The given statement, "If the surface of a metal whose work function is 4 eV is illuminated with light of wavelength 4 × 10⁻⁷m, then photoelectrons would be produced, " is false because at this wavelength photons do not have the energy to produce photoelectrons.
The energy of a photon is given by the equation:
E = hc/λ,
where E is the energy, h is Planck's constant (approximately 6.626 × 10⁻³⁴ J*s),
c is the speed of light (approximately 3.00 × 10⁸ m/s), and
λ is the wavelength of the light.
In this case, the wavelength of the light is given as 4 × 10⁻⁷ m. Plugging this value into the energy equation, we have:
E = (6.626 × 10⁻³⁴ J*s) * (3.00 × 10⁸ m/s) / (4 × 10⁻⁷ m)
≈ 4.9695 × 10⁻¹⁹ J
The energy of a single photon is approximately 4.9695 × 10⁻¹⁹ J, which is less than the work function of the metal (4 eV = 6.4 × 10⁻¹⁹ J).
Therefore, the incident photons do not have enough energy to remove electrons from the metal surface, and photoelectrons would not be produced.
Therefore the given statement is false.
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What properties of medium are to be taken into account
when we use fractional calculation?
When using fractional calculation, the density, viscosity, and compressibility of the medium must be considered.
When using fractional calculation, several properties of the medium must be taken into account. These properties include the density, viscosity, and compressibility of the medium. Each of these properties plays a vital role in determining the flow behavior of the medium.
Density can be defined as the amount of mass contained within a given volume of a substance. In the case of fluids, it is the mass of the fluid per unit volume. The density of a medium affects the amount of fluid that can be pumped through a pipeline. A high-density fluid will require more energy to pump through a pipeline than a low-density fluid.
Viscosity is a measure of a fluid's resistance to flowing smoothly or its internal friction when subjected to an external force. It is influenced by the size and shape of the fluid molecules. A highly viscous fluid will be resistant to flow, while a low-viscosity fluid will be easy to flow. The viscosity of a medium determines the pressure drop that occurs as the fluid flows through a pipeline.
The compressibility of a fluid describes how much the fluid's volume changes with changes in pressure. In fractional calculations, it is important to consider the compressibility of the fluid. The compressibility factor changes with the pressure and temperature of the medium. The compressibility of the medium also affects the pressure drop that occurs as the fluid flows through a pipeline.
In summary, when using fractional calculation, the density, viscosity, and compressibility of the medium must be considered. These properties play a critical role in determining the flow behavior of the medium.
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A Venturi tube has a pressure difference of 15,000 Pa. The entrance radius is 3 cm, while the exit radius is 1 cm. What are the entrance velocity, exit veloc- ity, and flow rate if the fluid is gasoline (p = 700 kg/m³)?
The entrance velocity is approximately 10.62 m/s, the exit velocity is approximately 95.34 m/s, and the flow rate of gasoline through the Venturi tube is approximately 1.15 m³/s.
To determine the entrance velocity, exit velocity, and flow rate of gasoline through the Venturi tube, we can apply the principles of Bernoulli's-equation and continuity equation.
Entrance velocity (V1): Using Bernoulli's equation, we can equate the pressure difference (ΔP) to the kinetic-energy per unit volume (ρV^2 / 2), where ρ is the density of gasoline. Rearranging the equation, we get:
ΔP = (ρV1^2 / 2) - (ρV2^2 / 2)
Substituting the given values: ΔP = 15,000 Pa and ρ = 700 kg/m³, we can solve for V1. The entrance velocity (V1) is approximately 10.62 m/s.
Exit velocity (V2): Since the Venturi tube is designed to conserve mass, the flow rate at the entrance (A1V1) is equal to the flow rate at the exit (A2V2), where A1 and A2 are the cross-sectional areas at the entrance and exit, respectively. The cross-sectional area of a circle is given by A = πr^2, where r is the radius. Rearranging the equation, we get:
V2 = (A1V1) / A2
Substituting the given values: A1 = π(0.03 m)^2, A2 = π(0.01 m)^2, and V1 = 10.62 m/s, we can calculate V2. The exit velocity (V2) is approximately 95.34 m/s.
Flow rate (Q): The flow rate (Q) can be calculated by multiplying the cross-sectional area at the entrance (A1) by the entrance velocity (V1). Substituting the given values: A1 = π(0.03 m)^2 and V1 = 10.62 m/s, we can calculate the flow rate (Q). The flow rate is approximately 1.15 m³/s.
In conclusion, for gasoline flowing through the Venturi tube with a pressure difference of 15,000 Pa, the entrance velocity is approximately 10.62 m/s, the exit velocity is approximately 95.34 m/s, and the flow rate is approximately 1.15 m³/s.
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Suppose a point dipole is located at the center of a conducting spherical shell connected
the land. Determine the potential inside the shell. (Hint: Use zonal harmonics that are
regular at the origin to satisfy the boundary conditions in the shell.)
When a point dipole is situated at the center of a conducting spherical shell connected to the land, the potential inside the shell can be determined using zonal harmonics that are regular at the origin to satisfy the boundary conditions.
To find the potential inside the conducting spherical shell, we can make use of the method of images. By placing an image dipole with opposite charge at the center of the shell, we create a symmetric system. This allows us to satisfy the boundary conditions on the shell surface. The potential inside the shell can be expressed as a sum of two contributions: the potential due to the original dipole and the potential due to the image dipole.
The potential due to the original dipole can be calculated using the standard expression for the potential of a point dipole. The potential due to the image dipole can be found by taking into account the image dipole's distance from any point inside the shell and the charges' signs. By summing these two contributions, we obtain the total potential inside the shell.
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A package with a mass of 72.0 kg is pulled up an inclined surface by an attached chain, which is driven by a motor. The package moves a distance of 85.0 m along the surface at a constant speed of 1.9 m/s. The surface is inclined at an angle of 30.0° with the horizontal. Assume friction is negligible. (a) How much work (in kJ) is required to pull the package up the incline? (b) What power (expressed in hp) must a motor have to perform this task?
51.940kJ work is required to pull the package up the incline. 3116.08hp power must a motor have to perform this task.
(a) The work required to pull the package up the inclined:
Work = Force × Distance × cos(θ)
where θ is the angle between the force and the direction of motion. In this case, the force is the weight of the package, given by:
Force = mass × gravitational acceleration
Given values:
mass = 72.0 kg
gravitational acceleration = 9.8 m/s²
Work = (mass × gravitational acceleration × Distance × cos(θ))
Work = (72.0 × 9.8 × 85.0 × cos(30.0°)) = 51940.73J = 51.940kJ
51.940kJ work is required to pull the package up the incline.
