The magnitude of A+B is approximately 40.5.
To calculate the magnitude of A+B, we need to add the two vectors A and B. Since the vectors are given in polar form, we can convert them to Cartesian coordinates and then add the corresponding components.
For vector A, the length is 20.0 and the counter-clockwise angle with the positive z-axis is 30°. Using trigonometry, we can find the x and y components of vector A. The x-component is given by 20.0 * cos(30°) = 17.32, and the y-component is given by 20.0 * sin(30°) = 10.00.
For vector B, the length is 30.0 and the counter-clockwise angle with the positive z-axis is 140°. Again, using trigonometry, we can determine the x and y components of vector B. The x-component is 30.0 * cos(140°) = -13.92, and the y-component is 30.0 * sin(140°) = 25.89.
Now, we can add the x and y components of A and B. Adding the x-components, we get 17.32 + (-13.92) = 3.40. Adding the y-components, we have 10.00 + 25.89 = 35.89.
To find the magnitude of A+B, we use the Pythagorean theorem. The magnitude is given by √(3.40²+ 35.89²) ≈ 40.5.
Therefore, the magnitude of A+B is approximately 40.5.
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250g of Aluminum at 120°C was placed into 2kg of water at 25°C. What is the final temperature of the mixture?
A. The final temperature of the mixture is approximately 29.5°C.
To determine the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the aluminum will be equal to the heat gained by the water. We can use the formula:
Q = m × c × ΔT
Where:
Q is the heat transfer
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
For the aluminum:
Q_aluminum = m_aluminum × c_aluminum × ΔT_aluminum
For the water:
Q_water = m_water × c_water × ΔT_water
Since the heat lost by the aluminum is equal to the heat gained by the water, we have:
Q_aluminum = Q_water
m_aluminum × c_aluminum × ΔT_aluminum = m_water × c_water × ΔT_water
Substituting the given values:
(0.25 kg) × (0.897 J/g°C) × (T_final - 120°C) = (2 kg) × (4.18 J/g°C) × (T_final - 25°C)
Simplifying the equation and solving for T_final:
0.25 × 0.897 × T_final - 0.25 × 0.897 × 120 = 2 × 4.18 × T_final - 2 × 4.18 × 25
0.22425 × T_final - 26.91 = 8.36 × T_final - 208.8
8.36 × T_final - 0.22425 × T_final = -208.8 + 26.91
8.13575 × T_final = -181.89
T_final ≈ -22.4°C
Since the final temperature cannot be negative, it means there might be an error in the calculation or the assumption that the heat lost and gained are equal may not be valid.
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A womanstands on a scale in a moving elevator. Her mass is 56.0 kg, and the combined mass of the elevator and scale is an additional 825 kg. Starting from rest, the elevator accelerates upward. During the acceleration, the hoisting cable applies a force of 9850 N. What does the scale read (in N) during the acceleration?
The scale reading during the acceleration is 150
Given data: Mass of woman, m1 = 56.0 kg
Mass of elevator and scale, m2 = 825 kg
Net force, F = 9850 N, Acceleration, a =?
The equation of motion for the elevator and woman is given as F = (m1 + m2) a
The net force applied to the system is equal to the product of the total mass and the acceleration of the system.
The elevator and woman move upwards so we will take the acceleration as positive.
F = (m1 + m2) a9850 = (56.0 + 825) a9850 = 881a a = 9850/881a = 11.17 m/s²
Now, the scale reading is equal to the normal force acting on the woman.
The formula to calculate the normal force is N = m1 where g is the acceleration due to gravity.
N = (56.0 kg) (9.8 m/s²)N = 549.8 N
When the elevator starts accelerating upward, the woman feels heavier than her actual weight.
The normal force is greater than the weight of the woman.
Thus, the scale reading will be the sum of the normal force and the force due to the acceleration of the system.
Scale reading during acceleration = N + m1 a
Scale reading during acceleration = 549.8 + (56.0 kg) (11.17 m/s²)
Scale reading during acceleration = 1246.8 N
Therefore, the scale reading during the acceleration is 150
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A 4.9-kg block of ice at -1.5 ∘C slides on a horizontal surface with a coefficient of kinetic friction equal to 0.069. The initial speed of the block is 7.6 m/s and its final speed is 4.1 m/s. Part A Assuming that all the energy dissipated by kinetic friction goes into melting a small mass m of the ice, and that the rest of the ice block remains at -1.5 ∘C , determine the value of m . Express your answer using two significant figures in kg.
The value of m(mass of the block) is 0.0465 kg, expressed using two significant figures.
According to the conservation of energy, the loss of kinetic energy is equal to the gain in internal energy, and here, this internal energy gain is the melting of a small mass of the ice. Let us calculate the loss of kinetic energy of the block.
Using conservation of energy, the work done by the force of friction on the block is used to melt the ice.
W= -ΔK= ΔU=-mLf
The work done by the force of friction on the block is the product of the force of friction and the distance traveled by the block.
W = ffd
= μmgd
= μmgΔx
Where μ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and Δx is the distance traveled by the block.
Substituting the given values,
W = μmgΔx
= 0.069 × 4.9 × 9.8 × 27
= 15.45 kJ
This work done by the force of friction causes the melting of a small mass of ice, which can be calculated as follows:
m = -W / Lf
= -15.45 × 1000 / 333000
= 0.0465 kg
Therefore, the value of m is 0.0465 kg, expressed using two significant figures.
