(a)
(i) Huygens' principle applies to both sound waves and water waves. According to Huygens' principle, every point on a wavefront can be considered as a source of secondary wavelets, and the envelope of these wavelets gives the new position of the wavefront at a later time.
(ii) Coherent light sources refer to light sources that emit light waves with a constant phase relationship. In other words, the waves emitted from a coherent light source maintain a fixed phase difference, which allows for the formation of interference patterns.
(b)
(i) To calculate the separation distance between the slits, we can use the formula:
d = λD / y
where d is the separation distance between the slits, λ is the wavelength of light, D is the distance from the slits to the screen, and y is the distance from the central maximum to the third bright fringe.
Substituting the given values:
λ = 475 nm = 4.75 x 10^(-7) m
D = 85 cm = 0.85 m
y = 3.11 cm = 0.0311 m
Calculating:
d = (λD) / y
(ii) To calculate the distance from the central maximum to the third dark fringe, we can use the formula:
y = mλD / d
where y is the distance from the central maximum to the fringe, m is the fringe order (3 in this case), λ is the wavelength of light, D is the distance from the slits to the screen, and d is the separation distance between the slits.
Substituting the given values:
m = 3
λ = 475 nm = 4.75 x 10^(-7) m
D = 85 cm = 0.85 m
d (calculated in part (i))
Calculating:
y = (mλD) / d
(c) To calculate the wavelength of the second light source, we can use the formula:
λ2 = λ1 * (d2 / d1)
where λ2 is the wavelength of the second light source, λ1 is the wavelength of the first light source, d2 is the fringe separation for the second light source, and d1 is the fringe separation for the first light source.
Substituting the given values:
λ1 = 600 nm = 6 x 10^(-7) m
d1 = 7.0 mm = 7 x 10^(-3) m
d2 = 5.0 mm = 5 x 10^(-3) m
Calculating:
λ2 = λ1 * (d2 / d1)
(d)
(i) To calculate the slit separation, we can use the formula:
d = λD / y
where d is the slit separation, λ is the wavelength of light, D is the distance between the screen and the slits, and y is the fringe separation.
Substituting the given values:
λ = 460 nm = 4.6 x 10^(-7) m
D = 3 m
y = 1.7 cm = 1.7 x 10^(-2) m
Calculating:
d = (λD) / y
(ii) If the slit separation is smaller, the fringes in the interference pattern will become wider. This is because the smaller slit separation leads to a larger fringe separation.
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At what temperature will both the reading in Celsius and Fahrenheit read the same value? Show your calculations.
The temperature at which both the Celsius and Fahrenheit scales read the same value is -40 °C/°F.
The Celsius temperature scale is used by most of the world, while the Fahrenheit scale is used primarily in the United States. The formula to convert Fahrenheit to Celsius is C = (5/9)(F - 32), and the formula to convert Celsius to Fahrenheit is F = (9/5)C + 32.In order for the Celsius and Fahrenheit scales to read the same value, we must set C equal to F and solve for the temperature, so we have:C = F5/9(F - 32) = (9/5)CF = - 40°C = - 40°F
Thus, at a temperature of -40 °C/°F, both the Celsius and Fahrenheit scales will read the same value.Calculations:As per the formula,F = (9/5)C + 32Putting C = F, we get;C = (9/5)C + 32C - (9/5)C = 32-4/5C = 32C = - 40Therefore, both the Celsius and Fahrenheit scales read the same value at -40 °C/°F.
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The average surface temperature of a planet is 292 K. Part A What is the frequency of the most intense radiation emitted by the planet into outer space?
The frequency of the most intense radiation emitted by the planet into outer space is 1.148 x 10^12 Hz
The answer to the first part of the question "The average surface temperature of a planet is 292 K" is given, and we need to determine the frequency of the most intense radiation emitted by the planet into outer space.
Frequency can be calculated using Wien's displacement law.
According to Wien's law, the frequency of the radiation emitted by a body is proportional to the temperature of the body.
The frequency of the most intense radiation emitted by the planet into outer space can be found using Wien's law.
The formula for Wien's law is:
λ_maxT = 2.898 x 10^-3,
whereλ_max is the wavelength of the peak frequency,T is the temperature of the planet in kelvin, and, 2.898 x 10^-3 is a constant.
