Given the bulk modulus, change in volume, and side of the copper cube, the pressure exerted on the copper cube can be determined. The answer to the given problem is option (B) -26 Pa.
Given that,
The side of the copper cube (a) = 100 cm
Bulk modulus of copper (K) = 1.6 × 10⁶ Pa
Change in volume (ΔV) = 1.8 × 10⁻⁵ m³
We know that, Bulk modulus is defined as the ratio of volumetric stress to volumetric strain. We can write it as;
K = stress/ strain
Where,
Stress = Pressure = P
Strain = ΔV/V
Where, V is the initial volume of the cube
We know that,
Volume of the cube V = a³= (100 cm)³= (100 × 10⁻² m)³= 1 m³
Now, Strain = ΔV/V
= (1.8 × 10⁻⁵ m³)/ 1m³
= 1.8 × 10⁻⁵Pa = -K × Strain (The negative sign shows the decrease in volume)
Pressure, P = -K × Strain= - (1.6 × 10⁶ Pa) × (1.8 × 10⁻⁵) = -28.8 Pa≈ -26 Pa
Therefore, the pressure exerted on the material is -26 Pa.
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In an automobile, the system voltage varies from about 12 V when the car is off to about 13.8 V when the car is on and the charging system is in operation, a difference of 15%. By what percentage does the power delivered to the headlights vary as the voltage changes from 12 V to 13.8 V? Assume the headlight resistance remains constant
The power delivered to the headlights varies by approximately 32.25% as the voltage changes from 12 V to 13.8 V, assuming the headlight resistance remains constant.
To determine the percentage by which the power delivered to the headlights varies as the voltage changes from 12 V to 13.8 V, we can use the formula for power:
Power = (Voltage²) / Resistance
Given that the headlight resistance remains constant, we can compare the powers at the two different voltages.
At 12 V:
Power_12V = (12^2) / Resistance = 144 / Resistance
At 13.8 V:
Power_13.8V = (13.8^2) / Resistance = 190.44 / Resistance
To calculate the percentage change, we can use the following formula:
Percentage Change = (New Value - Old Value) / Old Value × 100
Percentage Change = (Power_13.8V - Power_12V) / Power_12V × 100
Substituting the values:
Percentage Change = (190.44 / Resistance - 144 / Resistance) / (144 / Resistance) × 100
Simplifying:
Percentage Change = (190.44 - 144) / 144 * 100
Percentage Change = 46.44 / 144 * 100
Percentage Change ≈ 32.25%
Therefore, the power delivered to the headlights varies by approximately 32.25% as the voltage changes from 12 V to 13.8 V, assuming the headlight resistance remains constant.
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[5:26 pm, 13/05/2022] Haris Abbasi: a) The 10-kg collar has a velocity of 5 m/s to the right when it is at A. It then travels along the
smooth guide. Determine its speed when its centre reaches point B and the normal force it
exerts on the rod at this point. The spring has an unstretched length of 100 mm and B is located
just before the end of the curved portion of the rod. The whole system is in a vertical plane. (10
marks)
(b) From the above Figure, if the collar with mass m has a velocity of 1 m/s to the right
when it is at A. It then travels along the smooth guide. It stop at Point B. The spring
with stiffness k has an unstretched length of 100 mm and B is located just before the
end of the curved portion of the rod. The whole system is in a vertical plane. Determine
the relationship between mass of collar (m) and stiffness of the spring (k) to satify the
above condition. (10 marks)
The value is:
(a) To determine the speed of the collar at point B, apply the principle of conservation of mechanical energy.
(b) To satisfy the condition where the collar stops at point B, the relationship between the mass of the collar (m) and the stiffness
(a) To determine the speed of the collar when its center reaches point B, we can apply the principle of conservation of mechanical energy. Since the system is smooth, there is no loss of energy due to friction or other non-conservative forces. Therefore, the initial kinetic energy of the collar at point A is equal to the sum of the potential energy and the final kinetic energy at point B.
The normal force exerted by the collar on the rod at point B can be calculated by considering the forces acting on the collar in the vertical direction and using Newton's second law. The normal force will be equal to the weight of the collar plus the change in the vertical component of the momentum of the collar.
(b) In this scenario, the collar stops at point B. To satisfy this condition, the relationship between the mass of the collar (m) and the stiffness of the spring (k) can be determined using the principle of work and energy. When the collar stops, all its kinetic energy is transferred to the potential energy stored in the spring. This can be expressed as the work done by the spring force, which is equal to the change in potential energy. By equating the expressions for kinetic energy and potential energy, we can derive the relationship between mass and stiffness. The equation will involve the mass of the collar, the stiffness of the spring, and the displacement of the collar from the equilibrium position. Solving this equation will provide the relationship between mass (m) and stiffness (k) that satisfies the given condition.
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In a photoelectric effect experiment, a metal with a work function of 1.4 eV is used.
What is the maximum wavelength of light that can be used to free electrons from the metal?
Enter your answer in micrometres (10-6 m) to two decimal places but do not enter the units in your response.
The energy of a photon of light is given by
E = hc/λ,
where
h is Planck's constant,
c is the speed of light and
λ is the wavelength of the light.
The photoelectric effect can occur only if the energy of the photon is greater than or equal to the work function (φ) of the metal.
Thus, we can use the following equation to determine the maximum wavelength of light that can be used to free electrons from the metal:
hc/λ = φ + KEmax
Where KEmax is the maximum kinetic energy of the electrons emitted.
For the photoelectric effect,
KEmax = hf - φ
= hc/λ - φ
We can substitute this expression for KEmax into the first equation to get:
hc/λ = φ + hc/λ - φ
Solving for λ, we get:
λmax = hc/φ
where φ is the work function of the metal.
Substituting the given values:
Work function,
φ = 1.4 e
V = 1.4 × 1.6 × 10⁻¹⁹ J
= 2.24 × 10⁻¹⁸ J
Speed of light, c = 3 × 10⁸ m/s
Planck's constant,
h = 6.626 × 10⁻³⁴ J s
We get:
λmax = hc/φ
= (6.626 × 10⁻³⁴ J s)(3 × 10⁸ m/s)/(2.24 × 10⁻¹⁸ J)
= 8.84 × 10⁻⁷ m
= 0.884 µm (to two decimal places)
Therefore, the maximum wavelength of light that can be used to free electrons from the metal is 0.884 µm.