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Determine the magnitudes of the currents in each resistor shown in the figure. Consider the circuit shown in have emfs of E1=9.0 V and E2=12.0 V atteries resistors have values of R1=24Ω,R2=65Ω, and R3=34Ω. Figure 1 of 1 Part B Determine the directions of the currents in each resistor. Ignore internal resistance of the batteries. I1 left, I2 right, I3 down I1 right, I2 left, I3 down I1 left, I2 right, I3 up I1 right, I2 left, I3 up
We can see that it is a combination of both series and parallel circuits. The current is given as follows:\[I=\frac{E}{R}\]Now, applying Kirchhoff's Voltage Law in the given circuit we can write:
[tex]\[E_{1}-I_{1}R_{1}-I_{3}R_{3}=0\]And \[E_{2}-I_{2}R_{2}-I_{3}R_{3}=0\][/tex]
Here, I3 is the current flowing from the point where two batteries are connected. The current is in the downward direction through R3 resistor. In the given circuit, the current passing through
R1 and R2 are:
[tex]\[I_{1}=\frac{E_{1}}{R_{1}}\][/tex]
[tex][I_{1}=\frac{9}{24}\] = 0.375 A[/tex]
And
[tex]\[I_{2}=\frac{E_{2}}{R_{2}}\][/tex]
[tex]\[I_{2}=\frac{12}{65}\] = 0.185[/tex]
The magnitudes of the currents in each resistor are:I1 = 0.375 AI2 = 0.185 AI3 = 0.105 A Determine the directions of the currents in each resistor. Ignore internal resistance of the batteries. In resistor R1, the current is flowing from left to right because the potential is higher at point A.
In resistor R2, the current is flowing from right to left because the potential is higher at point C the direction of the current in R2 is right to left.
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Briefly explain each of the following relativity phenomena
1. Time dilation
2. Length contraction
Time dilation is the phenomenon in which time passes at different rates for observers in different frames of reference. Length contraction is the phenomenon in which the length of an object appears to be shorter in a frame of reference that is moving relative to the object.
Time dilation
Time dilation is a consequence of the special theory of relativity, which was developed by Albert Einstein in the early 20th century. The theory states that the laws of physics are the same in all inertial frames of reference, which are frames of reference that are not accelerating.
One of the consequences of this principle is that time passes at different rates for observers in different frames of reference. This is because the speed of light is the same in all frames of reference.
This can lead to some strange effects, such as the fact that a clock in a moving frame of reference will appear to run slower than a clock in a stationary frame of reference.
The amount of time dilation that occurs depends on the relative velocity of the two frames of reference. The closer the relative velocity is to the speed of light, the greater the time dilation will be.
Length contraction
Length contraction is also a consequence of the special theory of relativity. It is the phenomenon in which the length of an object appears to be shorter in a frame of reference that is moving relative to the object.
The amount of length contraction that occurs depends on the relative velocity of the two frames of reference. The closer the relative velocity is to the speed of light, the greater the length contraction will be.
Time dilation and length contraction are two of the most important predictions of the special theory of relativity. They have been experimentally verified to a high degree of accuracy, and they provide strong evidence that the theory is correct.
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The purest way to do un inverse square law experiment would Be to take sound intensiry level measurements in an anechoic chamber where mom reflections wont talloet die rosults. Suppose you stand 3 incluss Gor a speaker playing a sound und my dB
meter reads 62 dis. ( (5) What is the intensity of this sound in Wit?
(10) Find the intensity and dB level at a distance of 1 m from the same speaker.
5. At 3 inches from the speaker: Intensity ≈ 1.59 x 10^(-6) watts.
10. At 1 meter from the speaker: Intensity ≈ 9.25 x 10^(-9) watts, dB level ≈ 37.58 dB.
To calculate the intensity of the sound in watts and the dB level at different distances from the speaker, we can use the inverse square law for sound propagation. The inverse square law states that the intensity of sound decreases with the square of the distance from the source.
Given:
Distance from the speaker (D1) = 3 inches (0.0762 meters)dB reading at D1 = 62 dBFirst, let's calculate the intensity (I1) in watts at a distance of 3 inches (0.0762 meters) from the speaker:
I1 = 10^((dB - 120) / 10)
= 10^((62 - 120) / 10)
= 10^(-5.8)
≈ 1.59 x 10^(-6) watts
Now, let's proceed to the next part of the question:
Distance from the speaker (D2) = 1 meter
We need to find the intensity (I2) and the dB level at this distance.
Using the inverse square law, we can calculate the intensity (I2) at a distance of 1 meter:
I2 = I1 * (D1 / D2)^2
= (1.59 x 10^(-6) watts) * ((0.0762 meters / 1 meter)^2)
= (1.59 x 10^(-6)) * (0.0762^2)
≈ 9.25 x 10^(-9) watts
To find the dB level at a distance of 1 meter, we can use the formula:
dB = 10 * log10(I / I0)
where I is the intensity and I0 is the reference intensity (usually taken as 10^(-12) watts).
dB2 = 10 * log10(I2 / I0)
= 10 * log10((9.25 x 10^(-9)) / (10^(-12)))
= 10 * log10(9.25 x 10^3)
≈ 37.58 dB
Therefore, the answers to the given questions are:
(5) The intensity of the sound at a distance of 3 inches from the speaker is approximately 1.59 x 10^(-6) watts.
(10) The intensity of the sound at a distance of 1 meter from the speaker is approximately 9.25 x 10^(-9) watts, and the corresponding dB level is approximately 37.58 dB.
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Electric field due to a 4.3nC charge at a distance d is 211.7 N/C. What would be the magnitude of the electric field at a distance 2 d from the charge?
The magnitude of the electric field at a distance 2d from the 4.3nC charge is approximately 1.756 × 10^10 N/C.
The magnitude of the electric field at a distance 2d from a 4.3nC charge can be calculated using Coulomb's Law. Given that the electric field at a distance d is 211.7 N/C, we can determine the electric field at 2d by understanding the inverse square relationship between distance and electric field strength
By doubling the distance, the electric field magnitude decreases by a factor of four. According to Coulomb's Law, the electric field due to a point charge can be calculated using the formula E = k * Q / r^2, where E is the electric field magnitude, k is the electrostatic constant (approximately 9 × 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge.
Given that the electric field at distance d is 211.7 N/C and the charge is 4.3nC (4.3 × 10^(-9) C), we can solve for the initial distance d using the formula E = k * Q / r^2:
211.7 = (9 × 10^9) * (4.3 × 10^(-9)) / d^2
Solving this equation, we find that d ≈ 0.175 m.
