feather doesn't have much mass compared to the hammer, it takes more time for the feather to "push" itself through and overcome the opposite push from the air
Final answer:
On Earth, air resistance prevents a feather and a hammer from hitting the ground simultaneously when dropped from the same height. On the Moon, the absence of an atmosphere means no air resistance, allowing both objects to land at the same time in accordance with Galileo's principle of the universality of free fall.
Explanation:
If you drop a feather and a steel hammer at the same moment, they should hit the ground at the same time according to Galileo's principle of the universality of free fall. However, this does not occur on Earth due to the presence of air resistance. The feather experiences a significant amount of air resistance because of its shape and light weight, causing it to flutter and fall slower than the hammer.
On the Moon, where there is no atmosphere, there is no air resistance to act on the objects. When Apollo 15 astronaut David Scott conducted the experiment on the Moon, both the hammer and feather fell at the same acceleration and hit the lunar surface simultaneously. This specific demonstration was a perfect illustration of the universality of free fall in the absence of external forces besides gravity. On the Moon, the acceleration due to gravity is only 1.67 m/s², which is less than on Earth, but since it acts equally on all objects, both the feather and the hammer fell at the same rate.
How strong is the attractive force between a glass rod with a 0.700 μC charge and a silk cloth with a –0.600 μC charge, which are 12.0 cm apart, using the approximation that they act like point charges?
Answer:
[tex]F=0.26N[/tex]
Explanation:
Assuming that thet act like point charges, the attractive force is given by Coulomb's law:
[tex]F=\frac{kq_1q_2}{d^2}[/tex]
Where k is the Coulomb constant, [tex]q_1[/tex] and [tex]q_2[/tex] are the magnitudes of the point charges and d is the distance of separation between them. Thus, we replace the given values and get how strong is the attractive force between them:
[tex]F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(0.7*10^{-6}C)(-0.6*10^{-6}C)}{(12*10^{-2}m)^2}\\F=0.26N[/tex]
An ethernet cable is 3.80 m long and has a mass of 0.210 kg. A transverse pulse is produced by plucking one end of the taut cable. The pulse makes four trips down and back along the cable in 0.735 s. What is the tension in the cable?
Answer:
[tex]T=94.54N[/tex]
Explanation:
The tension in a cable is given by:
[tex]T=\mu v^2(1)[/tex]
Where [tex]\mu[/tex] is the mass density of the cable and v is the speed of the cable's pulse. These values are defined as:
[tex]\mu=\frac{m}{L}(2)\\v=\frac{d}{t}[/tex]
The pulse makes four trips down and back along the cable, so [tex]d=4(2L)[/tex]
[tex]v=\frac{8L}{t}(3)[/tex]
Replacing (2) and (3) in (1), we calculate the tension in the cable:
[tex]T=\frac{m}{L}(\frac{8L}{t})^2\\T=\frac{64mL}{t^2}\\T=\frac{64(0.21kg(3.80m))}{(0.735s)^2}\\T=94.54N[/tex]
A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , where x is in cm and t is in seconds. What is the frequency of the oscillation?
Answer:
1/2 Hz
Explanation:
A simple harmonic motion has an equation in the form of
[tex]x(t) = Acos(\omega t - \phi)[/tex]
where A is the amplitude, [tex]\omega = 2\pi f[/tex] is the angular frequency and [tex]\phi[/tex] is the initial phase.
Since our body has an equation of x = 5cos(π t + π/3) we can equate [tex]\omega = \pi[/tex] and solve for frequency f
[tex]2\pi f = \pi[/tex]
f = 1/2 Hz
0.5Hz
Explanation:The general equation of the displacement, x, of a body undergoing simple harmonic motion at a given point in time (t) is given by;
x = A cos (ωt ± ∅) --------------------------(i)
where;
A = amplitude of the wave
ω = angular velocity of the wave
∅ = phase constant of the wave
From the question;
x = 5cos(π t + π/3) -----------------------------(ii)
Comparing equations (i) and (ii), the following deductions among others can be made;
A = 5cm
ω = π
But the angular velocity (ω) of the wave is related to its frequency (f) as follows;
ω = 2 π f --------------------(iii)
Substitute the value of ω = π into equation (iii) as follows;
π = 2 π f
Divide through by π;
1 = 2f
Solve for f;
f = 1/2
f = 0.5
Frequency (f) is measured in Hz. Therefore, the frequency of the oscillation is 0.5Hz
Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?
To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.