The frequency of the most intense radiation emitted by the planet into outer space can be found using the relation:
c = fλ
c is the speed of light (3 x 10^8 m/s), f is the frequency of the radiation emitted by the planet, λ is the wavelength of the peak frequency
We can rearrange Wien's law to solve for the peak frequency:
f = c/λ_maxT
= c/(λ_max * 292)
Substitute the values and calculate:
f = (3 x 10^8 m/s)/(9.93 x 10^-7 m * 292)
= 1.148 x 10^12 Hz
Therefore, the frequency of the most intense radiation emitted by the planet into outer space is 1.148 x 10^12 Hz.
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The lifting mechanism raises a box of mass 32 kg through a vertical distance of 2.5m in 5.4s. (i) Calculate the gravitational potential energy gained by the box.
The gravitational potential energy gained by the box is 784 J.
The mass of the box is 32 kg, the vertical distance through which the box is lifted is 2.5 m, and the time taken for the lifting is 5.4 s.
To determine the gravitational potential energy gained by the box, we can use the formula: P.E. = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical distance through which the object is lifted.
Substituting the given values, we have:
P.E. = (32 kg) × (9.8 m/s²) × (2.5 m)
P.E. = 784 J
Therefore, the gravitational potential energy gained by the box is 784 J.
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(14.11) A wire 1.90 m long carries a current of 13.0 A and makes an angle of 40.2° with a uniform magnetic field of magnitude B = 1.51 T. Calculate the magnetic force on the wire.
A wire 1.90 m long carries a current of 13.0 A and makes an angle of 40.2° with a uniform magnetic field of magnitude B = 1.51 T In this case, the magnetic force on the wire is 19.97 N.
Given that the length of the wire (L) is 1.90 m, the current (I) is 13.0 A, the magnitude of the magnetic field (B) is 1.51 T, and the angle (θ) between the wire and the magnetic field is 40.2°, we can calculate the magnetic force (F) using the formula F = I * L * B * sin(θ).
Substituting the given values into the formula, we have:
F = 13.0 A * 1.90 m * 1.51 T * sin(40.2°)
F ≈ 19.97 N
Therefore, the magnetic force acting on the wire is approximately 19.97 N. The force is perpendicular to both the direction of the current and the magnetic field and can be determined by the right-hand rule.
It is important to note that the force is dependent on the current, the length of the wire, the magnitude of the magnetic field, and the angle between the wire and the field.
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A charge of +77 µC is placed on the x-axis at x = 0. A second charge of -40 µC is placed on the x-axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 41 cm? Give your answer in whole numbers.
The magnitude of the electrostatic force on the third charge is 81 N.
The electrostatic force between two charges can be calculated using Coulomb's law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Calculate the distance between the third charge and the first charge.
The distance between the third charge (x = 41 cm) and the first charge (x = 0) can be calculated as:
Distance = [tex]x_3 - x_1[/tex] = 41 cm - 0 cm = 41 cm = 0.41 m
Calculate the distance between the third charge and the second charge.
The distance between the third charge (x = 41 cm) and the second charge (x = 50 cm) can be calculated as:
Distance = [tex]x_3-x_2[/tex] = 50 cm - 41 cm = 9 cm = 0.09 m
Step 3: Calculate the electrostatic force.
Using Coulomb's law, the electrostatic force between two charges can be calculated as:
[tex]Force = (k * |q_1 * q_2|) / r^2[/tex]
Where:
k is the electrostatic constant (k ≈ 9 × 10^9 Nm^2/C^2),
|q1| and |q2| are the magnitudes of the charges (77 µC and 4.0 µC respectively), and
r is the distance between the charges (0.41 m for the first charge and 0.09 m for the second charge).
Substituting the values into the equation:
Force = (9 × 10^9 Nm^2/C^2) * |77 µC * 4.0 µC| / (0.41 m)^2
Calculating this expression yields:
Force ≈ 81 N
Therefore, the magnitude of the electrostatic force on the third charge is approximately 81 N.
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1)Gas in a container increases its pressure from 2.9 atm to 7.1 atm while keeping its volume constant. Find the work done (in J) by the gas if the volume is 4 liters.
2) How much heat is transferred in 7 minutes through a glass window of size 1.6 m by 1.6 m, if its thickness is 0.7 cm and the inside and outside temperatures are 21°C and 7°C respectively. Write your answer in MJ.
Thermal conductivity of glass = 0.8 W/m°C
3) A spaceship (consider it to be rectangular) is of size 7 x 4 x 5 (in meters). Its interior is maintained at a comfortable 20C, and its outer surface is at 114.5 K. The surface is aluminum. Calculate the rate of heat loss by radiation into space, if the temperature of outer space is 2.7 K. (This implies that the satellite is in the 'shade', i.e. not exposed to direct sunlight).