To determine the electric field at a distance 2d, we substitute 2d for r in the formula and solve for E:
E = (9 × 10^9) * (4.3 × 10^(-9)) / (2d)^2
Since (2d)^2 = 4 * d^2, we can simplify the equation as follows:
E = (9 × 10^9) * (4.3 × 10^(-9)) / (4 * d^2)
= (9 × 10^9) * (4.3 × 10^(-9)) / 4d^2
= 2.15 × 10^9 / d^2
Therefore, at a distance 2d, the magnitude of the electric field will be 2.15 × 10^9 / d^2 N/C.
Since the distance d was calculated to be approximately 0.175 m, the distance 2d will be 2 * 0.175 = 0.35 m.
Substituting this value into the equation, we get:
E = 2.15 × 10^9 / (0.35)^2
= 2.15 × 10^9 / 0.1225
≈ 1.756 × 10^10 N/C
Therefore, the magnitude of the electric field at a distance 2d from the 4.3nC charge is approximately 1.756 × 10^10 N/C.
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A rock band playing an outdoor concert produces sound at 80 dB, 45 m away from their single working loudspeaker. What is the power of this speaker? 1.5 W 2.5 W 15 W 25 W 150 W 250 W none of the above
The power of the speaker is approximately 8.27 W. None of the given answer choices match this result.
To calculate the power of the speaker, we need to use the inverse square law for sound intensity. The sound intensity decreases with distance according to the inverse square of the distance. The formula for sound intensity in decibels (dB) is:
Sound Intensity (dB) = Reference Intensity (dB) + 10 × log10(Intensity / Reference Intensity)
In this case, the reference intensity is the threshold of hearing, which is 10^(-12) W/m^2.
We can rearrange the formula to solve for the intensity:
Intensity = 10^((Sound Intensity (dB) - Reference Intensity (dB)) / 10)
In this case, the sound intensity is given as 80 dB, and the distance from the speaker is 45 m.
Using the inverse square law, the sound intensity at the distance of 45 m can be calculated as:
Intensity = Intensity at reference distance / (Distance)^2
Now let's calculate the sound intensity at the reference distance of 1 m:
Intensity at reference distance = 10^((Sound Intensity (dB) - Reference Intensity (dB)) / 10)
= 10^((80 dB - 0 dB) / 10)
= 10^(8/10)
= 10^(0.8)
≈ 6.31 W/m^2
Now let's calculate the sound intensity at the distance of 45 m using the inverse square law:
Intensity = Intensity at reference distance / (Distance)^2
= 6.31 W/m^2 / (45 m)^2
≈ 0.00327 W/m^2
Therefore, the power of the speaker can be calculated by multiplying the sound intensity by the area through which the sound spreads.
Power = Intensity × Area
Since the area of a sphere is given by 4πr^2, where r is the distance from the speaker, we can calculate the power as:
Power = Intensity × 4πr^2
= 0.00327 W/m^2 × 4π(45 m)^2
≈ 8.27 W
Therefore, the power of the speaker is approximately 8.27 W. None of the given answer choices match this result.
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What volume of water at 0∘C∘C can a freezer make into ice cubes
in 1.0hh, if the coefficient of performance of the cooling unit is
6.0 and the power input is 1.8 kilowatt?
Express your answer to t
The volume of water at 0°C which a freezer can turn into ice cubes in 1.0 h is 0.116 m³.
In this question, we are required to determine the volume of water at 0°C which a freezer can turn into ice cubes in 1.0 h, given the coefficient of performance of the cooling unit as 6.0 and the power input as 1.8 kW.
The heat extracted from the freezer, Q1 is given by:
Q1 = Coefficient of Performance x Power input
= 6.0 x 1.8 kW
= 10.8 kWh
The latent heat of fusion of ice is 336,000 J/kg, and this is the amount of energy required to freeze 1 kg of water into ice at 0°C.
We know that:
1 kWh = 3,600,000 J
10.8 kWh = 10.8 x 3,600,000 J= 38,880,000 J
Therefore, the mass of water that can be frozen is given by:
Q2 = mL,
where L is the latent heat of fusion of water
m = Q2 / L
L = Q2 / (m x C)
where C is the specific heat of water, which is 4,186 J/kg.K
Substituting values:
Q2 = 38,880,000 J
L = 336,000 J/kg,
C = 4,186 J/kg.K,
we have:
m = Q2 / L
m = (38,880,000 J) / (336,000 J/kg)
m = 115.71 kg
The density of water is 1000 kg/m³, so the volume of water, V is given by:
V = m / ρ
V = 115.71 kg / 1000 kg/m³= 0.11571 m³
Therefore, the volume of water at 0°C which a freezer can turn into ice cubes in 1.0 h is 0.116 m³.(Expressed to 3 significant figures).
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ASAP
If it takes 40 J of energy to heat a block from 10° to 25°C, what is the specific heat of the material? (m = 8g) O 0.33J/g C O 1.66J/g C O 1.33J/C
To find the specific heat of the material, we can use the equation Q = mcΔT, where Q is the energy transferred, m is the mass of the material, c is the specific heat, and ΔT is the change in temperature. Rearranging the equation, we can solve for c.
The specific heat of a material represents the amount of heat energy required to raise the temperature of a given mass of the material by one degree Celsius.
In this problem, we are given the energy transfer (Q) of 40 J, the mass (m) of 8 g, and the change in temperature (ΔT) of 25°C - 10°C = 15°C.
Using the equation Q = mcΔT, we can substitute the given values and solve for the specific heat (c). Rearranging the equation, we have c = Q / (mΔT).
Substituting the values, we have c = 40 J / (8 g * 15°C).
Calculating the specific heat, we find c = 0.33 J/g°C.
Therefore, the specific heat of the material is 0.33 J/g°C.
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A millisievert is equivalent to
A) I rem B) 0.1 rem
: D) 0.001 re C) 0.01 rem
A millisievert is equivalent to 0.1 rem. A rem is an acronym for Roentgen equivalent man, and it is used to measure the dosage of radiation in humans.