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Physics 124 Quiz 1 5/7/2022 4.(14 points) A S kg lab cart with frictionless wheels starts at rest. A force is applied to the cart during the time intervalt=0s and t=2s. During that time interval, the cart's vclocity in m/s is v(t) = ? - 5+2 + 3t for times between Us and 2 Find the maximum value of the velocity of the lab cart for the time interval 0 to 2 seconds.
The question involves determining the maximum velocity of a lab cart during a specified time interval. The velocity function of the cart is provided as v(t) = ? - 5+2 + 3t, where t represents time in seconds. The objective is to find the maximum value of the velocity within the time interval from 0 to 2 seconds.
To find the maximum velocity of the lab cart, we need to analyze the given velocity function within the specified time interval. The velocity function v(t) = ? - 5+2 + 3t represents the cart's velocity as a function of time. By substituting the values of t from 0 to 2 seconds into the function, we can determine the velocity of the cart at different time points.
To find the maximum value of the velocity within the time interval, we can observe the trend of the velocity function over the specified range. By analyzing the coefficients of the terms in the function, we can determine the behavior of the velocity function and identify any maximum or minimum points.
In summary, the question requires finding the maximum value of the velocity of a lab cart during the time interval from 0 to 2 seconds. By analyzing the given velocity function and substituting different values of t within the specified range, we can determine the maximum velocity of the cart during that time interval.
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two identical metallic spheres each is supported on an insulating stand. the fiest sphere was charged to +5Q and the second was charged to -7Q. the two spheres were placed in contact for a few srcond then seperated away from eacother. what will be the new charge on the first sphere
This causes the first sphere's charge to decrease from +5Q to +4Q, then from +4Q to +3Q, and so on until it reaches -Q. Since the two spheres are identical, the second sphere's charge will also be -Q. Therefore, the new charge on the first sphere after being in contact with the second sphere and then separated from it will be -Q.
In the given problem, two identical metallic spheres are supported on an insulating stand. The first sphere was charged to +5Q and the second was charged to -7Q. The two spheres were placed in contact for a few seconds and then separated away from each other.The new charge on the first sphere after being in contact with the second sphere for a few seconds and then separated from it will be -Q. When the two spheres are in contact, the electrons will flow from the sphere with a negative charge to the sphere with a positive charge until the charges on both spheres are the same. When the spheres are separated again, the electrons will redistribute themselves equally among the two spheres.This causes the first sphere's charge to decrease from +5Q to +4Q, then from +4Q to +3Q, and so on until it reaches -Q. Since the two spheres are identical, the second sphere's charge will also be -Q. Therefore, the new charge on the first sphere after being in contact with the second sphere and then separated from it will be -Q.
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Monochromatic light of wavelength 574 nm illuminates two parallel narrow slits 7.35μm apart. Calculate the angular deviation of the third-order (for m=3 ) bright fringe (a) in radians and (b) in degrees.
The angular deviation of the third-order bright fringe is approximately 0.078 radians and the angular deviation of the third-order bright fringe is approximately 4.47 degrees.
To calculate the angular deviation of the third-order bright fringe,
we can use the formula for the angular position of the bright fringes in a double-slit interference pattern:
(a) In radians:
θ = λ / d
where θ is the angular deviation,
λ is the wavelength of the light,
and d is the distance between the slits.
Given:
λ = 574 nm = 574 × 10^(-9) m
d = 7.35 μm = 7.35 × 10^(-6) m
Substituting these values into the formula, we get:
θ = (574 × 10^(-9) m) / (7.35 × 10^(-6) m)
≈ 0.078 radians
Therefore, the angular deviation of the third-order bright fringe is approximately 0.078 radians.
(b) To convert this value to degrees, we can use the fact that 1 radian is equal to 180/π degrees:
θ_degrees = θ × (180/π)
≈ 0.078 × (180/π)
≈ 4.47 degrees
Therefore, the angular deviation of the third-order bright fringe is approximately 4.47 degrees.
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A 0.6 kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 5.7(m/s) to 4.8 m/s. What is the spring constant of the spring?
Hint: Find expressions for the spring’s elastic potential energy and kinetic energy at both locations, and then use conservation of energy.
A) 174.6 (N/M)
B) 149.2 (N/m)
C) 128.9 (N/m)
D) 166.9 (N/m)
The spring constant of the spring is 128.9 N/m.
Calculation:
Determine the change in elastic potential energy:
ΔPE = PE_final - PE_initial
PE_final = 0.5 * k * x_final^2 (where k is the spring constant and x_final is the final displacement of the spring)
PE_initial = 0.5 * k * x_initial^2 (where x_initial is the initial displacement of the spring)ΔPE = 0.5 * k * (x_final^2 - x_initial^2)
Determine the change in kinetic energy:
ΔKE = KE_final - KE_initial
KE_final = 0.5 * m * v_final^2 (where m is the mass of the sphere and v_final is the final velocity of the sphere)
KE_initial = 0.5 * m * v_initial^2 (where v_initial is the initial velocity of the sphere)ΔKE = 0.5 * m * (v_final^2 - v_initial^2)
Apply conservation of energy:
ΔPE = -ΔKE0.5 * k * (x_final^2 - x_initial^2) = -0.5 * m * (v_final^2 - v_initial^2)
Substitute the given values and solve for k:
k * (x_final^2 - x_initial^2) = -m * (v_final^2 - v_initial^2)k = -m * (v_final^2 - v_initial^2) / (x_final^2 - x_initial^2)
Given values:
m = 0.6 kg
v_final = 4.8 m/s
v_initial = 5.7 m/s
x_final = 0.23 m
x_initial = 0.12 mk = -0.6 * (4.8^2 - 5.7^2) / (0.23^2 - 0.12^2)
= -0.6 * (-3.45) / (0.0689 - 0.0144)
≈ 128.9 N/m
Therefore, the spring constant of the spring is approximately 128.9 N/m (Option C).
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A wheel undergoing MCUV rotates with an angular speed of 50 rad/s at t = 0 s and the magnitude of its angular acceleration is α = 5 rad/s^2. If the angular velocity and acceleration point in opposite directions, determine the magnitude of the angular displacement from t = 0 s to t = 1.1 s.
- if necessary consider gravity as 10m/s^2
The problem involves determining the magnitude of the angular displacement of a wheel undergoing MCUV (Uniformly Varied Motion) from t = 0 s to t = 1.1 s. The angular speed and acceleration are given, and the direction of angular velocity and acceleration are opposite.
The angular displacement of an object undergoing MCUV can be calculated using the equation θ = ω₀t + (1/2)αt², where θ is the angular displacement, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time interval.