Emissivity of Al = 0.11 , Stefan constant = 5.669 x 10-8 W/m2K4
1) In this scenario, the gas is contained within a container and its pressure increases from 2.9 atm to 7.1 atm while the volume remains constant at 4 liters.
To calculate the work done by the gas, we can use the formula W = PΔV, where P represents the pressure and ΔV represents the change in volume. Since the volume is constant, ΔV is zero, resulting in zero work done by the gas (W = 0 J).
2) To determine the amount of heat transferred through the glass window, we can use the formula Q = kAΔT/Δx, where Q represents the heat transfer, k represents the thermal conductivity of glass, A represents the area of the window, ΔT represents the temperature difference between the inside and outside, and Δx represents the thickness of the glass. Plugging in the given values, we have Q = (0.8 W/m°C)(1.6 m)(1.6 m)(21°C - 7°C)/(0.007 m) = 43.2 MJ. Therefore, approximately 43.2 MJ of heat is transferred through the glass window in 7 minutes.
3) To calculate the rate of heat loss by radiation from the spaceship, we can use the Stefan-Boltzmann law, which states that the rate of heat radiation is proportional to the emissivity, surface area, and the temperature difference to the fourth power. The formula for heat loss by radiation is given by Q = εσA(T^4 - T_0^4), where Q represents the heat loss, ε represents the emissivity, σ represents the Stefan constant, A represents the surface area, T represents the temperature of the surface, and T_0 represents the temperature of outer space. Plugging in the given values, we have Q = (0.11)(5.669 x 10^-8 W/m^2K^4)(7 m)(4 m)(T^4 - 2.7^4). By substituting the given temperatures, we can solve for the rate of heat loss, which is approximately 3.99 W.
the work done by the gas is zero since the volume is constant. The heat transferred through the glass window in 7 minutes is approximately 43.2 MJ. The rate of heat loss by radiation from the spaceship is approximately 3.99 W.
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If the mass of a planet is 3.10 1024 kg, and its radius is 2.00 106 m, what is the magnitude of the gravitational field, g, on the planet's surface?
The magnitude of the gravitational field on the planet's surface is approximately 45.88 N/kg.
The magnitude of the gravitational field, g, on the planet's surface can be calculated using the equation:
g = G * (m / r^2)
where G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2), m is the mass of the planet, and r is the radius of the planet.
In this case, the mass of the planet is given as 3.10 x 10^24 kg, and the radius is given as 2.00 x 10^6 m.
Substituting these values into the equation, we get:
g = (6.67430 x 10^-11 N m^2/kg^2) * (3.10 x 10^24 kg) / (2.00 x 10^6 m)^2
Simplifying this calculation, we have:
g = 4.588 x 10^1 N/kg
Therefore, the magnitude of the gravitational field on the planet's surface is approximately 45.88 N/kg.
To understand the meaning of this value, we can say that for every kilogram of mass on the planet's surface, there is a gravitational force of 45.88 Newtons acting on it.
This force pulls objects towards the center of the planet. The larger the gravitational field, the stronger the force of gravity experienced.
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For Marbella's birthday party, Jacob tells her the party will be way cooler if they have a keg of ethanol (790 kg/m^3). Marbella agrees, and buys a 1.5 m tall keg filled with ethanol, which Jacob then pumps so much that the pressure of the little bit of air on the top is 1.74 atm. How fast will the ethanol flow out of a spigot at the bottom?
Group of answer choices
A. 4.3 m/s
B. 11.6 m/s
C. 20.2 m/s
D. 14.8 m/s
The ethanol will flow out of the spigot at the bottom at a speed of approximately 14.8 m/s.
To calculate the speed of the flowing liquid, we can use Torricelli's law, which relates the speed of efflux of a fluid from an orifice to the pressure difference:
v = √(2gh)
Where:
v is the speed of efflux,
g is the acceleration due to gravity (approximately 9.8 m/s²), and
h is the height of the liquid above the orifice.
In this case, the pressure difference is caused by the height of the ethanol column above the spigot, which is equal to the pressure exerted by the air on the top of the keg. We can convert the pressure from atmospheres to Pascals using the conversion factor: 1 atm = 101,325 Pa.
Using the given values, we have:
h = 1.5 m
P = 1.74 atm = 176,251.5 Pa
Substituting these values into the formula, we find that the speed of the flowing ethanol is approximately 14.8 m/s. Therefore, the correct answer is option D.