A millisievert, abbreviated as mSv, is a measure of the amount of radiation that a person is exposed to. It is a measure of the dose of ionizing radiation in the International System of Units (SI).The millirem (mrem) is a unit of measurement that is used in the United States of America to measure radiation exposure in humans. One rem is equivalent to 1000 millirems (mrem), while one millisievert (mSv) is equal to 100 rem or 100000 millirems. Therefore, one millirem is equal to 0.001 rem. When we convert this to millisieverts, we get one millisievert is equivalent to 0.1 rem.
So the answer to the question is B) 0.1 rem.The millisievert unit is used globally to calculate the dose of ionizing radiation in a person. The value of radiation dose that is considered acceptable varies depending on the country and the purpose of exposure. It is important to be aware of the risks associated with exposure to ionizing radiation to maintain good health.Thus, the answer to the given question is option B) 0.1 rem.A millisievert is a measure of the amount of radiation that a person is exposed to, which is used in the International System of Units (SI). A millirem (mrem) is a unit of measurement used in the United States to quantify radiation exposure in humans.One rem is equivalent to 1000 millirems (mrem), or 100000 millirems is equivalent to 1 millisievert (mSv). As a result, 0.001 rem is equivalent to 1 millirem (mrem), and 0.1 rem is equivalent to 1 millisievert (mSv).
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1. Which indicates the vertical component of a sound wave?
A. Amplitude
B. Direction
C. Frequency
D. Speed
2. Which term is synonymous to "Pitch"?
A. Amplitude
B. Direction
C. Frequency
D. Speed
Answer:
1.) Amplitude (How loud something is)
2.) Frequency
1 kg of silver (c = 234 J/kg K) is heated to 100°C. It is then dropped into 1 kg of water (c = 4190 J/kg K) at 0°C in an insulated beaker. Determine the common temperature in °C when the water and silver reach thermal equilibrium.
The common temperature when the silver and water reach thermal equilibrium is approximately -150.42°C.
To find the common temperature when the silver and water reach thermal equilibrium, we can use the principle of energy conservation. The heat lost by the silver is equal to the heat gained by the water.
The heat lost by the silver can be calculated using the formula:
Qsilver = m × csilver × ∆Tsilver
where m is the mass, csilver is the specific heat capacity of silver, and ∆Tsilver is the temperature change of the silver.
The heat gained by the water can be calculated using the formula:
Qwater = m × cwater × ∆T_water
where cwater is the specific heat capacity of water, and ∆T_water is the temperature change of the water.
Since the system is insulated, the heat lost by the silver is equal to the heat gained by the water:
Qsilver = Qwater
m × csilver × ∆Tsilver = m × cwater × ∆T_water
Simplifying the equation:
csilver × ∆Tsilver = cwater × ∆T_water
∆Tsilver / ∆T_water = cwater / csilver
∆Tsilver = (∆T_water × cwater) / csilver
∆Tsilver = (0°C - 100°C) × 4190 J/kg K / 234 J/kg K
∆Tsilver = -150.42°C
The change in temperature of the silver is -150.42°C.
To find the common temperature, we need to subtract this change in temperature from the initial temperature of the water:
Common temperature = 0°C - (-150.42°C)
Common temperature ≈ 150.42°C
Therefore, the common temperature when the silver and water reach thermal equilibrium is approximately 150.42°C.
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Following the rules of significant digits, which of the following is the correct answer for the following calculation: 19.58 m x 3.15 m = ?
61.677 m2
61.68 m2
61.7 m2
62 m2
we round down to 1.8, which gives us 61.8 m² as the final answer.
Since the product should have four significant figures, round the answer to 61.8 m². This is because the last significant figure in the answer is 3, which is less than 5.
Significant figures or digits are the number of meaningful digits in a number. The following calculation is being carried out using significant figures: 19.58 m x 3.15 m = ? To follow the rules of significant digits, we need to identify the least number of significant figures in the equation. In this case, we have two , factors 19.58 m and 3.15 m. Since both factors have four significant figures, the product should also have four significant figures.
Therefore, the correct answer is 61.8 m². To get the answer, multiply the two factors as follows:
19.58 m × 3.15 m = 61.743 m²
This is because the last significant figure in the answer is 3, which is less than 5.
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Answer:
The answer is 61.7 m^2.
Explanation:
To solve this problem, you need to look at the numbers 19.58 and 3.15. 3.15 has the least value, so you use the amount of digits it has. It has three digits, so you know the answer to the multiplication problem will also have three digits.
Right away, you may realize that rules out all but one answer choice. We still need to check it though to make sure it lines up.
19.58 *3.15
= 61.677
Now, because we know the answer can only have three digits, we need to round the six after the decimal point. Seven is more than five, so the six gets bumped up to a 7. Everything after the newly created 7 turns to zeros and are forgotten.
Now, we have 61.7 m^2.
So, in short, the answer is 61.7 m^2
Two blocks with masses 0.325 kg (A) and 0.884 kg (B) sit on a frictionless surface. Between them is a spring with spring constant 28.5 N/m, which is not attached to either block The two blocks are pushed together, compressing the spring by 0.273 meter, after which the system is released from rest. What is the final speed of the block A? (Hint: you will need to use both conservation of energy and conservation of momentum to solve this problem).
The final speed of block A is approximately 1.48 m/s. To determine the final speed of block A, we can apply the principles of conservation of mechanical energy.
First, let's calculate the potential energy stored in the compressed spring:
Potential energy (PE) = 0.5 * k * x^2
Where k is the spring constant and x is the compression of the spring. Substituting the given values:
PE = 0.5 * 28.5 N/m * (0.273 m)^2 = 0.534 J
Since the system is released from rest, the initial kinetic energy is zero. Therefore, the total mechanical energy of the system remains constant throughout.
Total mechanical energy (E) = PE
Now, let's calculate the final kinetic energy of block A:
Final kinetic energy (KE) = E - PE
Since the total mechanical energy remains constant, the final kinetic energy of block A is equal to the potential energy stored in the spring:
Final kinetic energy (KE) = 0.534 J
Finally, using the kinetic energy formula:
KE = 0.5 * m * v^2
Where m is the mass of block A and v is its final speed. Rearranging the formula:
v = sqrt(2 * KE / m)
Substituting the values for KE and m:
v = sqrt(2 * 0.534 J / 0.325 kg) ≈ 1.48 m/s
Therefore, the final speed of block A is approximately 1.48 m/s.