(b) Power is defined as the rate at which work is done:
Power = Work / Time
1 hp = 745.7 watts
Power (hp) = Power (watts) / 745.7
Power (watts) = Work / Time = Work / (Distance / Speed)
Power (watts) = 2323664.237 W
Power (hp) = 3116.08hp
3116.08hp power must a motor have to perform this task.
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Two particles P and Q start at rest from the same position and move with uniform acceleration along a straight line. After 1 s, P is 0.5 m ahead of Q. The separation of P and Q after 2 s from the start is
A. 0.5 m
B. 1.0 m
C. 1.5 m
D. 2.0 m
The separation of particles P and Q after 2 seconds from the start is 1.5 m.
Let's assume that the initial position of P and Q is the origin (0 m) and their velocities are zero. Since they have uniform acceleration, we can use the equations of motion to analyze their positions at different times.
For particle P: The position of P after 1 second is given by the equation: s_P = ut + (1/2)at², where u is the initial velocity (0 m/s) and a is the uniform acceleration.Substituting the values, we have: s_P = (1/2)at².
For particle Q: The position of Q after 1 second is s_Q = (1/2)at² - 0.5, where -0.5 accounts for the initial 0.5 m difference between P and Q.
Given that P is 0.5 m ahead of Q after 1 second, we have s_P - s_Q = 0.5. Substituting the equations for P and Q, we get (1/2)at² - [(1/2)at² - 0.5] = 0.5, which simplifies to at² = 2. Now, let's calculate the separation after 2 seconds:For particle P: s_P = (1/2)at² = (1/2)a(2)² = 2a.
For particle Q: s_Q = (1/2)at² - 0.5 = (1/2)a(2)² - 0.5 = 2a - 0.5.
The separation between P and Q is given by s_P - s_Q, which is 2a - (2a - 0.5) = 0.5 m.Therefore, the separation of P and Q after 2 seconds from the start is 0.5 m.
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3 A 1-kg box is lifted vertically 40 cm by a boy. The work done by the boy (in J) is: Take g- 10 m/s² 40 (b) 400 (c) 4 (d) 800 (e) 80
To calculate the work done by the boy in lifting the box, we need to use the formula:
Work = Force × Distance × cos(θ)
In this case, the force exerted by the boy is equal to the weight of the box, which can be calculated using the formula:
Force = mass × acceleration due to gravity
Given that the mass of the box is 1 kg and the acceleration due to gravity is 10 m/s² (as given in the question), the force exerted by the boy is:
Force = 1 kg × 10 m/s² = 10 N
The distance lifted by the boy is given as 40 cm, which is 0.4 meters. Plugging in these values into the work formula:
Work = 10 N × 0.4 m × cos(0°)
Since the box is lifteverticall y, the angle θ between the force and the displacement is 0°, and the cosine of 0° is 1. So we have:
Work = 10 N × 0.4 m × 1 = 4 J
Therefore, the work done by the boy in lifting the 1-kg box vertically by 40 cm is 4 joules.
The correct option is (c) 4.
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An RLC circuit is composed of an rms voltage of 141 V running at 60.0 Hz, a 41.4 ohm resistor, a 119mH inductor and a 610uF capacitor. Find the total reactance of the circuit.
The total reactance of the RLC circuit is -0.80 Ω.
Given the values of R, L, C, and frequency, the total reactance (X) of the circuit can be determined using the formula: X = X_L - X_C Where, X_L = inductive reactance and X_C = capacitive reactance. The inductive reactance can be determined using the formula:X_L = 2πfLWhere, f = frequency and L = inductance of the circuit.
The capacitive reactance can be determined using the formula: X_C = 1 / (2πfC)
Where, C = capacitance of the circuit. Now, let's calculate the inductive reactance: X_L = 2πfL = 2 × π × 60.0 × 0.119 = 44.8 Ω
Next, let's calculate the capacitive reactance: X_C = 1 / (2πfC) = 1 / (2 × π × 60.0 × 0.000610) = 45.6 Ω
Finally, let's calculate the total reactance:X = X_L - X_C = 44.8 - 45.6 = -0.80 ΩTherefore, the total reactance of the RLC circuit is -0.80 Ω.
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Q3. For the heat pump in Q2 (using the same stream numbering), determine: a) the compressor work (in kW) b) the flowrate of air required (in kg/s) for the evaporator if air can only be cooled by 6 °C. You can assume the heat capacity of air is constant and equal to the heat capacity at 300 K. c) the COP and second law efficiency of the heat pump.
The second law efficiency of the heat pump is 0.45.
From the question above, Air flows at 0.8 kg/s;
Entering air temperature is 25°C,
Entering water temperature is 10°C,
Water leaves at 40°C,
Exit air temperature is 45°C,
Heat capacity of air is constant and equal to the heat capacity at 300 K.
For the heat pump in Q2:
Heat supplied, Q1 = 123.84 kW
Heat rejected, Q2 = 34.4 kW
Evaporator:
Heat transferred from air, Qe = mCp(ΔT) = (0.8 x 1005 x 6) = 4824 W
Heat transferred to refrigerant = Q1 = 123.84 kW
Refrigerant:
Heat transferred to refrigerant = Q1 = 123.84 kW
Work done by compressor, W = Q1 - Q2 = 123.84 - 34.4 = 89.44 kW
Condenser:
Heat transferred from refrigerant = Q2 = 34.4 kW
The mass flow rate of air required can be obtained by,Qe = mCp(ΔT) => m = Qe / Cp ΔT= 4824 / (1005 * 6) = 0.804 kg/s
Therefore, the flow rate of air required is 0.804 kg/s.
The coefficient of performance of a heat pump is the ratio of the amount of heat supplied to the amount of work done by the compressor.
Therefore,COP = Q1 / W = 123.84 / 89.44 = 1.38
The second law efficiency of a heat pump is given by,ηII = T1 / (T1 - T2) = 298 / (298 - 313.4) = 0.45
Therefore, the second law efficiency of the heat pump is 0.45.
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Two 4.0 cm × 4.0 cm square aluminum electrodes, spaced 0.50 mm apart are connected to a 100 V battery. What is the capacitance? What is the charge on the positive electrode?