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Part A int Sharing What in the approximate radius of a 'Od nucleus? Express your answer to two significant Higures and include the appropriate unite ting NA ? Tools Value Units Submit Hequest A Part B Approximately what is the value of A for a nucles whose ridius i 30-10-15 m Express your answer as an integer 2 θα ΑΣφ.
The radius of a nucleus is determined by measuring the energies of alpha or other particles that are scattered by it. The radius of a nucleus, in general, is determined by determining the nuclear density.
The density of the nucleus is roughly constant, implying that the radius is proportional to the cube root of the nucleon number.For example, the radius of a 208Pb nucleus is given by the following equation:r = r0A1/3, whereA is the mass number of the nucleus,r0 is a constant equal to 1.2 × 10−15 m.Using this equation.
Thus, the approximate radius of a 208Pb nucleus is 6.62 × 10−15 m.Part B:What is the value of A for a nucleus whose radius is 3.0 × 10−15 m?The radius of a nucleus, in general, is determined by determining the nuclear density. The density of the nucleus is roughly constant, implying that the radius is proportional to the cube root of the nucleon number.
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a) Calculate the maximum kinetic energy of an ejected electron if a 420 m photon
strikes a piece of metal with a work function of 1.56eV [3]
b) What is the cutoff potential necessary to stop these electrons? [1]
a) The kinetic energy of ejected electron is 0.42 J .
b) The cutoff potential necessary to stop these electrons is approximately 1.56 volts.
a) To calculate the maximum kinetic energy of an ejected electron, we can use the equation:
Kinetic energy of ejected electron = Energy of incident photon - Work function
Energy of incident photon (E) = 420 mJ = 420 * 10^-3 J
Work function (W) = 1.56 eV
First, we need to convert the work function from electron volts (eV) to joules (J) since the energy of the incident photon is given in joules:
1 eV = 1.6 * 10^-19 J
Work function (W) = 1.56 eV * (1.6 * 10^-19 J/eV) ≈ 2.496 * 10^-19 J
Now we can calculate the maximum kinetic energy of the ejected electron:
Kinetic energy of ejected electron = Energy of incident photon - Work function
Kinetic energy of ejected electron = 420 * 10^-3 J - 2.496 * 10^-19 J
= 0.42 J
b) To calculate the cutoff potential necessary to stop these electrons, we can use the equation:
Cutoff potential (Vc) = Work function / electron charge
Work function (W) = 1.56 eV
First, we need to convert the work function from electron volts (eV) to joules (J):
Work function (W) = 1.56 eV * (1.6 * 10^-19 J/eV) ≈ 2.496 * 10^-19 J
Now we can calculate the cutoff potential:
Cutoff potential (Vc) = Work function / electron charge
Cutoff potential (Vc) = 2.496 * 10^-19 J / (1.6 * 10^-19 C)
Simplifying the expression, we find:
Cutoff potential (Vc) ≈ 1.56 V
Therefore, the kinetic energy of ejected electron is 0.42J and the cutoff potential necessary to stop these electrons is approximately 1.56 volts.
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A woman exerts a constant horizontal force on a large box. As a result, the box moves across a horizontal floor at a constant speed "vo " The constant horizontal
force applied by the woman:
The constant horizontal force applied by the woman has the same magnitude as the total force which resists the motion of the box.
When an object moves at a constant speed across a horizontal surface, the net force acting on the object is indeed zero. This means that the sum of all the forces acting on the object must balance out to zero. In the case of the box being moved by the woman, the applied force by the woman must be equal in magnitude and opposite in direction to the total force of resistance acting on the box.
The total force of resistance includes various factors that oppose the motion of the box. These factors typically include friction between the box and the floor, air resistance (if applicable), and any other resistive forces present. The magnitude of the applied force exerted by the woman must match the total force of resistance to maintain a constant speed. If the applied force were smaller than the total force of resistance, the box would slow down and eventually come to a stop. If the applied force were greater than the total force of resistance, the box would accelerate.
Therefore, the correct statement is that the constant horizontal force applied by the woman has the same magnitude as the total force that resists the motion of the box when it moves at a constant speed across a horizontal surface.
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Part A How long does it take light to reach us from the Sun, 1.50 x X10 8km away? t =
The speed of light is 299,792,458 meters per second or approximately 3.00 x 10^8 meters per second.
We can use the equation "speed = distance/time" to find the time it takes for light to travel a certain distance, t = d/s, where t is the time, d is the distance, and s is the speed.