The volume of a sphere can be expressed as
[tex]V = \frac{4}{3} \pi r^3[/tex]
Here r is the radius of the sphere and V is the volume of Sphere
Using the expression of the density we know that
[tex]\rho = \frac{m}{V} \rightarrow V = \frac{m}{\rho}[/tex]
The density is given as
[tex]\rho = (19.5g/cm^3)(\frac{10^3kg/m^3}{1g/cm^3})[/tex]
[tex]\rho = 19.5*10^3kg/m^3[/tex]
Now replacing the mass given and the actual density we have that the volume is
[tex]V = \frac{60kg}{19.5*10^3kg/m^3 }[/tex]
[tex]V = 3.0769*10^{-3} m ^3[/tex]
The radius then is,
[tex]V = \frac{4}{3} \pi r^3[/tex]
[tex]r = \sqrt[3]{\frac{3V}{4\pi}}[/tex]
Replacing,
[tex]r = \sqrt[3]{\frac{3(3.0769*10^{-3})}{4\pi}}[/tex]
The radius of a sphere made of this material that has a critical mass is 9.02 cm.
How far apart are two conducting plates that have an electric field strength of 4.5 × 103V/m between them, if their potential difference is 12.5 kV?
Answer:
Explanation:
Given
Electric Field Strength [tex]E=4.5\times 10^{3}\ V/m[/tex]
Potential Difference between Plates is given by [tex]V=12.5\ kV[/tex]
In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates
The potential difference is given by
[tex]\Delta V=Ed[/tex]
where E=Electric Field strength
d=distance between Plates
[tex]d=\frac{\Delta V}{E}[/tex]
[tex]d=\frac{12.5\times 10^3}{4.5\times 10^{3}}[/tex]
[tex]d=2.78\ m[/tex]
A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later, the bicyclist hops on his bike and accelerates at 2.11 m/s2 until he catches his friend. a. How much time does it take until he catches his friend?(b) How far has he traveled in this time? (c) What is his speed when he catches up?
Answer:
a. [tex]t=3.44s[/tex]
b. [tex]x=12.45m[/tex]
c. [tex]v_f=7.26\frac{m}{s}[/tex]
Explanation:
The bicyclist's friend moves with constant speed. So, we have:
[tex]x=vt[/tex]
Th bicyclist moves with constant acceleration and starts at rest ([tex]v_0=0[/tex]). So, we have:
[tex]x=v_0t+\frac{at^2}{2}\\x=\frac{at^2}{2}[/tex]
a. When he catches his friend, both travels the same distance, thus:
[tex]vt=\frac{at^2}{2}\\t=\frac{2v}{a}\\t=\frac{2(3.63\frac{m}{s})}{2.11\frac{m}{s^2}}\\t=3.44s[/tex]
b. We can use any of the distance equations, since both travels the same distance:
[tex]x=vt\\x=3.63\frac{m}{s}(3.44s)\\x=12.45m[/tex]
c. The bicyclist final speed is:
[tex]v_f=v_0+at\\v_f=at\\v_f=2.11\frac{m}{s^2}(3.44s)\\v_f=7.26\frac{m}{s}[/tex]
By what factor does the energy of a 1-nm X-ray photon exceed that of a 10-MHz radio photon? How many times more energy has a 1-nm gamma ray than a 10-MHz radio photon?
To solve this problem we will apply the concepts related to the relationship between energy and frequency, from the latter we will obtain similar expressions that relate to the wavelength to find the two energy states between the given values. Finally we will make the comparative radius between the two. The relation between energy and frequency is given as,
[tex]E = hf[/tex]
Here,
E = Energy
h = Planck's constant
The relation between the speed of the electromagnetic waves (c), frequency (f) and wavelength ([tex]\lambda[/tex] ) is,
[tex]c = f\lambda[/tex]
Rearrange the above equation for frequency f as follows
[tex]f = \frac{c}{\lambda}[/tex]
Substitute,
[tex]E = h\frac{c}{\lambda}[/tex]
The wavelength x-ray or gamma ray photon is
[tex]\lambda = 1.0nm (\frac{1nm}{10^{9}nm})[/tex]
[tex]\lambda = 10^{-9} m[/tex]
Therefore the energy would be,
[tex]E_1 = \frac{hc}{\lambda}[/tex]
[tex]E_1 = \frac{(6.63*!0^{-34}J\cdo s)(3*10^{8}m/s)}{10^{-9}m}[/tex]
[tex]E_1 = 19.89*10^{-17} J[/tex]
The frequency is given as,
[tex]f = 10MHz (\frac{10^6z}{1.0MHz})[/tex]
[tex]f = 10^7Hz[/tex]
Now the second energy would be
[tex]E_2 = hf[/tex]
[tex]E_2 = (6.63*10^{-27}J\cdot s)(10^7Hz)[/tex]
[tex]E_2 = 6.63*10^{-27}J[/tex]
Therefore the ratio between them is
[tex]\frac{E_1}{E_2} = \frac{19.89*10^{-17}J}{6.63*10^{-27}J}[/tex]
[tex]\frac{E_1}{E_2} = 3*10^{20}[/tex]
Therefore the energy of 1nm x ray or gamma ray photon is [tex]3*10^{20}[/tex] times more than energy of 10MHz radio photon
A lightbulb with an intrinsic resistance of 270 \OmegaΩ is hooked up to a 12-volt battery. How much power is output by the lightbulb? Give your answer out to the thousandths place in units of watts (W).