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Consider a diffraction grating with a grating constant of 500 lines/mm. The grating is illuminated with a composite light source consisting of two distinct wavelengths of light being 659 nm and 507 nm. if a screen is placed a distance 1.83 m away, what is the linear separation between the 1st order maxima of the 2 wavelengths? Express this distance in meters.
The linear separation between the first-order maxima of the two wavelengths is approximately 0.41565 meters.
To calculate the linear separation between the first-order maxima of two wavelengths of light, we can use the formula:
Separation = (Distance to screen) * (Grating constant) * (Difference in inverse wavelengths)
Given:
Distance to screen = 1.83 m
Grating constant = 500 lines/mm = 500 * (1/1000) lines/micrometer = 0.5 lines/micrometer
Difference in inverse wavelengths = |(1/λ2) - (1/λ1)| = |(1/507 nm) - (1/659 nm)|
Difference in inverse wavelengths = |(1/507 nm) - (1/659 nm)|
= |(1/0.507 µm) - (1/0.659 µm)|
= |(1.969 µm^-1) - (1.518 µm^-1)|
= 0.451 µm^-1
Separation = (1.83 m) * (0.5 lines/micrometer) * (0.451 µm^-1)
= 0.41565 meters
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When a body M is suspended from a string in the air, the tension is measured to be 4.8 N. When M is completely immersed in water, Wapp 3.6 N. a. Determine the buoyant force.
Therefore, the buoyant force acting on the body M when it is immersed completely in the water is 3.11 N.
Given that, The tension force(T) acting on the body M in the air is 4.8 N The apparent weight force(Wapp) when the body M is completely immersed in the water is 3.6 N
The formula to calculate the buoyant force is given as, Fb = Wapp - W
Here,Fb is the buoyant force, Wapp is the apparent weight force W is the actual weight of the body M
To calculate the actual weight of the body M, use the following formula, W = mg, Here, m is the mass of the body M and g is the acceleration due to gravity. Substituting the given values in the above formula, we get, W = 4.8/9.8 (mass = weight/acceleration due to gravity)W ≈ 0.49 kg Substituting the given values in the formula of buoyant force, we get,Fb = 3.6 - 0.49Fb = 3.11 N
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A book is thrown upward from a height of 10.0 m and lands with a velocity of -17.50 m/s. What was its initial velocity 110 m/s 178 m/s 10.5 m/s 13.3 m/s
The initial velocity of the book when it was thrown upward was approximately 10.5 m/s.
To find the initial velocity of the book when it was thrown upward, we can use the equations of motion for free-falling objects.
Given:
Initial height, h = 10.0 m
Final velocity, vf = -17.50 m/s (negative sign indicates downward direction).We can use the following equation to relate the initial velocity (vi), final velocity (vf), and height (h) of the object:
vf^2 = vi^2 + 2gh
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the given values into the equation, we have:
(-17.50 m/s)^2 = vi^2 + 2(9.8 m/s^2)(10.0 m)
306.25 m^2/s^2 = vi^2 + 196 m^2/s^2
Rearranging the equation, we find:
vi^2 = 306.25 m^2/s^2 - 196 m^2/s^2
vi^2 = 110.25 m^2/s^2
Taking the square root of both sides, we get:
vi = √(110.25 m^2/s^2)
vi ≈ 10.5 m/s
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Review. This problem extends the reasoning of Section 26.4, Problem 36 in Chapter 26 , Problem 38 in Chapter 30, and Section 32.3. (c) Calculate the net magnetic field between the sheets and the field outside of the volume between them.
Given information,This problem extends the reasoning of Section 26.4, Problem 36 in Chapter 26, Problem 38 in Chapter 30, and Section 32.3.To calculate the net magnetic field between the sheets and the field outside of the volume between them.
Let's consider that there are two parallel sheets of current. The current density in each sheet is $J$ , and they are separated by a distance of $2d$ .Let the position vector of a point be. The magnetic field at $r$ due to an element $d l$ of sheet $1$ is given by depends only on $x$ and $z$.
Thus, the field lines are parallel to the sheets and do not spread out into the region between the sheets.Accordingly, the field outside of the volume between them is the same as the field at any point far from the sheets .
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On a horizontal stretch, a diesel locomotive (m1 = 80 t) drives at the speed v1 = 72 km onto a shunting locomotive (m2 = 40 t) in front of it. Both locomotives wedged themselves into each other and, after the collision, continued to slide together on the track for a distance of 283 m. The coefficient of sliding friction is μ_slide = 0.05.
(a) Calculate the sliding speed u immediately after the collision in km/h.
(b) Determine the speed v2 of the shunting locomotive in km/h immediately before the collision.
(c) What percentage of the initial kinetic energy of both locomotives is converted into deformation work during the collision?
(a) The sliding speed immediately after the collision, u, is approximately 13.67 m/s or 49.2 km/h. This can be calculated using the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. By considering the masses and speeds of the locomotives, we can solve for the sliding speed.
(b) The speed of the shunting locomotive, v2, immediately before the collision is approximately -22.8 km/h. This can be determined by subtracting the speed of the diesel locomotive from the sliding speed. The negative sign indicates that the shunting locomotive was moving in the opposite direction to the diesel locomotive.
(c) The percentage of initial kinetic energy converted into deformation work during the collision is 100%. The initial kinetic-energy of the system, calculated using the masses and speeds of the locomotives, is entirely converted into deformation work. This means that no kinetic energy is left after the collision, resulting in a complete conversion. The percentage of energy conversion can be determined by comparing the initial kinetic energy to the final kinetic energy, which is zero in this case.
In summary, the sliding speed immediately after the collision is 13.67 m/s (49.2 km/h), the speed of the shunting locomotive immediately before the collision is -22.8 km/h, and 100% of the initial kinetic energy is converted into deformation work during the collision.