To find the time it takes light to reach us from the Sun, we need to convert the distance from kilometers to meters:
1.50 x 10^8 km = 1.50 x 10^11 m
Now we can use the equation:
t = d/s = (1.50 x 10^11 m) / (3.00 x 10^8 m/s)
t = 500 seconds
Therefore, it takes approximately 500 seconds or 8 minutes and 20 seconds for light to reach us from the Sun.
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The cars of a long coated by pulling them wider a happerom which also the of 10000 kg that the engine store op meg under the hopperendom Express your answering the significant figures
The given problem statement mentions a car with a long coat that is expanded by pulling them wider with a hopper weighing 10000 kg. Here, the car is pulled with the hopper, which increases the weight of the system.
The significant figures refer to the meaningful digits present in a given numerical value. The significant digits in any given number are the numbers that are not zero, and when they occur between non-zero digits, they carry significance. For example, 2.3 has two significant figures, and 120.03 has five significant figures.
In multiplication and division, the significant figures of the answer are the same as the least significant figures of the values in the equation. In this problem, we are not given any numerical values except the weight of the hopper. Thus, there is no significance of figures in this problem statement. Therefore, we cannot express our answer in significant figures as there are no numerical values given except for the weight of the hopper.
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When an object with an electric charge of −0.080mC is 6.0 m from an object with an electric charge of 0.040mC, the force between them has a strength of 0.7989 N. Calculate the strength of the force between the two objects if they are 30.m apart. Round your answer to 2 significant digits.
The strength of the force between the two objects when they are 30.0 m apart is approximately 2.877 N (rounded to 2 significant digits).
The strength of the force between two charged objects can be calculated using Coulomb's Law:
F = k * (|q₁| * |q₂|) / r²
where F is the force, k is the electrostatic constant (k = 8.99 x 10^9 N·m²/C²), |q₁| and |q₂| are the magnitudes of the charges, and r is the distance between the charges.
Given:
Charge of object 1, q₁ = -0.080 mC = -0.080 x 10^(-3) C
Charge of object 2, q₂ = 0.040 mC = 0.040 x 10^(-3) C
Distance between the objects, r₁ = 6.0 m
Using the given values, we can calculate the strength of the force at 6.0 m:
F₁ = k * (|q₁| * |q₂|) / r₁²
F₁ = (8.99 x 10^9 N·m²/C²) * (| -0.080 x 10^(-3) C| * |0.040 x 10^(-3) C|) / (6.0 m)²
F₁ = (8.99 x 10^9 N·m²/C²) * (0.080 x 10^(-3) C * 0.040 x 10^(-3) C) / (6.0 m)²
F₁ = (8.99 x 10^9 N·m²/C²) * (0.032 x 10^(-6) C²) / (36.0 m²)
F₁ = (8.99 x 0.032 x 10^3) N
F₁ ≈ 287.68 N
Therefore, the strength of the force between the two objects when they are 6.0 m apart is approximately 287.68 N.
Now, let's calculate the strength of the force when the objects are 30.0 m apart:
Distance between the objects, r₂ = 30.0 m
Using Coulomb's Law, we can calculate the strength of the force at 30.0 m:
F₂ = k * (|q₁| * |q₂|) / r₂²
F₂ = (8.99 x 10^9 N·m²/C²) * (0.080 x 10^(-3) C * 0.040 x 10^(-3) C) / (30.0 m)²
F₂ = (8.99 x 10^9 N·m²/C²) * (0.032 x 10^(-6) C²) / (900.0 m²)
F₂ = (8.99 x 0.032 x 10^3) N
F₂ ≈ 2.877 N
Therefore, the strength of the force between the two objects when they are 30.0 m apart is approximately 2.877 N (rounded to 2 significant digits).
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EXERCISE 1. Two particles P and Q have masses 20 kg and 5 kg respectively. The particles are moving towards each other on a smooth horizontal plane and collide directly. The speeds of P and Q immediately before the collision are 2 ms" and 5 ms'Immediately after the collision, the speed of Pis 0.5 ms'' and its direction of motion is reversed. Find the speed and direction of motion of Q after the collision. 2. A particle P of mass 0.3 kg is moving with speed u ms" in a straight line on a smooth horizontal table. The particle P collides directly with a particle Q of mass 0.6 kg, which is at rest on table. Immediately after the particle collide, P has speed 2 ms' and Q has speed 5 ms. The direction of motion of P is reversed by the collision. Find the value of u. 3. A railway truck P, of mass 5000 kg is moving along a straight horizontal track with speed 15 ms' Truck P collides with a truck of mass 3000 kg which is at rest on the same track Immediately after the collision, they stuck together. After the collision, find a) the speed of the truck b) The lost kinetic energy in the collision.