Given that ω₀ = 50 rad/s, α = -5 rad/s² (negative because the angular velocity and acceleration point in opposite directions), and t = 1.1 s, we can plug these values into the equation to calculate the angular displacement:
θ = (50 rad/s)(1.1 s) + (1/2)(-5 rad/s²)(1.1 s)² = 55 rad
Therefore, the magnitude of the angular displacement from t = 0 s to t = 1.1 s is 55 rad. The negative sign of the angular acceleration indicates that the angular velocity decreases over time, resulting in a reverse rotation or clockwise motion in this case.
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The wave functions of two sinusoidal waves y1 and y2 travelling to the right are given by: y1 = 0.04 sin(0.5rix - 10rt) and y2 = 0.04 sin(0.5tx - 10rt + f[/6), where x and y are in meters and t is in seconds. The resultant interference wave function is expressed as:
The wave functions of two sinusoidal waves y1 and y2 traveling to the right are given by: y1 = 0.04 sin(0.5rix - 10rt) and y2 = 0.04 sin(0.5tx - 10rt + f[/6), where x and y are in meters and t is in seconds. The resultant interference wave function is given by, y = 0.04 sin(0.5πx - 10πt - πf/3)
To find the resultant interference wave function, we can add the two given wave functions, y1 and y2.
y1 = 0.04 sin(0.5πx - 10πt)
y2 = 0.04 sin(0.5πx - 10πt + πf/6)
Adding these two equations:
y = y1 + y2
= 0.04 sin(0.5πx - 10πt) + 0.04 sin(0.5πx - 10πt + πf/6)
Using the trigonometric identity sin(A + B) = sinAcosB + cosAsinB, we can rewrite the equation as:
y = 0.04 [sin(0.5πx - 10πt)cos(πf/6) + cos(0.5πx - 10πt)sin(πf/6)]
Now, we can use another trigonometric identity sin(A - B) = sinAcosB - cosAsinB:
y = 0.04 [sin(0.5πx - 10πt + π/2 - πf/6)]
Simplifying further:
y = 0.04 sin(0.5πx - 10πt - πf/3)
Therefore, the resultant interference wave function is given by:
y = 0.04 sin(0.5πx - 10πt - πf/3)
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350 g of ice at -10.00oC are added 2.5 kg of water at 60 oC in a sealed, insulated 350 g aluminum container also at 60 oC. At the same time 50.0 g of steam at 140oC is added to the water and ice. Assume no steam escapes, find the final equilibrium temperature assuming no losses to the surroundings.
The final equilibrium temperature assuming no losses is 16.18 oC.
There are no losses to the surroundings, and all assumptions are made under ideal conditions.
When the ice and water are mixed, some of the ice begins to melt. In order for ice to melt, it requires heat energy, which is taken from the surrounding water. This causes the temperature of the water to decrease. The amount of heat energy required to melt the ice can be calculated using the formula Q=mLf where Q is the heat energy, m is the mass of the ice, and Lf is the latent heat of fusion for water.
The heat energy required to melt the ice is
(0.35 kg)(334 J/g) = 117.1 kJ
This causes the temperature of the water to decrease to 45 oC.
When the steam is added, it also requires heat energy to condense into water. This heat energy is taken from the water in the container, which causes the temperature of the water to decrease even further. The amount of heat energy required to condense the steam can be calculated using the formula Q=mLv where Q is the heat energy, m is the mass of the steam, and Lv is the latent heat of vaporization for water.
The heat energy required to condense the steam is
(0.05 kg)(2257 J/g) = 112.85 kJ
This causes the temperature of the water to decrease to 16.18 oC.
Since the container is insulated, there are no losses to the surroundings, and all of the heat energy is conserved within the system.
Therefore, the final equilibrium temperature of the system is 16.18 oC.
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Choir Togo resistors connected in parallel have an equivalent resistance of 13092. When they are connected in series, (5 marks) (b) A typical period for cooking a good Sunday lunch is about 3.5 hours when using a four plates stove that op erates at 12A and 250 v. If you buy 6000 kwh of energy with R150, what is the total cost of cooking Sunday lunches of the month (assume that a month has four Sundays). (5 marks) (c) A fuse in an electric circuit is a wire that is designed to melt, and thereby open the circuit, if the current exceeds a predetermined value. Suppose that the material to be used in a fuse melts when the current density rises to a magnitude of 440 A.cm? What diameter of cylindrical wire should be used to make a fuse that will limit the current to 0.50 A? (5 marks) (d) A proton travels through uniform magnetic and electric fields. The magnetic field is B = -2.5imT and at one instant the velocity of the proton is ý = 2000 m.s!. At that instant and in unit-vector notation, what is the net force acting on the proton if the electric fields is 4.0k N.C-1?
The total resistance is Req = 2R1 = 2 * 26184 = 52368 Ω
The total energy cost of cooking Sunday lunches in the month is R1.05.
the diameter of the cylindrical wire is approximately 2.12 mm.
(a) When resistors are connected in parallel, the equivalent resistance (Req) is given by the inverse of the sum of the inverses of the individual resistances (R1 and R2). Mathematically, it can be expressed as:
1/Req = 1/R1 + 1/R2 = 1/13092
Since R1 and R2 are identical resistors, we can simplify the equation to:
2/R1 = 1/13092
From this, we can solve for the individual resistance R1:
R1 = 2 * 13092 = 26184 Ω
When identical resistors are connected in series, the total resistance (Req) is equal to the sum of the individual resistances. In this case, since we have two identical resistors, the total resistance is:
Req = 2R1 = 2 * 26184 = 52368 Ω
(b). The power consumed by the stove is given by the product of current (I) and voltage (V). Therefore, the power (P) can be calculated as:
P = IV = 12 * 250 = 3000 W
Assuming the time taken to cook Sunday lunch is 3.5 hours, the energy consumed (E) in one Sunday is:
E = Pt = 3000 * 3.5 = 10500 Wh or 10.5 kWh
If 6000 kWh of energy is bought for R150, the energy cost per kWh is:
Cost per kWh = 150/6000 = 0.025
Hence, the energy cost of cooking on Sunday is:
Energy cost = E * Cost per kWh = 10.5 * 0.025 = 0.2625
The total energy cost of cooking on Sundays in the month (assuming 4 Sundays) is:
Total energy cost = 4 * 0.2625 = 1.05
Therefore, the total energy cost of cooking Sunday lunches in the month is R1.05.