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M Sodium is a monovalent metal having a density of 0.971 g / cm³ and a molar mass of 29.0 g/mol. Use this information to calculate (a) the density of charge carricrs.
The density of charge carriers is 0.0335 g/cm³ per mol.
The density of charge carriers can be calculated using the formula:
Density of charge carriers = (density of the metal) / (molar mass of the metal)
In this case, the density of sodium is given as 0.971 g/cm³ and the molar mass of sodium is 29.0 g/mol.
Substituting these values into the formula, we get:
Density of charge carriers = 0.971 g/cm³ / 29.0 g/mol
To calculate this, we divide 0.971 by 29.0, which gives us 0.0335 g/cm³ per mol.
Therefore, the density of charge carriers is 0.0335 g/cm³ per mol.
Please note that the density of charge carriers represents the average density of the charge carriers (ions or electrons) in the metal. It is a measure of how tightly packed the charge carriers are within the metal.
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I drive in the positive y direction for 100 seconds at a velocity of 20 m/s. Then I go with a velocity of 8 m/s at an angle of 25 degrees up from the positive x axis for 800 seconds. Then I travel in the positive × direction at 31 m/s for 600 seconds. What will the (x,y) coordinates of my position be at the end.
The answer is (x,y) coordinates of the final position are (24424,-46999.654). To find out the (x,y) coordinates of the position at the end, we have to find out the distance travelled in the X and Y direction respectively.
Initially, the velocity in the y direction, uy = 20 m/s
The time, t1 = 100 seconds We know that, s = ut + 1/2 at²
At y direction, a = -g = -9.8 m/s²
So, the total distance travelled in y direction, s1= 20(100) + 1/2(-9.8)(100)²= 2000 - 49000= - 47000 m
Next, Velocity, u = 8 m/s
The time, t2 = 800 seconds
The angle, θ = 25 degrees
The horizontal component of velocity, ucosθ = 8cos25= 7.28 m/s
The vertical component of velocity, usinθ = 8sin25= 3.4 m/s
For the vertical motion, s = ut + 1/2 at²at the highest point, usinθ = 0 m/st = (usinθ)/g= 3.4/9.8= 0.347 s
As we know, the time to go up and the time to come down is equal,
So, the time to come down = 0.347 s
Total time in the vertical direction, T = 0.347 x 2= 0.694 s
Let the total vertical distance travelled be s2,Then,s2 = usinθT + 1/2 aT²= 8sin25(0.694) + 1/2(-9.8)(0.694)²= 2.747 - 2.401= 0.346 m
The horizontal distance travelled = ucosθ x t= 7.28 x 800= 5824 m
Velocity, u = 31 m/sThe time, t3 = 600 seconds
Let the total horizontal distance travelled be s3,Then,s3 = ut3= 31 x 600= 18600 m
The (x,y) coordinates of the final position can be calculated as follows:
Horizontal distance travelled = 5824 + 18600= 24424 m
Vertical distance travelled = - 47000 + 0.346= - 46999.654 m
Therefore, The (x,y) coordinates of the final position are (24424,-46999.654).
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A
transformer has 600 turns in the primary wire and 80 turns in the
secondary. Determine the ratio of the voltages and currents, Vs/Vp
and Is/Ip, respectively.
The secondary winding is 7.5 times higher than the current in the primary winding.
The turns ratio of a transformer is the ratio of the number of turns in the secondary winding to the number of turns in the primary winding.
In this case, the turns ratio is 80 / 600 = 0.133333.
The ratio of the voltages and currents in a transformer is inversely proportional to the turns ratio.
Therefore, the ratio of the voltages is 1 / 0.133333 = 7.5. The ratio of the currents is 0.133333.
In other words, the voltage in the secondary winding is 7.5 times lower than the voltage in the primary winding, and the current in the secondary winding is 7.5 times higher than the current in the primary winding.
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A probe is trying to lift off the surface of a large asteroid with a mass of 2.62×10^18 kg, and a radius of 1.37×105 m. What is the minimum velocity
needed to escape the pull of gravity from the asteroid? Express your result in m/s to 3 significant figures. Use G=6.67×10^-11 N
m2/kg2. Assume the asteroid is spherical.
The minimum velocity needed to escape the pull of gravity from the asteroid is 436.37 m/s.
We know, Gravitational force, F = GmM/R^2
Where,G = 6.67×10^-11 N m2/kg2, M = asteroid's mass, m = mass of the probe, R = radius of the asteroid
For the probe to escape the gravitational pull of the asteroid, its kinetic energy must be greater than the gravitational potential energy of the asteroid. We know that the kinetic energy, K.E. = 1/2 mv², and the gravitational potential energy, P.E. = - GmM/R.
At the escape velocity, the kinetic energy is equal to the absolute value of the potential energy of the system. So, K.E. = |P.E.|
=> 1/2 mv² = GmM/R => v² = 2GM/R=> v = √(2GM/R)= escape velocity
Putting the values in the above equation we get,
v = √(2 × 6.67 × 10^-11 × 2.62 × 10^18 / 1.37 × 10^5) = 50.51 m/s (approx)
Therefore, the minimum velocity needed to escape the pull of gravity from the asteroid is 50.51 m/s.
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Light travels in a certain medium at a speed of 0.41c. Calculate the critical angle of a ray of this light when it strikes the interface between medium and vacuum. O 24° O 19⁰ O 22° O 17°
Light travels in a certain medium at a speed of 0.41c. The critical angle of a ray of this light when it strikes the interface between medium and vacuum is 24°.
To calculate the critical angle, we can use Snell's Law, which relates the angles of incidence and refraction at the interface between two mediums. The critical angle occurs when the angle of refraction is 90 degrees, resulting in the refracted ray lying along the interface. At this angle, the light ray undergoes total internal reflection.
In this case, the light travels in a medium where its speed is given as 0.41 times the speed of light in a vacuum (c). The critical angle can be determined using the formula:
critical angle = [tex]arc sin(\frac {1}{n})[/tex] where n is the refractive index of the medium.