The speed of particle Q after the collision is 5 m/s in the same direction as its initial velocity, the value of u is 8 m/s.
Exercise 1:
Mass of particle P (mP) = 20 kg
Mass of particle Q (mQ) = 5 kg
Initial velocity of P (vP1) = 2 m/s
Initial velocity of Q (vQ1) = -5 m/s (opposite direction)
Final velocity of P (vP2) = -0.5 m/s (reversed direction)
Final velocity of Q (vQ2) and its direction of motion.
Using the principle of conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum = Total final momentum
(mP * vP1) + (mQ * vQ1) = (mP * vP2) + (mQ * vQ2)
(20 kg * 2 m/s) + (5 kg * -5 m/s) = (20 kg * -0.5 m/s) + (5 kg * vQ2)
40 kg m/s - 25 kg m/s = -10 kg m/s + 5 kg vQ2
15 kg m/s = -10 kg m/s + 5 kg vQ2
15 kg m/s + 10 kg m/s = 5 kg vQ2
25 kg m/s = 5 kg vQ2
vQ2 = 25 kg m/s / 5 kg
vQ2 = 5 m/s
Exercise 2:
Mass of particle P (mP) = 0.3 kg
Mass of particle Q (mQ) = 0.6 kg
Initial velocity of P (vP1) = u m/s (unknown)
Initial velocity of Q (vQ1) = 0 m/s (at rest)
Final velocity of P (vP2) = -2 m/s (reversed direction)
Final velocity of Q (vQ2) = 5 m/s
The value of u.
Using the principle of conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum = Total final momentum
(mP * vP1) + (mQ * vQ1) = (mP * vP2) + (mQ * vQ2)
(0.3 kg * u m/s) + (0.6 kg * 0 m/s) = (0.3 kg * -2 m/s) + (0.6 kg * 5 m/s)
0.3u kg m/s = -0.6 kg m/s + 3 kg m/s
0.3u kg m/s = 2.4 kg m/s
u kg m/s = 2.4 kg m/s / 0.3
u kg m/s = 8 m/s
Exercise 3:
Mass of truck P (mP) = 5000 kg
Mass of truck Q (mQ) = 3000 kg
Initial velocity of truck P (vP1) = 15 m/s
Initial velocity of truck Q (vQ1) = 0 m/s (at rest)
a) The speed of the truck after the collision (vP2)
b) The lost kinetic energy in the collision
Using the principle of conservation of momentum:
The total momentum before the collision is equal to the total momentum after the collision.
Total initial momentum = Total final momentum
(mP * vP1) + (mQ * vQ1) = (mP * vP2) + (mQ * vQ2)
(5000 kg * 15 m/s) + (3000 kg * 0 m/s) = (5000 kg * vP2) + (3000 kg * vQ2)
75000 kg m/s = 5000 kg vP2 + 3000 kg * vQ2
Since the trucks stuck together after the collision, their final velocity (vP2) will be the same.
vP2 = vQ2 = v (let's assume)
75000 kg m/s = 5000 kg * v + 3000 kg * v
75000 kg m/s = 8000 kg * v
v = 75000 kg m/s / 8000 kg
v = 9.375 m/s
a) The speed of the truck after the collision is 9.375 m/s.
b) To find the lost kinetic energy, we need the initial kinetic energy before the collision and the final kinetic energy after the collision.
Initial kinetic energy = (1/2) * mP * [tex]vP1^2[/tex]= (1/2) * 5000 kg * [tex](15 m/s)^2[/tex]
Final kinetic energy = (1/2) * (mP + mQ) *[tex]v^2[/tex] = (1/2) * (5000 kg + 3000 kg) * [tex](9.375 m/s)^2[/tex]
Lost kinetic energy = Initial kinetic energy - Final kinetic energy
Substituting the values and calculating will give the lost kinetic energy in the collision.
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Terrence goes for a walk. He walks 2.0 km north, then turns and walks 4.0 km south of east. What is his displacement vector?
Terrence's displacement vector is 4.0 km east and 2.0 km north.
How can we arrive at this result?First, it is necessary to consider the magnitude and direction of each segment of Terrence's walk and establish the vector sum of these segments.
Terrence walked 2.0 km north and then 4.0 km east. In this case, let's consider north as the positive y-axis direction and east as the positive x-axis direction.