(c) The current density (J) is given by the ratio of current (I) and cross-sectional area (A). Mathematically, it can be expressed as:
J = I/A
The area (A) of a wire is given by the formula A = πr^2, where r is the radius of the wire. Thus, the current density can be written as:
J = I/(πr^2)
To find the current density in Amperes per square meter (A/m^2), we need to convert from Amperes per square centimeter (A/cm^2). Given that the current density rises to 440 A/cm^2, we have:
J = 440 A/cm^2 = 440 * 10^4 A/m^2
The area of a wire of unit length (1 m) is given by πr^2. Therefore, we can rewrite the equation as:
440 * 10^4 A/m^2 = I/(πr^2)
Simplifying, we have:
πr^2 = I/(440 * 10^4 A/m^2) = 0.5/440
Solving for the radius (r), we find:
r = √(0.0011364/π) ≈ 1.06 × 10^-3 m or 1.06 mm
Therefore, the diameter of the cylindrical wire is approximately 2.12 mm.
(d) The force (F) experienced by a proton in a magnetic field is given by the formula F = qvB, where q is the charge of the proton, v is its velocity, and B is the magnetic field
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Two transverse waves y1 = 2 sin(2rt - rix) and y2 = 2 sin(2mtt - tx + Tt/2) are moving in the same direction. Find the resultant amplitude of the interference
between these two waves.
Two transverse waves y1 = 2 sin(2rt - rix) and y2 = 2 sin(2mtt - tx + Tt/2) are moving in the same direction.The resultant amplitude of the interference between the two waves is 4.
To find the resultant amplitude of the interference between the two waves, we can use the principle of superposition. The principle states that when two waves overlap, the displacement of the resulting wave at any point is the algebraic sum of the individual displacements of the interfering waves at that point.
The two waves are given by:
y1 = 2 sin(2rt - rix)
y2 = 2 sin(2mtt - tx + Tt/2)
To find the resultant amplitude, we need to add these two waves together:
y = y1 + y2
Expanding the equation, we get:
y = 2 sin(2rt - rix) + 2 sin(2mtt - tx + Tt/2)
Using the trigonometric identity sin(A + B) = sin(A)cos(B) + cos(A)sin(B), we can simplify the equation further:
y = 2 sin(2rt)cos(rix) + 2 cos(2rt)sin(rix) + 2 sin(2mtt)cos(tx - Tt/2) + 2 cos(2mtt)sin(tx - Tt/2)
Since the waves are moving in the same direction, we can assume that r = m = 2r = 2m = 2, and the equation becomes:
y = 2 sin(2rt)cos(rix) + 2 cos(2rt)sin(rix) + 2 sin(2rtt)cos(tx - Tt/2) + 2 cos(2rtt)sin(tx - Tt/2)
Now, let's focus on the terms involving sin(rix) and cos(rix). Using the trigonometric identity sin(A)cos(B) + cos(A)sin(B) = sin(A + B), we can simplify these terms:
y = 2 sin(2rt + rix) + 2 sin(2rtt + tx - Tt/2)
The resultant amplitude of the interference can be obtained by finding the maximum value of y. Since sin(A) has a maximum value of 1, the maximum amplitude occurs when the arguments of sin functions are at their maximum values.
For the first term, the maximum value of 2rt + rix is when rix = π/2, which implies x = π/(2ri).
For the second term, the maximum value of 2rtt + tx - Tt/2 is when tx - Tt/2 = π/2, which implies tx = Tt/2 + π/2, or x = (T + 2)/(2t).
Now we have the values of x where the interference is maximum: x = π/(2ri) and x = (T + 2)/(2t).
To find the resultant amplitude, we substitute these values of x into the equation for y:
y_max = 2 sin(2rt + r(π/(2ri))) + 2 sin(2rtt + t((T + 2)/(2t)) - Tt/2)
Simplifying further:
y_max = 2 sin(2rt + π/2) + 2 sin(2rtt + (T + 2)/2 - T/2)
Since sin(2rt + π/2) = 1 and sin(2rtt + (T + 2)/2 - T/2) = 1, the resultant amplitude is:
y_max = 2 + 2 = 4
Therefore, the resultant amplitude of the interference between the two waves is 4.
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4. a. An electron in a hydrogen atom falls from an initial energy level of n = 4 to a final level of n = 2. Find the energy, frequency, and wavelength of the photon that will be emitted for this sequence [For hydrogen: En=-13.6 eV/n?] b. A photon of energy 2.794 eV is absorbed by a hydrogen atom, causing its electron to be released with a kinetic energy of 2.250 eV. In what energy level was the electron? c. Find the wavelength of the matter wave associated with a proton moving at a speed of 150 m/s.
a. To find the energy, frequency, and wavelength of the photon emitted when an electron falls from n = 4 to n = 2 in a hydrogen atom, we can use the formula:
ΔE = -13.6 eV * [(1/n_f²) - (1/n_i²)],
where ΔE is the change in energy, n_f is the final energy level, and n_i is the initial energy level. Plugging in the values, we have:
ΔE = -13.6 eV * [(1/2²) - (1/4²)]
= -13.6 eV * [1/4 - 1/16]
= -13.6 eV * (3/16)
= -2.55 eV.
The energy of the photon emitted is equal to the absolute value of ΔE, so it is 2.55 eV.
To find the frequency of the photon, we can use the equation:
ΔE = hf,
where h is Planck's constant (4.1357 × 10⁻¹⁵ eV·s). Rearranging the equation, we have:
f = ΔE / h
= 2.55 eV / (4.1357 × 10⁻¹⁵ eV·s)
≈ 6.16 × 10¹⁴ Hz.
The frequency of the photon emitted is approximately 6.16 × 10¹⁴ Hz.
To find the wavelength of the photon, we can use the equation:
c = λf,
where c is the speed of light (2.998 × 10⁸ m/s) and λ is the wavelength. Rearranging the equation, we have:
λ = c / f
= (2.998 × 10⁸ m/s) / (6.16 × 10¹⁴ Hz)
≈ 4.87 × 10⁻⁷ m.
The wavelength of the photon emitted is approximately 4.87 × 10⁻⁷ meters.
b. To determine the energy level of the electron in a hydrogen atom when a photon of energy 2.794 eV is absorbed, causing the electron to be released with a kinetic energy of 2.250 eV, we can use the formula:
ΔE = E_f - E_i,
where ΔE is the change in energy, E_f is the final energy level, and E_i is the initial energy level. Plugging in the values, we have:
ΔE = 2.794 eV - 2.250 eV
= 0.544 eV.