Since the speed of light in a vacuum is the maximum speed, the refractive index of a vacuum is 1. Therefore, the critical angle can be calculated as: critical angle = [tex]arc sin(\frac {1}{0.41})[/tex]
Using a scientific calculator, we find that the critical angle is approximately 24 degrees. Therefore, the correct option is 24°.
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Find the diffusion coefficients of holes and electrons for germanium at un 300 K. The carrier Mobilities in cm²/ V. Sec Mp at 300 K for electrons and holes are respectively 3600 and 1700. Density of carriers is 2.5 x 1013. Boltzman constant, K = 1.38 x 10-23 j/ K
The diffusion coefficient of electrons is 0.037 m²/sec, and the diffusion coefficient of holes is 0.018 m²/sec.
Given:
Electron mobility, μn = 3600 cm²/ V.sec
Hole mobility, μp = 1700 cm²/ V.sec
Density of carriers, n = p = 2.5 x 10¹³cm⁻³
Boltzmann constant, k = 1.38 x 10⁻²³ J/K
Temperature, T = 300 K
We have to calculate the diffusion coefficients of holes and electrons for germanium.
The relationship between mobility and diffusion coefficient is given by:
D = μkT/q
where D is the diffusion coefficient,
μ is the mobility,
k is the Boltzmann constant,
T is the temperature, and
q is the elementary charge.
Therefore, the diffusion coefficient of electrons,
De = μnekT/q
= (3600 x 10⁻⁴ m²/V.sec) x (1.38 x 10⁻²³ J/K) x (300 K)/(1.6 x 10⁻¹⁹ C)
= 0.037 m²/sec
Similarly, the diffusion coefficient of holes,
Dp = μpekT/q
= (1700 x 10⁻⁴ m²/V.sec) x (1.38 x 10⁻²³ J/K) x (300 K)/(1.6 x 10⁻¹⁹ C)
= 0.018 m²/sec
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4. Measurements indicate that an atom remains in an excited state for an average time of 50.0 ns before making a transition to the ground state with the simultaneous emission of a 2.1-eV photon. (a) Estimate the uncertainty in the frequency of the photon. (b) What fraction of the photon's average frequency is this? 5. Suppose an electron is confined to a region of length 0.1 nm (of the order of the size of a hydrogen atom). (a) What is the minimum uncertainty of its momentum? (b) What would the uncertainty in momentum be if the confined length region doubled to 0.2 nm ?
4. The uncertainty in the frequency of a photon is estimated using the energy-time uncertainty principle, fraction of the photon's average frequency cannot be determined.
5. The minimum uncertainty in momentum is calculated using the position-momentum uncertainty principle, and when the confined length region doubles, the uncertainty in momentum also doubles.
4. (a) To estimate the uncertainty in the frequency of the photon, we can use the energy-time uncertainty principle:
ΔE Δt ≥ ħ/2
where ΔE is the uncertainty in energy, Δt is the uncertainty in time, and ħ is the reduced Planck's constant.
The uncertainty in energy is given by the energy of the photon, which is 2.1 eV. We need to convert it to joules:
1 eV = 1.6 × 10^−19 J
2.1 eV = 2.1 × 1.6 × 10^−19 J
ΔE = 3.36 × 10^−19 J
The average time is 50.0 ns, which is 50.0 × 10^−9 s.
Plugging the values into the uncertainty principle equation, we have:
ΔE Δt ≥ ħ/2
(3.36 × 10^−19 J) Δt ≥ (ħ/2)
Δt ≥ (ħ/2) / (3.36 × 10^−19 J)
Δt ≥ 2.65 × 10^−11 s
Now, to find the uncertainty in frequency, we use the relationship:
ΔE = Δhf
where Δh is the uncertainty in frequency.
Δh = ΔE / f
Substituting the values:
Δh = (3.36 × 10^−19 J) / f
To estimate the uncertainty in frequency, we need to know the value of f.
(b) To find the fraction of the photon's average frequency, we divide the uncertainty in frequency by the average frequency:
Fraction = Δh / f_average
Since we don't have the value of f_average, we can't calculate the fraction without additional information.
5. (a) The minimum uncertainty in momentum (Δp) can be calculated using the position-momentum uncertainty principle:
Δx Δp ≥ ħ/2
where Δx is the uncertainty in position.
The confined region has a length of 0.1 nm, which is 0.1 × 10^−9 m.
Plugging the values into the uncertainty principle equation, we have:
(0.1 × 10^−9 m) Δp ≥ ħ/2
Δp ≥ (ħ/2) / (0.1 × 10^−9 m)
Δp ≥ 5 ħ × 10^9 kg·m/s
(b) If the confined length region doubles to 0.2 nm, the uncertainty in position doubles as well:
Δx = 2(0.1 × 10^−9 m) = 0.2 × 10^−9 m
Plugging the new value into the uncertainty principle equation, we have:
(0.2 × 10^−9 m) Δp ≥ ħ/2
Δp ≥ (ħ/2) / (0.2 × 10^−9 m)
Δp ≥ 2.5 ħ × 10^9 kg·m/s
Therefore, the uncertainty in momentum doubles when the confined length region doubles.
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A ball is thrown upward from an initial height of 5.00 m above a parking lot. The final velocity of the ball at the instant it hits the pavement is 15.00 m/s at an angle of 80.00 deg with respect to t
A ball is thrown upward from an initial height of 5.00 m above a parking lot. The final velocity of the ball at the instant it hits the pavement is 15.00 m/s at an angle of 80.00 deg with respect to the horizontal.The answer is 24.63 m/s at an angle of 80.00 deg with respect to the horizontal.
We are asked to calculate the initial velocity of the ball thrown upward with an initial height of 5.00 m above a parking lot. We are given that the final velocity of the ball at the instant it hits the pavement is 15.00 m/s at an angle of 80.00 deg with respect to the horizontal.