Answer:
P = 0.533 W
Explanation:
given,
Resistance of the bulb, R = 270 Ω
Potential of the battery, V= 12 V
Power output of the bulb = ?
we know,
P = I² R
also, V = IR
[tex]P = \dfrac{V^2}{R}[/tex]
[tex]P =\dfrac{12^2}{270}[/tex]
[tex]P =\dfrac{144}{270}[/tex]
P = 0.533 W
Hence, the Power delivered by the bulb is equal to 0.533 W.
The atomic radii of a divalent cation and a monovalent anion are 0.35 nm and 0.129 nm, respectively.(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).Enter your answer for part (a) in accordance to the question statement N(b) What is the force of repulsion at this same separation distance?
Answer:
a) The force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another) is - 2.01 × 10⁻⁹ N
b) The force of repulsion at this same separation distance is 2.01 × 10⁻⁹ N
Explanation:
F = kq₁q₂/r²
r = 0.35 + 0.129 (since the ions are just touching each other)
r = 0.479 nm = 4.79 × 10⁻¹⁰ m
Since the first ion is a divalent cation, Z₁ = +2 and the monovalent anion, Z₂ = -1
q = Ze; e = 1.602 × 10⁻¹⁹ C
K = 8.99 × 10⁹ Nm²/C²
F = (8.99 × 10⁹)(1.602 × 10⁻¹⁹)²(2)(-1)/(4.79 × 10⁻¹⁰)² = - 2.01 × 10⁻⁹ N
b) At equilibrium,
Force of attraction + Force of repulsion = 0
Force of repulsion = -(Force of attraction) = 2.01 × 10⁻⁹ N
Two point charges, +2.20 μC and -8.00 μC, are separated by 2.60 m. What is the electric potential midway between them? Number Units
Answer:
Electric potential, [tex]V=-4.01\times 10^4\ volts[/tex]
Explanation:
Given that,
Charge 1, [tex]q_1=2.2\ \mu C[/tex]
Point charge 2, [tex]q_2=-8\ \mu C[/tex]
Distance between charges, d = 2.6 m
We need to find the electric potential midway between them. The electric potential is given by :
[tex]V=\dfrac{kq}{r}[/tex]
In this case, r = 1.3 m (midway between charges)
[tex]V=\dfrac{kq_1}{r}-\dfrac{kq_2}{r}[/tex]
[tex]V=\dfrac{k}{r}(q_1-q_2)[/tex]
[tex]V=\dfrac{9\times 10^9}{1.3}(2.2\times 10^{-6}-8\times 10^{-6})[/tex]
[tex]V=-40153.84\ volts[/tex]
or
[tex]V=-4.01\times 10^4\ volts[/tex]
So, the electric potential midway between the charges is [tex]V=-4.01\times 10^4\ volts[/tex]. Hence, this is the required solution.
Final answer:
The electric potential midway between two point charges of +2.20 μC and -8.00 μC, separated by 2.60 m, is calculated separately for each charge using the formula V = kq/r and summed up. The total electric potential at the midpoint is -4.00 × 10⁴ V.
Explanation:
To find the electric potential midway between two point charges, we need to consider the contribution from each charge separately and then sum them up.
The electric potential V due to a single point charge q at a distance r is given by the formula:
V = kq/r
where k is the Coulomb's constant (k ≈ 8.99 × 109 Nm²/C²).
In this case, we have two charges, +2.20 μC and -8.00 μC, and they are separated by 2.60 m. So the distance from the midpoint to each charge is 1.30 m (half of 2.60 m).
Calculating the potential due to the +2.20 μC charge:
V₁ = (8.99 × 109)(+2.20 × 10⁻⁶) / 1.30 = 1.53 × 10⁴ V
And for the -8.00 μC charge:
V₂ = (8.99 × 10⁹)(-8.00 × 10⁻⁶) / 1.30 = -5.53 × 10⁴ V
The total electric potential at the midpoint is the sum of V₁ and V₂:
Vtotal = V₁ + V₂ = 1.53 × 10⁴ V - 5.53 × 10⁴ V = -4.00 × 10⁴V
The electric potential midway between the two charges is -4.00 × 10⁴V.