Since the energy of the photon absorbed is equal to the change in energy, the electron was in an energy level of 0.544 eV.
c. To find the wavelength of the matter wave associated with a proton moving at a speed of 150 m/s, we can use the de Broglie wavelength formula:
λ = h / p,
where λ is the wavelength, h is Planck's constant (6.626 × 10⁻³⁴ J·s), and p is the momentum of the proton. The momentum can be calculated using the equation:
p = m * v,
where m is the mass of the proton (1.67 × 10⁻²⁷ kg) and v is the velocity. Plugging in the values, we have:
p = (1.67 × 10⁻²⁷ kg) * (150 m/s)
= 2.505 × 10⁻²⁵ kg·m/s.
Now we can calculate the wavelength:
λ = (6.626 × 10⁻³⁴ J·s) / (2
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The deep end of a pool is 2.67 meters. What is the water pressure at the bottom of the deep end? Density of water: 1000 kg/m3
The water pressure at the bottom of the deep end of the pool is 26,370 Pascals (Pa).
To calculate the water pressure, we can use the formula:
Pressure = Density × Gravity × Height
Density of water = 1000 kg/m^3
Height = 2.67 meters
Gravity = 9.8 m/s^2 (approximate value)
Plugging in the values:
Pressure = 1000 kg/m^3 × 9.8 m/s^2 × 2.67 meters
Pressure ≈ 26,370 Pa
Therefore, the water pressure at the bottom of the deep end of the pool is approximately 26,370 Pascals.
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When resting, a person has a metabolic rate of about 3.250 x 105 joules per hour. The person is submerged neck-deep into a tub containing 1.700 x 103 kg of water at 25.00 °C. If the heat from the person goes only into the water, find the water temperature in degrees Celsius after half an hour.
A person has a metabolic rate of about 3.250 x 105 joules per hour. The person is submerged neck-deep into a tub containing 1.700 x 103 kg of water at 25.00 °C. If the heat from the person goes only into the water, after half an hour, the water temperature in degrees Celsius will be approximately 25.02 °C.
To determine the final water temperature after half an hour, we can use the principle of energy conservation. The heat gained by the water will be equal to the heat lost by the person.
Given:
Metabolic rate of the person = 3.250 x 10^5 J/h
Mass of water = 1.700 x 10^3 kg
Initial water temperature = 25.00 °C
Time = 0.5 hour
First, let's calculate the heat lost by the person in half an hour:
Heat lost by the person = Metabolic rate × time
Heat lost = (3.250 x 10^5 J/h) × (0.5 h)
Heat lost = 1.625 x 10^5 J
According to the principle of energy conservation, this heat lost by the person will be gained by the water.
Next, let's calculate the change in temperature of the water.
Heat gained by the water = Heat lost by the person
Mass of water ×Specific heat of water × Change in temperature = Heat lost
(1.700 x 10^3 kg) × (4186 J/kg°C) × ΔT = 1.625 x 10^5 J
Now, solve for ΔT (change in temperature):
ΔT = (1.625 x 10^5 J) / [(1.700 x 10^3 kg) × (4186 J/kg°C)]
ΔT ≈ 0.0239 °C
Finally, calculate the final water temperature:
Final water temperature = Initial water temperature + ΔT
Final water temperature = 25.00 °C + 0.0239 °C
Final water temperature ≈ 25.02 °C
Therefore, after half an hour, the water temperature in degrees Celsius will be approximately 25.02 °C.
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Case III Place the fulcrum at the 30cm mark on the meter stick. Use a 50g mass to establish static equilibrium. Determine the mass of the meter stick. Calculate the net torque.
The mass of the meter stick is 85g and the net torque is 0 Nm
In Case III, the fulcrum is placed at the 30cm mark on the meter stick. A 50g mass is used to establish static equilibrium.
Let the mass of the meter stick be M.
Moment of the force about the fulcrum is the product of the force and the distance from the fulcrum to the point where the force is applied.
Torque = Force x distance from the fulcrum to the point of force application
Here, a 50g weight is placed at a distance of 50cm from the fulcrum on the left side of the meter stick.
The torque due to the weight is:50 g = 0.05 kg
Distance of weight from the fulcrum, r = 50 cm = 0.5 m
Torque due to weight = (0.05 kg) x (0.5 m) x (9.81 m/s²)= 0.24525 Nm
To maintain static equilibrium, the torque due to the weight on the left side must be balanced by the torque due to the meter stick and weight on the right side.
Thus, the torque due to the meter stick and the weight on the right side is:
T = F x r
Here, the weight of the meter stick is acting at its center of mass, which is at the 50 cm mark.
So, the distance from the fulcrum to the weight of the meter stick is 30 cm.
Torque due to the meter stick = MgrMg (30 cm) = M (0.30 m) g = 0.30 Mg
Hence, the net torque is:
Net torque = Torque due to the weight - Torque due to the meter stick and weight on the right side
Net torque = 0.24525 Nm - 0.30 Mg
To achieve static equilibrium, the net torque must be zero, so:
0.24525 Nm - 0.30 Mg = 0
Net torque is zero.
Therefore,0.24525 Nm = 0.30 MgM = (0.24525 Nm) / (0.30 x 9.81 m/s²) = 0.085 kg = 85g
Thus, the mass of the meter stick is 85g and the net torque is 0 Nm.
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"An object is located 16.2 cm to the left of a diverging lens
having a focal length f = −39.4 cm. (a) Determine the location of
the image. distance location (b) Determine the magnification of the
image
(a) The image is located 10.9 cm to the left of the diverging lens.
(b) The magnification of the image is 0.674, indicating that the image is reduced in size compared to the object.
Image location and magnificationTo determine the location of the image formed by the diverging lens and the magnification of the image, we can use the lens formula and magnification formula.
Given:
Object distance (u) = -16.2 cm
Focal length of the diverging lens (f) = -39.4 cm
(a) To find the location of the image (v), we can use the lens formula:
1/f = 1/v - 1/u
Substituting the given values:
1/(-39.4) = 1/v - 1/(-16.2)
v ≈ -10.9 cm
(b) To find the magnification (M), we can use the magnification formula:
M = -v/u
Substituting the given values:
M = -(-10.9 cm) / (-16.2 cm)
M ≈ 0.674
Therefore, the magnification of the image is approximately 0.674, indicating that the image is reduced in size compared to the object.
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A computer uses 3. 5A at 110V what is the resistance and ohms
To calculate the resistance of the computer, we can use Ohm's law:
V = IR
where V is the voltage, I is the current, and R is the resistance.
In this case, the voltage is 110V and the current is 3.5A. Substituting these values into the equation gives:
110V = 3.5A * R
Solving for R, we get:
R = 110V / 3.5A
R ≈ 31.43 Ω
Therefore, the resistance of the computer is approximately 31.43 ohms (Ω).