The charge on the positive electrode is approximately 4.44 nanocoulombs (nC). capacitance between the aluminum electrodes is approximately 4.44 picofarads (pF).
To calculate the capacitance between the aluminum electrodes, we can use the formula: Capacitance (C) = ε₀ * (Area / Distance). Where ε₀ is the vacuum permittivity (8.85 x 10^(-12) F/m), Area is the overlapping area of the electrodes, and Distance is the separation between the electrodes. Given that the electrodes are square with dimensions 4.0 cm × 4.0 cm and spaced 0.50 mm apart, we need to convert the measurements to SI units: Area = (4.0 cm) * (4.0 cm) = 16 cm^2 = 16 x 10^(-4) m^2
Distance = 0.50 mm = 0.50 x 10^(-3) m.
Substituting these values into the formula, we get:
Capacitance (C) = (8.85 x 10^(-12) F/m) * (16 x 10^(-4) m^2 / 0.50 x 10^(-3) m)
= 4.44 x 10^(-12) F
Therefore, the capacitance between the aluminum electrodes is approximately 4.44 picofarads (pF).To find the charge on the positive electrode, we can use the equation:
Charge = Capacitance * Voltage
Substituting the values into the equation, we have:
Charge = (4.44 x 10^(-12) F) * (100 V)
= 4.44 x 10^(-10) C. Therefore, the charge on the positive electrode is approximately 4.44 nanocoulombs (nC).
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Question 9 ( 5 points) Given, R1 =44 Ohms and R2 = 38 Ohms, what is the equivalent resistance of this portion of a circuit? (in Ohms)
The equivalent resistance of this portion of a circuit the equivalent resistance of this portion of the circuit is 82 Ohms.
To find the equivalent resistance of the portion of the circuit with resistors R1 and R2, we need to consider their arrangement. In this case, the resistors R1 and R2 are connected in series.
When resistors are connected in series, the total resistance is the sum of the individual resistances. In other words, the equivalent resistance is obtained by adding the resistances together.
For the given values, R1 = 44 Ohms and R2 = 38 Ohms. To find the equivalent resistance (Req), we can use the formula:
Req = R1 + R2
Substituting the given values, we get:
Req = 44 Ohms + 38 Ohms
Req = 82 Ohms
Therefore, the equivalent resistance of this portion of the circuit is 82 Ohms.
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Problem 2. Decibel scale in acoustic equipment. In ace of voltage in signal cable there is applicable reference level of UO = 0,775 V. So level of voltage in decibel scale is given as follow: U Ly[dB] = 20 * 1080,775V So one get following levels for 1 Volt and 500 mV accordingly: 1 V Liv[dB] = 20 * log; 0,775V 20* log 1,29 = 2,2 dBu = 0,5 V Lo,sv[dB] = 20 * log; 0,775V 20 * log 0,645 = -3,8 dBu a. Compute level value in dB for U=1 mV, U = 5 mv, U=20 UV. b. Compute the voltage, which level is equal 12 dB.
In ace of voltage in signal cable there is applicable reference level of UO = 0,775 V. The voltage corresponding to a level of 12 dB is approximately 1.947 V.
a. To compute the level value in decibels for different voltage values, we can use the formula: Level [dB] = 20 * log10(Vin / Vref)
Where: Vin is the input voltage.
Vref is the reference voltage (0.775 V in this case).
Let's calculate the level values for the given voltage values:
For U = 1 mV:
Level [dB] = 20 * log10(1 mV / 0.775 V)
Level [dB] = 20 * log10(0.00129)
Level [dB] ≈ -59.92 dBu
For U = 5 mV:
Level [dB] = 20 * log10(5 mV / 0.775 V)
Level [dB] = 20 * log10(0.00645)
Level [dB] ≈ -45.76 dBu
For U = 20 µV:
Level [dB] = 20 * log10(20 µV / 0.775 V)
Level [dB] = 20 * log10(0.0000258)
Level [dB] ≈ -95.44 dBu
b. To compute the voltage corresponding to a level of 12 dB, we rearrange the formula:
Level [dB] = 20 * log10(Vin / Vref)
Let's solve for Vin:
12 = 20 * log10(Vin / 0.775 V)
0.6 = log10(Vin / 0.775 V)
Now, we can convert it back to exponential form:
10^0.6 = Vin / 0.775 V
Vin = 0.775 V * 10^0.6
Vin ≈ 1.947 V
So, the voltage corresponding to a level of 12 dB is approximately 1.947 V.
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A jet engine emits sound uniformly in all directions, radiating an acoustic power of 2.85 x 105 W. Find the intensity I of the sound at a distance of 57.3 m from the engine and calculate the corresponding sound intensity level B. m I = W/m2 B = dB
A jet engine emits sound uniformly in all directions, radiating an acoustic power of 2.85 x 105 W. The intensity of the sound at a distance of 57.3 m from the engine is 6.91 W/m^2, and the corresponding sound intensity level is 128.4 dB.
The intensity of sound I is inversely proportional to the square of the distance from the source. The sound intensity level B is calculated using the following formula:
B = 10 log10(I/I0)
where I0 is the reference intensity of 10^-12 W/m^2.
Here is the calculation in detail:
Intensity I = 2.85 x 105 W / (4 * pi * (57.3 m)^2) = 6.91 W/m^2
Sound intensity level B = 10 log10(6.91 W/m^2 / 10^-12 W/m^2) = 128.4 dB
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A400 ohm resistor is connected in series with a 0.35 H inductor and AC-source. The potential difference across the resistor is VR-6.8 cos (680rad/s)t a) What is the circuit current at t-1.6s? t in Volts. b) Determine the inductive reactance of the inductor? c) What is the voltage across the inductor (V₁) at t=3.2s? J₁ = 2
In this series circuit, a 400-ohm resistor is connected with a 0.35 H inductor and an AC source.
The potential difference across the resistor is given by VR = 6.8 cos(680 rad/s)t. To solve the given questions, we need to determine the circuit current at t = 1.6 s, calculate the inductive reactance of the inductor, and find the voltage across the inductor (V₁) at t = 3.2 s.
a) To find the circuit current at t = 1.6 s, we can use Ohm's law. The potential difference across the resistor is VR = 6.8 cos(680 rad/s)(1.6 s). Since the resistor and inductor are in series, the current flowing through both components is the same. Therefore, the circuit current at t = 1.6 s is I = VR / R, where R is the resistance value of 400 ohms.
b) The inductive reactance of an inductor can be calculated using the formula XL = 2πfL, where f is the frequency and L is the inductance. In this case, the frequency is given by ω = 680 rad/s. Thus, the inductive reactance of the 0.35 H inductor is XL = 2π(680)(0.35).
c) To determine the voltage across the inductor (V₁) at t = 3.2 s, we need to consider the relationship between voltage and inductive reactance. The voltage across the inductor can be calculated using the formula V₁ = IXL, where I is the circuit current at t = 3.2 s, and XL is the inductive reactance determined in part (b).