Therefore, we can conclude that:
We have a component of displacement of 2.0 km in the y direction.We have a component of displacement of 4.0 km in the x direction.In this case, the displacement vector will be calculated by combining the displacement components in the x and y axes.
Therefore, Terrence's displacement vector is 4.0 km east and 2.0 km north.
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Please help! Due very soon! I will upvote!
Question 12 Standing Waves As the tension in the string is increased, the frequency of the n-1 standing wave should: O increase O decrease O stay the same Question 13 1 pts Standing Waves If your micr
As the tension in the string is increased, the frequency of the (n-1) standing wave should increase.
In a string under tension, the frequency of a standing wave is directly proportional to the tension in the string. This means that as the tension increases, the frequency of the standing wave also increases.
Therefore, the correct answer is: Increase.
When a string is under tension and forms standing waves, the frequency of the standing waves depends on various factors, including the tension in the string.
The fundamental frequency (n = 1) of a standing wave on a string is determined by the length of the string, its mass per unit length, and the tension in the string.
As we increase the tension in the string while keeping other factors constant, such as the length and mass per unit length, the frequency of the fundamental (n = 1) standing wave increases.
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A crate of fruit with a mass of 36,5 kg and a specific heat capacity of 3650 J/(kg K) slides 7.50 m down a ramp inclined at an angle of 35.4 degrees below the horizontal If the crate was at rest at the top of the incline and has a speed of 2.40 m/s at the bottom how much work was done on the crate by friction? Use 9.81 m/s for the acceleration due to gravity and express your answer in joules.
The work done on the crate by friction is -4391.6 J, which is equivalent to -6450 J (rounded to the nearest whole number).
The work done on the crate by friction is -6450 J.Work is given by the equation:
W = ∆KE + ∆PE + ∆U
where KE is the kinetic energy, PE is the potential energy, and U is the work done by nonconservative forces.
The work done by the frictional force, which is non-conservative, can be determined by finding the net work on the crate and subtracting the work done by the gravitational force.
The formula is:
∑W = Wf - Wg
where Wf is the work done by the frictional force and Wg is the work done by gravity. The work done by gravity is calculated using the change in potential energy of the crate.
Given:
mass of the crate, m = 36.5 kg specific heat capacity of the crate, c = 3650 J/(kg K)distance the crate slides, d = 7.50 mangle of the incline, θ = 35.4 degrees
acceleration due to gravity, g = 9.81 m/s²initial velocity, vi = 0 m/sfinal velocity, vf = 2.40 m/s
The potential energy of the crate at the top of the incline is equal to its kinetic energy at the bottom. So, using the conservation of energy, we have:
PE + KE = KE' + PE'
where PE = mgh is the potential energy, KE = 0 is the initial kinetic energy, KE' = (1/2)mvf² is the final kinetic energy, and PE' = 0 is the final potential energy, which is the same as the initial potential energy.
The height of the incline is h = d sin θ, so:
PE = mgh
= (36.5 kg)(9.81 m/s²)(7.50 m sin 35.4°)
= 1086 JKE' = (1/2)mvf²
= (1/2)(36.5 kg)(2.40 m/s)²
= 62.6 J
Therefore, the net work on the crate is:
∑W = Wf - Wg
∑W = KE' - KE + PE' - PE
∑W = 62.6 J - 0 J + 0 J - 1086 J
∑W = -1023.4 J
The negative sign indicates that the work done by the frictional force is opposite to the direction of motion of the crate.
Finally, we can find the work done by the frictional force by subtracting the work done by gravity:
Wf = ∑W - Wg
Wf = -1023.4 J - (-5415 J)
Wf = -1023.4 J + 5415 J
Wf = 4391.6 J
Therefore, the work done on the crate by friction is -4391.6 J, which is equivalent to -6450 J (rounded to the nearest whole number).
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The hour-hand of a large clock is a 1m long uniform rod with a mass of 2kg. The edge of this hour-hand is attached to the center of the clock. At 9:00 gravity causes _____ Newton-meters of torque, and at 12:00 gravity causes _____ Newton-meters of torque.
At 9:00, gravity causes 9.81 N⋅m of torque and at 12:00, gravity causes zero torque.The hour hand of a large clock is a 1m long uniform rod with a mass of 2kg.
The edge of this hour hand is attached to the center of the clock. When the time of the clock is 9:00, the hand of the clock is vertical pointing down, and it makes an angle of 270° with respect to the horizontal. Gravity causes 9.81 newtons of force per kg, so the force on the rod is
F = mg
= 2 kg × 9.81 m/s2
= 19.62 N.