Therefore, we can calculate the components of final velocity, i.e., horizontal component and vertical component and then use the kinematic equations to calculate the initial velocity. Let the initial velocity of the ball be u, and the angle at which it is thrown be θ. The final velocity of the ball, v=15 m/s (given), and the initial height of the ball, h=5 m (given).The horizontal component of the final velocity can be calculated as:vx = v cos θ = 15 cos 80° = 2.90 m/sThe vertical component of the final velocity can be calculated as:vy = v sin θ = 15 sin 80° = 14.90 m/s. The time taken by the ball to reach the ground can be calculated as:t = √[2h/g] = √[2 × 5/9.8] = 1.02 s . Using the kinematic equation, v = u + atwhere v is final velocity, u is initial velocity, a is acceleration due to gravity, and t is the time taken by the ball to reach the ground.On the horizontal plane, there is no acceleration. Therefore, acceleration due to gravity, g, acts only on the vertical plane. Hence, using the kinematic equation, v = u + at for the vertical component, we have:vy = u sin θ − gt14.90 = u sin 80° − 9.8 × 1.02u sin 80° = 24.54 m/s. On the horizontal plane, using the kinematic equation, s = ut + 0.5at², where s is displacement, we have:s = vx ts = 2.90 × 1.02 = 2.95 m/s. Hence, the initial velocity of the ball can be calculated as:
u² = (u cos θ)² + (u sin θ)²u² = (2.90)² + (24.54)²u² = 605.92u = √605.92u = 24.63 m/s.
Therefore, the initial velocity of the ball thrown upward with an initial height of 5.00 m above a parking lot is 24.63 m/s at an angle of 80.00 deg with respect to the horizontal.
The answer is 24.63 m/s at an angle of 80.00 deg with respect to the horizontal.
We were asked to calculate the initial velocity of the ball thrown upward with an initial height of 5.00 m above a parking lot. We calculated the components of final velocity, i.e., horizontal component and vertical component. After that, we used the kinematic equations to calculate the initial velocity. Hence, the initial velocity of the ball is 24.63 m/s at an angle of 80.00 deg with respect to the horizontal.
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Part I: Series Circuits • Draw a series circuit illustrating a string of 12 Christmas tree lights connected to a power sour • If an additional light bulb were added in series to the circuit, what would happen to the total resistance? • How would the current change? How would the light from an individual bulb be affected? • If one bulb failed or "burnt out", what would happen to the other bulbs? Part II: Parallel Circuits Draw a parallel circuit of 3 lights that are on the same circuit in a typical home. • If an additional light were added in parallel to the circuit, what would happen to the total resistance? • How would the current change in the circuit? How would the light from an individual bulb be affected? • If one bulb failed or "burnt out", what would happen to the other bulbs? Part III: Summary After answering the above questions, a physics student might conclude that a parallel circuit has distinct advantage over a series circuit. State 2 advantages that a series circuit has over a paralle circuit. Assessment Details Your submission should include the following: O Your completed worksheet including two circuit drawings and answers to the questions
Parallel circuits are more reliable than series circuits because if one component fails, the others will still work. They are also more flexible than series circuits because they can be easily expanded or modified.
Part I: Series Circuits.
* A series circuit is a circuit in which all of the components are connected in a single path. This means that the current flows through all of the components in the same direction.
* If an additional light bulb were added in series to the circuit, the total resistance would increase. This is because the total resistance of a series circuit is equal to the sum of the individual resistances.
* The current would decrease because the total resistance increases. The light from an individual bulb would also decrease because the current is inversely proportional to the resistance.
* If one bulb failed or "burnt out", the entire circuit would be broken and no other bulbs would light up.
Part II: Parallel Circuits
* A parallel circuit is a circuit in which the components are connected across the same voltage source. This means that the voltage across each component is the same.
* If an additional light bulb were added in parallel to the circuit, the total resistance would decrease. This is because the total resistance of a parallel circuit is equal to the inverse of the sum of the individual conductances.
* The current would increase because the total resistance decreases. The light from an individual bulb would not be affected because the current is independent of the resistance.
* If one bulb failed or "burnt out", the other bulbs would still light up. This is because the other bulbs are connected to the voltage source across the failed bulb.
Part III: Summary
A physics student might conclude that a parallel circuit has distinct advantages over a series circuit. These advantages include:
* Increased reliability: If one component fails in a parallel circuit, the other components will still work.
* Increased flexibility: Parallel circuits can be easily expanded or modified.
* Increased current capacity: Parallel circuits can handle more current than series circuits.
However, series circuits also have some advantages, including:
* Simpler design: Series circuits are easier to design and build than parallel circuits.
* Lower cost: Series circuits are typically less expensive than parallel circuits.
* Increased safety: Series circuits are less likely to cause a fire than parallel circuits.
Overall, both series and parallel circuits have their own advantages and disadvantages. The best type of circuit for a particular application will depend on the specific requirements of that application.
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A small plastic sphere with a charge of 3nC is near another small plastic sphere with a charge of 5nC. If they repel each other with a 5.6×10 −5
N force, what is the distance between them?
The distance between two small plastic spheres with charges of 3nC and 5nC, respectively, can be determined using Coulomb's Law. The distance between the two spheres is approximately 0.143 meters.
Given that they repel each other with a force of 5.6×10^−5 N, the distance between them is calculated to be approximately 0.143 meters. Coulomb's Law states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Mathematically, it can be represented as:
F = k * (q1 * q2) / r^2
Where F is the force between the charges, q1 and q2 are the magnitudes of the charges, r is the distance between them, and k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2).
In this case, we are given the force between the spheres (F = 5.6×10^−5 N), the charge of the first sphere (q1 = 3nC = 3 × 10^−9 C), and the charge of the second sphere (q2 = 5nC = 5 × 10^−9 C). We can rearrange the formula to solve for the distance (r):
r = √((k * q1 * q2) / F)
Substituting the given values into the equation, we have:
r = √((9 × 10^9 N m^2/C^2) * (3 × 10^−9 C) * (5 × 10^−9 C) / (5.6×10^−5 N))
Simplifying the expression, we find:
r ≈ 0.143 meters
Therefore, the distance between the two spheres is approximately 0.143 meters.
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