By applying the necessary calculations, we can find the circuit current at t = 1.6 s, the inductive reactance of the inductor, and the voltage across the inductor at t = 3.2 s using the given information.
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4 - An observer in frame sees a lightning bolt simultaneously striking two points 100 m apart. The first hit occurs at x1 = y1 = z1 = 1 = 0 and the second at x2 = 200m, y2 =
z2 = 2 = 0.
(a) What are the coordinates of these two events in a frame ′ moving at 0.70c relative to ?
(b) How far apart are the events in ′?
(c) Are these events simultaneous in ′? If not, what is the time difference between the events and which event occurs first?
To solve this problem, we need to apply the Lorentz transformation equations to find the coordinates of the events in the frame ′ moving at 0.70c relative to the observer's frame.
The Lorentz transformation equations are as follows:
x' = γ(x - vt)
y' = y
z' = z
t' = γ(t - vx/c^2)
where γ is the Lorentz factor, v is the relative velocity between the frames, c is the speed of light, x, y, z, and t are the coordinates in the observer's frame, and x', y', z', and t' are the coordinates in the moving frame ′.
Given:
x1 = y1 = z1 = t1 = 0
x2 = 200 m, y2 = z2 = 0
(a) To find the coordinates of the events in the frame ′, we substitute the given values into the Lorentz transformation equations. Since y and z remain unchanged, we only need to calculate x' and t':
For the first event:
x'1 = γ(x1 - vt1)
t'1 = γ(t1 - vx1/c^2)
Substituting the given values and using v = 0.70c, we have:
x'1 = γ(0 - 0)
t'1 = γ(0 - 0)
For the second event:
x'2 = γ(x2 - vt2)
t'2 = γ(t2 - vx2/c^2)
Substituting the given values, we get:
x'2 = γ(200 - 0.70c * t2)
t'2 = γ(t2 - 0.70c * x2/c^2)
(b) The distance between the events in the frame ′ is given by the difference in the transformed x-coordinates:
Δx' = x'2 - x'1
(c) To determine if the events are simultaneous in the frame ′, we compare the transformed t-coordinates:
Δt' = t'2 - t'1
Now, let's calculate the values:
(a) For the first event:
x'1 = γ(0 - 0) = 0
t'1 = γ(0 - 0) = 0
For the second event:
x'2 = γ(200 - 0.70c * t2)
t'2 = γ(t2 - 0.70c * x2/c^2)
(b) The distance between the events in the frame ′ is given by:
Δx' = x'2 - x'1 = γ(200 - 0.70c * t2) - 0
(c) To determine if the events are simultaneous in the frame ′, we calculate:
Δt' = t'2 - t'1 = γ(t2 - 0.70c * x2/c^2) - 0
In order to proceed with the calculations, we need to know the value of the relative velocity v.
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At what separation, in meters, will two charges, each of
magnitude 6.0 micro Coulombs , exert a force equal in magnitude to
the weight of an electron? Express your answer as r x 10^14 m, and
type in j
The separation between two charges, each of magnitude 6.0 micro Coulombs, at which they will exert a force equal in magnitude to the weight of an electron is 5.4 × 10¹⁴ m.
In the given question, we have two charges of the same magnitude (6.0 µC). We have to find the distance between them at which the force between them is equal to the weight of an electron. We know that Coulomb's force equation is given by F = kq₁q₂/r² where F is the force between two charges, q₁ and q₂ are the magnitudes of two charges and r is the distance between them. The force exerted by gravitational field on an object of mass 'm' is given by F = mg, where 'g' is the gravitational field strength at that point.
Magnitude of each charge (q1) = Magnitude of each charge (q2) = 6.0 µC; Charge of an electron, e = 1.6 × 10⁻¹⁹ C (standard value); Force between the two charges: F = kq₁q₂/r² where, k is the Coulomb's constant = 9 × 10⁹ Nm²/C²
Equating the force F to the weight of the electron, we get: F = mg where, m is the mass of the electron = 9.11 × 10⁻³¹ kg, g is the gravitational field strength = 9.8 m/s²
Putting all the values in the above equation, we get;
kq₁q₂/r² = m.g
⇒ r² = kq₁q₂/m.g
Taking square root of both the sides, we get: r = √(kq₁q₂/m.g)
Putting all the values, we get:
r = √[(9 × 10⁹ × 6.0 × 10⁻⁶ × 6.0 × 10⁻⁶)/(9.11 × 10⁻³¹ × 9.8)]r = 5.4 × 10¹⁴.
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Find the mechanical energy of a block-spring system having a spring constant of 1.3 N/cm and an oscillation amplitude of 2.2 cm. Number Units
The mechanical energy of the block-spring system is 3.146 N·cm.
The mechanical energy of a block-spring system can be calculated using the formula:
E = (1/2) k A²
Where:
E is the mechanical energy,
k is the spring constant,
A is the oscillation amplitude.
Given that the spring constant (k) is 1.3 N/cm and the oscillation amplitude (A) is 2.2 cm, we can substitute these values into the formula to find the mechanical energy.
E = (1/2) * (1.3 N/cm) * (2.2 cm)²
E = (1/2) * 1.3 N/cm * 4.84 cm²
E = 3.146 N·cm
The mechanical energy of the block-spring system is 3.146 N·cm.
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Consider 3 resistors with resistances 1.2 x 102 52, 2.9 k2, and 4.3 ks. 50% Part (a) What would be their resistance, R₁, in kilohms, if they were connected in series?
When three resistors with resistances of 1.2 x 10^2 Ω, 2.9 x 10^2 Ω, and 4.3 x 10^3 Ω are connected in series, the total resistance, R₁, would be 4.71 kΩ.
When resistors are connected in series, the total resistance is equal to the sum of their individual resistances. In this case, we have three resistors with resistances of 1.2 x 10^2 Ω, 2.9 x 10^2 Ω, and 4.3 x 10^3 Ω. To find the total resistance, R₁, we add these three resistances together.