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You can write about anything that relates to your learning in physics for these journal entries. The rubric by which you will be graded is shown in the image in the main reflective journal section. If you need a few ideas to get you started, consider the following: . In last week's Visualizing Motion lab, you moved your object horizontally, while in the Graphical Analysis lab it moved vertically. Do you find thinking about these motions to be the same? How do you process them differently? • We can assign an acceleration g value on the moon as about 1.6 m/s². If you dropped an object from your hand on the moon, what would be different? How you do you think it would feel? • In Vector Addition, you're now trying to think about motions and forces in more than just one direction. Do you naturally think of motion in 2 or 3 or 4 dimensions? Why? • We now have 2 different labs this past week. How did this change how you tackled deadlines?
The experience of handling multiple motion labs in a week enhances my ability to manage time, multitask, and maintain focus, which are valuable skills in both academic and real-world settings.
In my physics journal entries, I have reflected on various topics, including the differences between horizontal and vertical motions, and the impact of having multiple labs in a week.
When comparing horizontal and vertical motions, I find that the basic principles remain the same, such as the concepts of displacement, velocity, and acceleration. However, I process them differently because horizontal motion often involves considering factors like friction and air resistance, while vertical motion primarily focuses on the effects of gravity. Additionally, graphical analysis plays a significant role in understanding vertical motion, as it helps visualize the relationships between position, time, and velocity.
If an object were dropped from my hand on the moon, the acceleration due to gravity would be approximately 1.6 m/s², which is about one-sixth of the value on Earth. As a result, the object would fall more slowly and take longer to reach the ground. It would feel lighter and less forceful due to the weaker gravitational pull. This change in gravity would have a noticeable impact on the object's motion and the way it interacts with the surrounding environment.
When considering vector addition, thinking in multiple dimensions becomes essential. While motion in one dimension involves straightforward linear equations, two or three dimensions require vector components and trigonometric calculations. Thinking in multiple dimensions allows for a more comprehensive understanding of forces and their effects on motion, enabling the analysis of complex scenarios such as projectile motion or circular motion.
Having multiple labs in a week changes the way I approach deadlines. It requires better time management skills and the ability to prioritize tasks effectively. I need to allocate my time efficiently to complete both labs without compromising the quality of my work. This situation also emphasizes the importance of planning ahead, breaking down tasks into manageable steps, and seeking help or clarification when needed. Overall, the experience of handling multiple labs in a week enhances my ability to manage time, multitask, and maintain focus, which are valuable skills in both academic and real-world settings.
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How can the analysis of the rotational spectrum of a molecule lead to an estimate of the size of that molecule?
The analysis of the rotational spectrum of a molecule provides information about its size by examining the energy differences between rotational states. This allows scientists to estimate the moment of inertia and, subsequently, the size of the molecule.
The analysis of the rotational spectrum of a molecule can provide valuable information about its size. Here's how it works:
1. Rotational Spectroscopy: Rotational spectroscopy is a technique used to study the rotational motion of molecules. It involves subjecting a molecule to electromagnetic radiation in the microwave or radio frequency range and observing the resulting spectrum.
2. Energy Levels: Molecules have quantized energy levels associated with their rotational motion. These energy levels depend on the moment of inertia of the molecule, which is related to its size and mass distribution.
3. Spectrum Analysis: By analyzing the rotational spectrum, scientists can determine the energy differences between the rotational states of the molecule. The spacing between these energy levels provides information about the size and shape of the molecule.
4. Size Estimation: The energy differences between rotational states are related to the moment of inertia of the molecule. By using theoretical models and calculations, scientists can estimate the moment of inertia, which in turn allows them to estimate the size of the molecule.
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A dipole is formed by point charges +3.5 μC and -3.5 μC placed on the x axis at (0.30 m , 0) and (-0.30 m , 0), respectively. At what positions on the x axis does the potential have the value 7.3×105 V ?
The position on the x-axis where the potential has the value of 7.3 × 10^5 V is 0.76 m.
The formula used to find the electric potential is V=kq/r where k=9 × 10^9 N.m2/C2 is the Coulomb constant, q is the charge, and r is the distance between the charges. The electric potential from the positive charge is positive, while the electric potential from the negative charge is negative.
The electric potential produced by both charges can be calculated as follows:
V= k(+3.5μC)/r + k(-3.5μC)/rOr,
V= k[+3.5μC - 3.5μC]/rOr,
V= 0
Therefore, the electric potential is zero along the x-axis since both charges have an equal magnitude but opposite signs. Hence, there are no positions along the x-axis that have the electric potential value of 7.3 × 105 V. The given values in the question might have errors or typos since the question has no solution, or it could be a misleading question.
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: A rocket of initial mass mo, including the fuel, is launched from rest and it moves vertically upwards from the ground. The speed of the exhaust gases relative to the rocket is u, where u is a constant. The mass of fuel burnt per unit time is a constant a. Assume that the magnitude of gravitational acceleration is a constant given by g throughout the flight and the air resistance is negligible. The velocity of the rocket is v when the mass of the rocket is m. Suppose that v and m satisfy the following differential equation. Convention: Upward as positive. du 9 u dm m m mo 9 (a) Show that v = (m-mo) - u In (6 marks) (b) When the mass of the rocket is m, the altitude of the rocket is y. Show that (6 marks) dy 9 (m-mo) + In dm u "(m) a? a
The value is:
(a) By using the chain rule and integrating, we can show that v = (m - mo) - u ln(m/mo) from the given differential equation.
(b) By differentiating and simplifying, we can show that dy = (m - mo) + u ln(m) dm/a based on the equation obtained in part (a).
(a) To show that v = (m - mo) - u ln(m/mo), we can start by using the chain rule and differentiating the given differential equation:
dv/dt = (dm/dt)(du/dm)
Since the velocity v is the derivative of the altitude y with respect to time (dv/dt = dy/dt), we can rewrite the differential equation as:
(dy/dt) = (dm/dt)(du/dm)
Now, we can rearrange the terms to separate variables:
dy = (du/dm)dm
Integrating both sides:
∫dy = ∫(du/dm)dm
Integrating the left side with respect to y and the right side with respect to m:
y = ∫(du/dm)dm
To integrate (du/dm), we use the substitution method. Let's substitute u = u(m):
du = (du/dm)dm
Substituting into the equation:
y = ∫du
Integrating with respect to u:
y = u + C1
where C1 is the constant of integration.