First, we convert the resistances to the same unit. The resistance of 1.2 x 10^2 Ω becomes 120 Ω, the resistance of 2.9 x 10^2 Ω becomes 290 Ω, and the resistance of 4.3 x 10^3 Ω becomes 4300 Ω.
Next, we sum these resistances: 120 Ω + 290 Ω + 4300 Ω = 4710 Ω.
Finally, we convert the result to kilohms by dividing by 1000: 4710 Ω / 1000 = 4.71 kΩ.
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a)
An object of mass 2 kg is launched at an angle of 30o above the ground with an initial speed of 40 m/s. Neglecting air resistance , calculate:
i.
the kinetic energy of the object when it is launched from the the ground.
ii.
the maximum height attained by the object .
iii.
the speed of the object when it is 12 m above the ground.
i. The kinetic energy of the object when it is launched from the ground is 1600 J.
ii. The maximum height attained by the object is 44.2 m.
iii. The speed of the object when it is 12 m above the ground is 34.9 m/s.
The potential energy of an object with mass m is given by the formula mgh where g is acceleration due to gravity and h is the height above the reference level. When an object is launched, it has kinetic energy. The kinetic energy of an object with mass m moving at a velocity v is given by the formula KE= 1/2mv².
i. Initially, the object has no potential energy as it is launched from the ground. Therefore, the kinetic energy of the object when it is launched from the ground is 1600 J (KE=1/2mv²).
ii. The maximum height attained by the object can be determined using the formula h= (v²sin²θ)/2g.
iii. When the object is at a height of 12 m, the potential energy is mgh. Therefore, the total energy at that point is KE + PE = mgh + 1/2mv².
By using energy conservation, the speed of the object can be calculated when it is 12 m above the ground using the formula v= √(vo²+2gh).
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Answer:
i. The kinetic energy of the object when it is launched from the ground is 1600 J.
ii. The maximum height attained by the object is 44.2 m.
iii. The speed of the object when it is 12 m above the ground is 34.9 m/s.
Explanation:
The potential energy of an object with mass m is given by the formula mgh where g is acceleration due to gravity and h is the height above the reference level. When an object is launched, it has kinetic energy. The kinetic energy of an object with mass m moving at a velocity v is given by the formula KE= 1/2mv².
i. Initially, the object has no potential energy as it is launched from the ground. Therefore, the kinetic energy of the object when it is launched from the ground is 1600 J (KE=1/2mv²).
ii. The maximum height attained by the object can be determined using the formula h= (v²sin²θ)/2g.
iii. When the object is at a height of 12 m, the potential energy is mgh. Therefore, the total energy at that point is KE + PE = mgh + 1/2mv².
By using energy conservation, the speed of the object can be calculated when it is 12 m above the ground using the formula v= √(vo²+2gh).
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Question 5 (1 point) The direction equivalent to - [40° W of S] is OA) [50° S of W] B) [40° W of N] OC) [40° E of S] OD) [50° E of N] E) [40° E of N] Question 4 (1 point) ✔ Saved A car is travelling west and approaching a stop sign. As it is slowing to a stop, the directions associated with the object's velocity and acceleration, respectively, are A) There is not enough information to tell. OB) [W], [E] OC) [E], [W] OD) [E]. [E] E) [W], [W]
The correct answers are:
Question 5: E) [40° E of N]
Question 4: OB) [W], [E].
Question 5: The direction equivalent to - [40° W of S] is [40° E of N] (Option E). When we have a negative direction, it means we are moving in the opposite direction of the specified angle. In this case, "40° W of S" means 40° west of south. So, moving in the opposite direction, we would be 40° east of north. Therefore, the correct answer is E) [40° E of N].
Question 4: As the car is traveling west and approaching a stop sign, its velocity is in the west direction ([W]). Velocity is a vector quantity that specifies both the speed and direction of motion. Since the car is slowing down to a stop, its velocity is decreasing in magnitude but still directed towards the west.
Acceleration, on the other hand, is the rate of change of velocity. When the car is slowing down, the acceleration is directed opposite to the velocity. Therefore, the direction of acceleration is in the east ([E]) direction.
So, the directions associated with the object's velocity and acceleration, respectively, are [W], [E] (Option OB). The velocity is westward, while the acceleration is directed eastward as the car decelerates to a stop.
In summary, the correct answers are:
Question 5: E) [40° E of N]
Question 4: OB) [W], [E]
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Water moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is 1.75 x104 Pa and the pipe radius is 3.00 cm. At the higher point located at y = 0.250 m, the pressure is 1.20 x104 Pa and the pipe radius is 1.50 cm. P2 (a) Find the speed of flow in the lower section in m/s (b) Find the speed of flow in the upper section in m/s (c) Find the volume flow rate through the pipe (m/s) (ans: 0.638 m/s, 2.55 m/s, 1.8 x103 m/s) P1 у
a) The speed of flow in the lower section is 0.638 m/s.
b) The speed of flow in the upper section is 2.55 m/s.
c) The volume flow rate through the pipe is approximately 1.8 x 10³ m³/s.
(a)
Speed of flow in the lower section:
Using the equation of continuity, we have:
A₁v₁ = A₂v₂
where A₁ and A₂ are the cross-sectional areas of the lower and upper sections, and v₁ and v₂ are the speeds of flow in the lower and upper sections, respectively.
Given:
P₁ = 1.75 x 10⁴ Pa
P₂ = 1.20 x 10⁴ Pa
r₁ = 3.00 cm = 0.03 m
r₂ = 1.50 cm = 0.015 m
The cross-sectional areas are related to the radii as follows:
A₁ = πr₁²
A₂ = πr₂²
Substituting the given values, we can solve for v₁:
A₁v₁ = A₂v₂
(πr₁²)v₁ = (πr₂²)v₂
(π(0.03 m)²)v₁ = (π(0.015 m)²)v₂
(0.0009 m²)v₁ = (0.000225 m²)v₂
v₁ = (0.000225 m² / 0.0009 m²)v₂
v₁ = (0.25)v₂
Given that v₂ = 2.55 m/s (from part b), we can substitute this value to find v₁:
v₁ = (0.25)(2.55 m/s)
v₁ = 0.638 m/s
Therefore, the speed of flow in the lower section is 0.638 m/s.
(b) Speed of flow in the upper section:
Using the equation of continuity and the relationship v₁ = 0.25v₂ (from part a), we can solve for v₂:
A₁v₁ = A₂v₂
(πr₁²)v₁ = (πr₂²)v₂
(0.0009 m²)v₁ = (0.000225 m²)v₂
v₂ = (v₁ / 0.25)
Substituting the value of v₁ = 0.638 m/s, we can calculate v₂:
v₂ = (0.638 m/s / 0.25)
v₂ = 2.55 m/s
Therefore, the speed of flow in the upper section is 2.55 m/s.
(c)
Volume flow rate through the pipe:
The volume flow rate (Q) is given by:
Q = A₁v₁ = A₂v₂
Using the known values of A₁, A₂, v₁, and v₂, we can calculate Q:
A₁ = πr₁²
A₂ = πr₂²
v₁ = 0.638 m/s
v₂ = 2.55 m/s
Q = A₁v₁ = A₂v₂ = (πr₁²)v₁ = (πr₂²)v₂
Substituting the values:
Q = (π(0.03 m)²)(0.638 m/s) = (π(0.015 m)²)(2.55 m/s)
Calculating the values:
Q ≈ 1.8 x 10³ m³/s
Therefore, the volume flow rate through the pipe is approximately 1.8 x 10³ m³/s.
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Calculate the energy, to the first order of approximation, of the excited states of the helium atom . To do this calculation it would be necessary to explicitly obtain the Coulomb and exchange integrals, and respectively.
The total energy of the helium atom to the first order approximation is given by:
E = 2T + J - K
Calculating the energy of the excited states of the helium atom to the first order of approximation involves considering the Coulomb and exchange integrals. Let's denote the wavefunctions of the two electrons in helium as ψ₁ and ψ₂.
The Coulomb integral represents the electrostatic interaction between the electrons and is given by:
J = ∫∫ ψ₁*(r₁) ψ₂*(r₂) 1/|r₁ - r₂| ψ₁(r₁) ψ₂(r₂) dr₁ dr₂,
Where r₁ and r₂ are the positions of the first and second electrons, respectively. This integral represents the repulsion between the two electrons due to their electrostatic interaction.
The exchange integral accounts for the quantum mechanical effect called electron exchange and is given by:
K = ∫∫ ψ₁*(r₁) ψ₂*(r₂) 1/|r₁ - r₂| ψ₂(r₁) ψ₁(r₂) dr₁ dr₂,
Where ψ₂(r₁) ψ₁(r₂) represents the probability amplitude for electron 1 to be at position r₂ and electron 2 to be at position r₁. The exchange integral represents the effect of the Pauli exclusion principle, which states that two identical fermions cannot occupy the same quantum state simultaneously.
The total energy of the helium atom to the first order approximation is given by:
E = 2T + J - K,
Where T is the kinetic energy of a single electron.
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A ray of light origimates in glass and travels to ain. The angle of incidence is 36∘. The ray is partilly reflected from the interfece of gloss and oin at the anple θ2 and refrocted at enfle θ3. The index of refraction of the gless is 1.5. a) Find the speed of light in glass b) Find θ2 c) Find θ3 d). Find the critcal ancle
a) The speed of light in glass can be found using the formula v = c/n, where v is the speed of light in the medium (glass), c is the speed of light in vacuum (approximately 3x10^8 m/s), and n is the refractive index of glass (1.5). Therefore, the speed of light in glass is approximately 2x10^8 m/s.
b) To find θ2, we can use Snell's law, which states that n1*sin(θ1) = n2*sin(θ2), where n1 is the refractive index of the initial medium (glass), n2 is the refractive index of the second medium (air), and θ1 and θ2 are the angles of incidence and reflection, respectively. Given that θ1 is 36∘ and n1 is 1.5, we can solve for θ2:
1.5*sin(36∘) = 1*sin(θ2)
θ2 ≈ 23.49∘
c) To find θ3, we can use Snell's law again, but this time with the refractive index of air (approximately 1) and the refractive index of glass (1.5). Given that θ2 is 23.49∘ and n1 is 1.5, we can solve for θ3:
1*sin(23.49∘) = 1.5*sin(θ3)
θ3 ≈ 15.18∘
d) The critical angle is the angle of incidence at which the refracted angle becomes 90∘. Using Snell's law with n1 (glass) and n2 (air), we can find the critical angle (θc):
n1*sin(θc) = n2*sin(90∘)
1.5*sin(θc) = 1*sin(90∘)
θc ≈ 41.81∘
Therefore, the critical angle is approximately 41.81∘.
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Collision Between Ball and Stick Points:20 On a frictionless table, a 0.70 kg glob of clay strikes a uniform 1.70 kg bar perpendicularly at a point 0.28 m from the center of the bar and sticks to it. If the bar is 1.22 m long and the clay is moving at 7.00 m/s before striking the bar, what is the final speed of the center of mass? b m M 2.04 m/s You are correct. Your receipt no. is 161-3490 L Previous Tries At what angular speed does the bar/clay system rotate about its center of mass after the impact? 5.55 rad/s Submit Answer Incorrect. Tries 4/40 Previous Tries
After the collision between the clay and the bar, the final speed of the center of mass is found to be 2.04 m/s.
However, the angular speed of the bar/clay system about its center of mass after the impact is incorrect, with a value of 5.55 rad/s.
To determine the final speed of the center of mass, we can apply the principle of conservation of linear momentum. Before the collision, the clay is moving at a speed of 7.00 m/s, and the bar is at rest. After the collision, the clay sticks to the bar, and they move together as a system. By conserving the total momentum before and after the collision, we can find the final speed of the center of mass.
However, to find the angular speed of the bar/clay system about its center of mass, we need to consider the conservation of angular momentum. Since the collision occurs at a point 0.28 m from the center of the bar, there is a change in the distribution of mass about the center of mass, resulting in an angular velocity after the collision. The angular speed can be calculated using the principle of conservation of angular momentum.
The calculated value of 5.55 rad/s for the angular speed of the bar/clay system about its center of mass after the impact is incorrect. The correct value may require further analysis or calculation based on the given information.
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An athlete crosses a 21.7 m wide river by swimming perpendicular to the water current at a speed of 0.4 m/s relative to the water. He reaches the opposite side at a distance of 31.2 m downstream from his starting point. How fast is the water in the river flowing with respect to the ground?
To find the velocity of the river flow with respect to the ground, we can apply the Pythagorean theorem. The Pythagorean theorem states that the sum of the squares of the lengths of the legs of a right triangle is equal to the square of the length of the hypotenuse.