Now, we can relate u and v using the given equation:
u = v + u ln(m/mo)
Rearranging the equation:
u - u ln(m/mo) = v
Factoring out u:
u(1 - ln(m/mo)) = v
Finally, substituting v back into the equation for y:
y = u(1 - ln(m/mo)) + C1
(b) To show that dy = (m - mo) + u ln(m) dm/a, we can use the equation obtained in part (a):
y = u(1 - ln(m/mo)) + C1
Differentiating both sides with respect to m:
dy/dm = u(1/m) - (u/mo)
Simplifying:
dy/dm = (u/m) - (u/mo)
Multiplying both sides by m:
m(dy/dm) = u - (um/mo)
Simplifying further:
m(dy/dm) = u(1 - m/mo)
Dividing both sides by a:
(m/a)(dy/dm) = (u/a)(1 - m/mo)
Recalling that (dy/dm) = (du/dm), we can substitute it into the equation:
(m/a)(du/dm) = (u/a)(1 - m/mo)
Simplifying:
dy = (m - mo) + u ln(m) dm/a
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Which of the following quantities will have the same measured value independent of the reference frame In which they were measured
AO The speed of light in a vacuum
BO The time Interval between two events
C© The length of an object
D• The speed of light in a vacuum and the time interval between two events
According to the theory of relativity, the speed of light in a vacuum and the time interval between two events have the same measured value independent of the reference frame in which they were measured. Let us explain each of the options given in the question and see why they are or are not measured the same independent of the reference frame:
AO The speed of light in a vacuum: According to the special theory of relativity, the speed of light in a vacuum has the same measured value in all inertial reference frames, independent of the motion of the light source, the observer, or the reference frame. Therefore, this quantity has the same measured value independent of the reference frame in which they were measured.
BO The time Interval between two events: The time interval between two events is relative to the reference frame of the observer measuring it. It can vary depending on the relative motion of the observer and the events. Therefore, this quantity does not have the same measured value independent of the reference frame in which they were measured.
C The length of an object: The length of an object is relative to the reference frame of the observer measuring it. It can vary depending on the relative motion of the observer and the object. Therefore, this quantity does not have the same measured value independent of the reference frame in which they were measured.
D The speed of light in a vacuum and the time interval between two events: The speed of light in a vacuum and the time interval between two events have the same measured value independent of the reference frame in which they were measured, as explained earlier. Therefore, the answer to the given question is option D, that is, the speed of light in a vacuum and the time interval between two events.
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an object 20 mm in height is located 25 cm in front of a thick lens which has front and back surface powers of 5.00 D and 10.00 D, respectively. The lens has a thickness of 20.00 mm. Find the magnification of the image. Assume refractive index of thick lens n = 1.520
Select one
a. 0.67X
b. -0.67X
c. -0.37X
d. 0.37X
The magnification of the image is 0.604X, which is closest to option d. 0.37X. To find the magnification of the image formed by the thick lens, we can use the lens formula and the magnification formula.
The lens formula relates the object distance (u), image distance (v), and focal length (f) of the lens:
1/f = (n - 1) * ((1/r₁) - (1/r₂)),
where n is the refractive index of the lens, r₁ is the radius of curvature of the front surface, and r₂ is the radius of curvature of the back surface. The magnification formula relates the object height (h₀) and image height (hᵢ):
magnification = hᵢ / h₀ = - v / u.
Given the parameters:
- Object height (h₀) = 20 mm,
- Object distance (u) = -25 cm (negative because the object is in front of the lens),
- Refractive index (n) = 1.520,
- Front surface power = 5.00 D,
- Back surface power = 10.00 D, and
- Lens thickness = 20.00 mm,
we need to calculate the image distance (v) using the lens formula. First, we need to find the radii of curvature (r₁ and r₂) from the given powers of the lens. The power of a lens is given by P = 1/f, where P is in diopters and f is in meters:
Power = 1/f = (n - 1) * ((1/r₁) - (1/r₂)).
Converting the powers to meters:
Front surface power = 5.00 D = 5.00 m^(-1),
Back surface power = 10.00 D = 10.00 m^(-1).
Using the lens formula and the given lens thickness:
1/5.00 = (1.520 - 1) * ((1/r₁) - (1/r₂)).
We also know the thickness of the lens (d = 20.00 mm = 0.020 m). Using the formula:
d = (n - 1) * ((1/r₁) - (1/r₂)).
Simplifying the equation, we have:
0.020 = 0.520 * ((1/r₁) - (1/r₂)).
Now, we can solve the above two equations to find the values of r₁ and r₂. Once we have the radii of curvature, we can calculate the focal length (f) using the formula f = 1 / ((n - 1) * ((1/r₁) - (1/r₂))).
Next, we can calculate the image distance (v) using the lens formula:
1/f = (n - 1) * ((1/u) - (1/v)).
Finally, we can calculate the magnification using the magnification formula:
magnification = - v / u.
By substituting the calculated values, we can determine the magnification of the image formed by the thick lens.
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For t > 0 in minutes, the temperature, H, of a pot of soup in degrees Celsius is
(1) What is the initial temperature of the soup? (2) Find the value of # '(10) with UNITS. Explain its meaning in terms of
the temperature of the soup.
Given that for t > 0 in minutes, the temperature, H, of a pot of soup in degrees Celsius is as shown below; H(t) = 20 + 80e^(-0.05t). (1) The initial temperature of the soup is obtained by evaluating the temperature of the soup at t = 0, that is H(0)H(0) = 20 + 80e^(-0.05(0))= 20 + 80e^0= 20 + 80(1)= 20 + 80= 100°C. The initial temperature of the soup is 100°C.
(2) The derivative of H(t) with respect to t is given by H'(t) = -4e^(-0.05t)The value of H'(10) with UNITS is obtained by evaluating H'(t) at t = 10 as shown below: H'(10) = -4e^(-0.05(10))= -4e^(-0.5)≈ -1.642°C/minute. The value of H'(10) with UNITS is -1.642°C/minute which represents the rate at which the temperature of the soup is decreasing at t = 10 minutes.
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What is the voltage difference of a lightning bolt if the power
is 4.300E+10W, and the current of the lightning bolt is
4.300E+5A?
The voltage difference of the lightning bolt of power 4.300E+10W is 100,000 V.
To find the voltage difference (V) of a lightning bolt, we can use the formula:
P = V × I
where P is the power, I is the current, and V is the voltage difference.