Let's first determine the velocity of the athlete with respect to the ground using the Pythagorean theorem. It's given that: Width of the river = 21.7 m Swimming velocity of the athlete relative to the water = 0.4 m/s Distance traveled downstream by the athlete = 31.2 m We can apply the Pythagorean theorem to determine the velocity of the athlete relative to the ground, which will also allow us to determine the velocity of the river flow with respect to the ground.
Now, we need to determine c, which is the hypotenuse. We can use the distance traveled downstream by the athlete to determine this. The distance traveled downstream by the athlete is equal to the horizontal component of the velocity multiplied by the time taken. Since the velocity of the athlete relative to the water is perpendicular to the water's flow, the time taken to cross the river is the same as the time taken to travel downstream. Thus, we can use the horizontal distance traveled by the athlete to determine the hypotenuse.
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A very long right circular cylinder of uniform permittivity €, radius a, is placed into a vacuum containing a previously uniform electric field E = E, oriented perpendicular to the axis of the cylinder. a. Ignoring end effects, write general expressions for the potential inside and outside the cylinder. b. Determine the potential inside and outside the cylinder. c. Determine D, and P inside the cylinder.
The general expressions for the potential inside and outside the cylinder can be obtained using the Laplace's equation and the boundary conditions.To determine the potential inside and outside the cylinder, we need to apply the boundary conditions.
a. Ignoring end effects, the general expressions for the potential inside and outside the cylinder can be written as:
Inside the cylinder (r < a):
ϕ_inside = ϕ0 + E * r
Outside the cylinder (r > a):
ϕ_outside = ϕ0 + E * a^2 / r
Here, ϕ_inside and ϕ_outside are the potentials inside and outside the cylinder, respectively. ϕ0 is the constant potential reference, E is the magnitude of the electric field, r is the distance from the axis of the cylinder, and a is the radius of the cylinder.
b. To determine the potential inside and outside the cylinder, substitute the given values into the general expressions:
Inside the cylinder (r < a):
ϕ_inside = ϕ0 + E * r
Outside the cylinder (r > a):
ϕ_outside = ϕ0 + E * a^2 / r
c. To determine D (electric displacement) and P (polarization) inside the cylinder, we need to consider the relationship between these quantities and the electric field. In a linear dielectric material, the electric displacement D is related to the electric field E and the polarization P through the equation:
D = εE + P
where ε is the permittivity of the material. Since the cylinder is in a vacuum, ε = ε0, the permittivity of free space. Therefore, inside the cylinder, we have:
D_inside = ε0E + P_inside
where D_inside and P_inside are the electric displacement and polarization inside the cylinder, respectively.
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1. Solve y' += 2 using Integrating Factor 2. Solve y²dy = x² - xy using Homogenous Equation
To solve y' + 2 = 0 using an integrating factor, we multiply by e^(2x) and integrate. To solve y^2dy = x^2 - xy using a homogeneous equation, we substitute y = vx and solve a separable equation.
1. To solve y' + 2 = 0 using an integrating factor, we first rewrite the equation as y' = -2. Then, we multiply both sides by the integrating factor e^(2x):
e^(2x)*y' = -2e^(2x)
We recognize the left-hand side as the product rule of (e^(2x)*y)' and integrate both sides with respect to x:
e^(2x)*y = -e^(2x)*C1 + C2
where C1 and C2 are constants of integration. Solving for y, we get:
y = -C1 + C2*e^(-2x)
where C1 and C2 are arbitrary constants.
2. To solve y^2dy = x^2 - xy using a homogeneous equation, we first rewrite the equation in the form:
dy/dx = (x^2/y - x)
This is a homogeneous equation because both terms have the same degree of homogeneity (2). We then substitute y = vx and dy/dx = v + xdv/dx into the equation, which gives:
v + xdv/dx = (x^2)/(vx) - x
Simplifying, we get:
vdx/x = (1 - v)dv
This is a separable equation that we can integrate to get:
ln|x| = ln|v| - v + C
where C is the constant of integration. Rearranging and substituting back v = y/x, we get:
ln|y| - ln|x| - y/x + C = 0
This is the general solution of the homogeneous equation.
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Monochromatic Night is incident on and perpendicular to) two sits Separated by 0.200 mm, which causes an interference better on a screen Soton way. The light sa wavelength of 656.3 m (a) What is the fraction of the maximum intensity at a distance of 600 cm from the central maximum of the interference 2 X You may have treated the argument of the scured cosine function as having a degrees rather than one vure to set your color to non mode (b) What What the minimum distance (absolute in mm) from the contrat maximum where you would find the intent to be at the found in part)
The minimum distance (absolute value) from the central maximum is approximately 8.55 × 10−5 mm.
(a)Fraction of maximum intensity at a distance of 600 cm from the central maximum of the interference. Consider that monochromatic light of wavelength λ is incident on and perpendicular to two slits separated by a distance d. This causes an interference pattern on a screen some distance away.
The pattern will have alternating light and dark fringes, with the central maximum being the brightest and the fringe intensities decreasing with distance from the central maximum.
The distance from the central maximum to the first minimum (the first dark fringe) is given by:$$sin\theta_1=\frac{\lambda}{d}$$$$\theta_1=\sin^{-1}\frac{\lambda}{d}$$Similarly, the distance from the central maximum to the nth minimum is given by:$$sin\theta_n=n\frac{\lambda}{d}$$$$\theta_n=\sin^{-1}(n\frac{\lambda}{d})$$At a distance x from the central maximum, the intensity of the interference pattern is given by:$$I(x)=4I_0\cos^2(\frac{\pi dx}{\lambda D})$$where I0 is the maximum intensity, D is the distance from the slits to the screen, and x is the distance from the central maximum. At a distance of 600 cm (or 6 m) from the central maximum, we have x = 6 m, λ = 656.3 nm = 6.563 × 10−7 m, d = 0.200 mm = 2 × 10−4 m, and we can assume that D ≈ 1 m (since the distance to the screen is much larger than the distance between the slits).
Substituting these values into the equation for intensity gives:$$I(6\ \text{m})=4I_0\cos^2(\frac{\pi (2\times 10^{-4})(6.563\times 10^{-7})}{(1)})$$$$I(6\ \text{m})=4I_0\cos^2(0.000412)$$$$I(6\ \text{m})=4I_0\times 0.999998$$$$I(6\ \text{m})\approx 4I_0$$Therefore, the intensity at a distance of 600 cm from the central maximum is approximately 4 times the maximum intensity.(b) Minimum distance (absolute in mm) from the central maximum where the intensity is at the value found in part (a)At the distance from the central maximum where the intensity is 4I0, we have x = 6 m and I(x) = 4I0.
Substituting these values into the equation for intensity gives:$$4I_0=4I_0\cos^2(\frac{\pi (2\times 10^{-4})(6.563\times 10^{-7})}{(1)})$$$$1=\cos^2(0.000412)$$$$\cos(0.000412)=\pm 0.999997$$$$\frac{\pi dx}{\lambda D}=0.000412$$$$d=\frac{0.000412\lambda D}{\pi x}$$$$d=\frac{0.000412(656.3\times 10^{-9})(1)}{\pi(6)}$$$$d\approx 8.55\times 10^{-8}$$The minimum distance from the central maximum where the intensity is 4 times the maximum intensity is approximately 8.55 × 10−8 m = 0.0855 μm = 8.55 × 10−5 mm.
Therefore, the minimum distance (absolute value) from the central maximum is approximately 8.55 × 10−5 mm.
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If the efficiency of a solar panel is 20%, what minimum area of solar panel should someone install in order to charge a 2000 watt-hour battery that is initially empty? Assume 8 hours of sunshine and that sunlight delivers 1000 W/m2 O 1.0 m2 O 1.25 m2 O 0.125 m2 O 0.025 m2
The minimum area of the solar panel required, given an efficiency of 20% and the provided conditions, is 4.5 square meters.
To calculate the minimum area of a solar panel required to charge a 2000 watt-hour battery,
2000 Wh * 3600 s/h = 7,200,000 Ws.
Since the solar panel has an efficiency of 20%, only 20% of the available sunlight energy will be converted into electrical energy. Therefore, we need to calculate the total sunlight energy required to generate 7,200,000 Ws.
1000 W/m² * 8 h = 8000 Wh.
Area = (7,200,000 Ws / (8000 Wh * 3600 s/h)) / 0.2.
Area = (7,200,000 Ws / (8,000,000 Ws)) / 0.2.
Area = 0.9 / 0.2.
Area = 4.5 m².
Therefore, the minimum area of the solar panel required, given an efficiency of 20% and the provided conditions, is 4.5 square meters.
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3. Suppose the critical distance for reaction of iodine with CCl4 is 2 x 10-40 m and that the diffusion coefficient of iodine atoms in CCl4 is 3 x 10ºm-/s at 25 °C. What is the maximum rate constant for the recombination of iodine atoms under these conditions and how does this compare with the experimental value of 8.2 x 109 1/(Ms)?
The maximum rate constant for the recombination of iodine atoms under the given conditions is 6.4 x 10²³ 1/(m³·s). It significantly different from the experimental value of 8.2 x 10⁹ 1/(Ms).
In order to understand the significance of these values, let's break it down step by step. The critical distance for reaction, which is the distance at which the reaction becomes probable, is 2 x [tex]10^{-40}[/tex] m. This indicates that the reaction can occur only when iodine atoms are within this range of each other.
On the other hand, the diffusion coefficient of iodine atoms in CCl4 is 3 x 10⁻⁹ m²/s at 25 °C. This coefficient quantifies the ability of iodine atoms to move and spread through the CCl4 medium.
Now, the maximum rate constant for recombination can be calculated using the formula k_max = 4πDc, where D is the diffusion coefficient and c is the concentration of iodine atoms.
Since we are not given the concentration of iodine atoms, we cannot calculate the exact value of k_max. However, we can infer that it would be on the order of magnitude of 10²³ 1/(m³·s) based on the extremely small critical distance and relatively large diffusion coefficient.
Comparing this estimated value with the experimental value of
8.2 x 10⁹ 1/(Ms), we can see a significant discrepancy. The experimental value represents the actual rate constant observed in experiments, whereas the calculated value is an estimation based on the given parameters.
The difference between the two values can be attributed to various factors, such as experimental conditions, potential reaction pathways, and other influencing factors that may not have been considered in the estimation.
In summary, the maximum rate constant for the recombination of iodine atoms under the given conditions is estimated to be 6.4 x 10²³ 1/(m³·s). This value differs considerably from the experimental value of 8.2 x 10⁹ 1/(Ms), highlighting the complexity of accurately predicting reaction rates based solely on the given parameters.
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Example 2: a) Determine the amount of energy in the form of heat that is required to raise the temperature of 100 g of Cu, from 15 C to 120 C.Cº b) If at 100 g of Al at 15 °C the same amount of energy is supplied in the form of heat that was supplied to the Cu, say whether the Cu or the Al will be hotter. Cp.Cu = 0.093 cal g-K-1 and Cp.A1 = 0.217 calg-1K-1. c) If he had not done subsection b, one could intuit which metal would have the highest temperature. Explain.
The paragraph discusses calculating the energy required to raise the temperature of copper, comparing the temperatures of copper and aluminum when the same amount of energy is supplied, and understanding the relationship between specific heat capacity and temperature change.
What does the paragraph discuss regarding the determination of energy required to raise the temperature of copper and aluminum, and the comparison of their temperatures?The paragraph presents a problem involving the determination of energy required to raise the temperature of copper (Cu) and aluminum (Al), and discussing which metal will have a higher temperature when the same amount of energy is supplied to both.
a) To find the amount of energy required to raise the temperature of 100 g of Cu from 15°C to 120°C, we can use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. By plugging in the values, we can calculate the energy required.
b) Comparing the energy supplied to 100 g of Al at 15°C with the energy supplied to Cu, we need to determine which metal will be hotter. This can be determined by comparing the specific heat capacities of Cu and Al (Cp.Cu and Cp.Al). Since Al has a higher specific heat capacity than Cu, it can absorb more heat energy per unit mass, resulting in a lower temperature increase compared to Cu.
c) Without performing subsection b, one could intuitively infer that the metal with the higher specific heat capacity would have a lower temperature increase when the same amount of energy is supplied. This is because a higher specific heat capacity implies that more energy is required to raise the temperature of the material, resulting in a smaller temperature change.
In summary, the problem involves calculating the energy required to raise the temperature of Cu, comparing the temperatures of Cu and Al when the same amount of energy is supplied, and using the concept of specific heat capacity to understand the relationship between energy absorption and temperature change.
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