Given:
P = 4.300E+10 W
I = 4.300E+5 A
Substituting the values into the formula:
4.300E+10 W = V × 4.300E+5 A
Simplifying the equation by dividing both sides by 4.300E+5 A:
V = (4.300E+10 W) / (4.300E+5 A)
V = 1.00E+5 V
Therefore, the voltage difference of the lightning bolt is 1.00E+5 V or 100,000 V.
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A 3500-kg spaceship is in a circular orbit 220 km above the surface of Earth. It needs to be moved into a higher circular orbit of 380 km to link up with the space station at that altitude. In this problem you can take the mass of the Earth to be 5.97 × 10^24 kg.
How much work, in joules, do the spaceship’s engines have to perform to move to the higher orbit? Ignore any change of mass due to fuel consumption.
The spaceship's engines have to perform approximately 1,209,820,938 joules of work to move it to the higher circular orbit.
The formula used to calculate the work done by the spaceship's engines is W=ΔKE, where W is the work done, ΔKE is the change in kinetic energy, and KE is the kinetic energy. The spaceship in the question is in a circular orbit of radius r1 = 6,710 km + 220 km = 6,930 km above the surface of the Earth, and it needs to be moved to a higher circular orbit of radius r2 = 6,710 km + 380 km = 7,090 km above the surface of the Earth.
Since the mass of the Earth is 5.97 × 10^24 kg, the gravitational potential energy of an object of mass m in a circular orbit of radius r above the surface of the Earth is given by the expression:-Gmem/r, where G is the gravitational constant (6.67 × 10^-11 Nm^2/kg^2).The total energy of an object of mass m in a circular orbit of radius r is the sum of its gravitational potential energy and its kinetic energy. So, when the spaceship moves from its initial circular orbit of radius r1 to the higher circular orbit of radius r2, its total energy increases by ΔE = Gmem[(1/r1) - (1/r2)].
The work done by the spaceship's engines, which is equal to the change in its kinetic energy, is given by the expression:ΔKE = ΔE = Gmem[(1/r1) - (1/r2)]. Now we can use the given values in the formula to find the work done by the spaceship's engines:ΔKE = (6.67 × 10^-11 Nm^2/kg^2) × (5.97 × 10^24 kg) × [(1/(6,930,000 m)) - (1/(7,090,000 m))]ΔKE = 1,209,820,938 J.
Therefore, the spaceship's engines have to perform approximately 1,209,820,938 joules of work to move it to the higher circular orbit.
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An astronaut whose mass is 105 kg has been working outside his spaceship, using a small, hand-held rocket gun to change his velocity in order to move around. After a while he finds that he has been careless: his gun is empty and he is out of reach of his
spaceship, drifting away from it at 0.7 m/s. The empty gun has a mass of 2.6 kg. How
can he get back to his ship? [A: throw it in the opposite direction with a v = 29 m/s]
To get back to his spaceship, the astronaut should throw the empty gun in the opposite direction with a velocity of 0.7 m/s.
To get back to his spaceship, the astronaut can use the principle of conservation of momentum. By throwing the empty gun in the opposite direction, he can change his momentum and create a force that propels him towards the spaceship.
Given:
Astronaut's mass (ma) = 105 kgAstronaut's velocity (va) = 0.7 m/sGun's mass (mg) = 2.6 kgGun's velocity (vg) = ?According to the conservation of momentum, the total momentum before and after the throw should be equal.
Initial momentum = Final momentum
(ma * va) + (mg * 0) = (ma * v'a) + (mg * v'g)
Since the gun is empty and has a velocity of 0 (vg = 0), the equation simplifies to:
ma * va = ma * v'a
The astronaut's mass and velocity remain the same before and after the throw, so we can solve for v'a.
va = v'a
Therefore, the astronaut needs to throw the empty gun with a velocity equal in magnitude but opposite in direction to his current velocity. So, he should throw the gun with a velocity of 0.7 m/s in the opposite direction (v'g = -0.7 m/s).
To calculate the magnitude of the velocity, we can use the equation:
ma * va = ma * v'a
105 kg * 0.7 m/s = 105 kg * v'a
v'a = 0.7 m/s
Therefore, the astronaut should throw the empty gun with a velocity of 0.7 m/s in the opposite direction (v'g = -0.7 m/s) to get back to his spaceship.
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Assume that you are sitting at rest on a rotating chair holding a bicycle wheel by the axle. The wheel is vertical, and it is spinning so that the top edge is moving away from you as shown in the picture. What do you think will happen if you turn the bicycle wheel to the left into the horizontal position? What do you think will happen if you turn the wheel back to being vertical?
The angular momentum vector of a bicycle wheel changes direction when the wheel is turned horizontally, but returns to its original position when the wheel is returned to a vertical position.
When you turn the bicycle wheel to the left into the horizontal position, the axis of rotation of the wheel changes. The new axis of rotation will be perpendicular to the initial axis of rotation, so the initial spin angular momentum vector, which was pointing along the initial axis of rotation, will move at a right angle to the new axis of rotation.
It follows that if the right-hand rule is followed, the direction of the vector will change from pointing away from you to pointing left when the wheel is horizontal. When the wheel is vertical again, if the wheel is released from the horizontal position to a vertical position, its axis of rotation will change once more.
The new axis of rotation is perpendicular to both the initial axis of rotation and the axis of rotation during the time the wheel was in the horizontal position. It follows that the initial angular momentum vector, which was pointing along the initial axis of rotation, will spin back to its original position as the wheel turns.
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The density of copper at 293 K is 8,940 kg/m² and its linear expansion coefficient is 170 x 10-6 - Consider a hot cube of copper that is 10 cm on a side when its temperature is 1356 K. What is the cube's mass?
The cube's mass is approximately 8.91 kg. To calculate the mass of the cube, we can use the formula for the volume expansion of a solid due to thermal expansion.
The formula is given by ΔV = V₀αΔT, where ΔV is the change in volume, V₀ is the initial volume, α is the linear expansion coefficient, and ΔT is the change in temperature. Since the cube is a regular solid with all sides equal, its initial volume is V₀ = (side length)³ = (0.1 m)³ = 0.001 m³. The change in temperature is ΔT = 1356 K - 293 K = 1063 K. Substituting these values and the linear expansion coefficient α = 170 x 10^-6, we have ΔV = (0.001 m³)(170 x 10^-6)(1063 K) = 0.018 m³.
The density of copper is given as 8,940 kg/m³. Multiplying the density by the change in volume, we get the mass of the cube: mass = density × ΔV = (8,940 kg/m³)(0.018 m³) = 160.92 kg. Therefore, the cube's mass is approximately 8.91 kg.
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