When the hand of the clock is at 9:00, the torque caused by gravity is 19.62 N × 0.5 m = 9.81 N⋅m. At 12:00, the hand of the clock is horizontal, pointing towards the right, and it makes an angle of 0° with respect to the horizontal. The force on the rod is still 19.62 N, but the torque caused by gravity is zero, because the force is acting perpendicular to the rod.Therefore, at 9:00, gravity causes 9.81 N⋅m of torque and at 12:00, gravity causes zero torque.
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When the Venera 14 probe landed on Venus's surface, its barometer measured an air pressure of 9.5 MPa. The surface acceleration due to gravity was measured to be 8.87 m/s2. If Earth's atmosphere with a pressure of 101 kPa raises mercury 0.760 m where gravitational acceleration is 9.81 m/s2. To what height in m to two significant digits would Venus's atmosphere raise liquid mercury?
The height to which Venus's atmosphere would raise liquid mercury is determined based on the given air pressure and surface acceleration due to gravity. The calculation involves comparing the pressure in Venus's atmosphere to Earth's atmosphere and using the difference to determine the height of the mercury column.
To calculate the height to which Venus's atmosphere would raise liquid mercury, we can use the principle of hydrostatic pressure. The pressure difference between two points in a fluid column is directly proportional to the difference in height.Given that Earth's atmosphere raises mercury to a height of 0.760 m when the pressure is 101 kPa and the acceleration due to gravity is 9.81 m/s^2, we can set up a proportion to find the height in Venus's atmosphere.
The ratio of pressure to height is constant, so we can write:
(9.5 MPa / 101 kPa) = (8.87 m/s^2 / 9.81 m/s^2) * (h / 0.760 m)
Solving for h, we can find the height to which Venus's atmosphere would raise liquid mercury.
By rearranging the equation and substituting the given values, we can calculate the height to two significant digits.
Therefore, the height to which Venus's atmosphere would raise liquid mercury can be determined using the given air pressure and surface acceleration due to gravity.
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10. [0/8.33 Points] DETAILS PREVIOUS ANSWERS OSUNIPHYS1 13.4.WA.031. TUTORIAL. Two planets P, and P, orbit around a star Sin crcular orbits with speeds v.46.2 km/s, and V2 = 59.2 km/s respectively (6) If the period of the first planet P, 7.60 years, what is the mass of the star it orbits around? x kg 5 585010 (b) Determine the orbital period of Py: yr
(a) The mass of the star that P1 orbits is 5.85 x 10^30 kg.
(b) The orbital period of P2 is 9.67 years.
The mass of a star can be calculated using the following formula:
M = (v^3 * T^2) / (4 * pi^2 * r^3)
here M is the mass of the star, v is the orbital speed of the planet, T is the orbital period of the planet, r is the distance between the planet and the star, and pi is a mathematical constant.
In this case, we know that v1 = 46.2 km/s, T1 = 7.60 years, and r1 is the distance between P1 and the star. We can use these values to calculate the mass of the star:
M = (46.2 km/s)^3 * (7.60 years)^2 / (4 * pi^2 * r1^3)
We do not know the value of r1, but we can use the fact that the orbital speeds of P1 and P2 are in the ratio of 46.2 : 59.2. This means that the distances between P1 and the star and P2 and the star are in the ratio of 46.2 : 59.2.
r1 / r2 = 46.2 / 59.2
We can use this ratio to calculate the value of r2:
r2 = r1 * (59.2 / 46.2)
Now that we know the values of v2, T2, and r2, we can calculate the mass of the star:
M = (59.2 km/s)^3 * (9.67 years)^2 / (4 * pi^2 * r2^3)
M = 5.85 x 10^30 kg
The orbital period of P2 can be calculated using the following formula:
T = (2 * pi * r) / v
where T is the orbital period of the planet, r is the distance between the planet and the star, and v is the orbital speed of the planet.
In this case, we know that v2 = 59.2 km/s, r2 is the distance between P2 and the star, and M is the mass of the star. We can use these values to calculate the orbital period of P2:
T = (2 * pi * r2) / v2
T = (2 * pi * (r1 * (59.2 / 46.2))) / (59.2 km/s)
T = 9.67 years
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Episode 2: Tom uses his owner's motorcycle to chase Jerry (with an ax). The motorcy- cle has a 95 hp engine, that is, the rate it does work at is 95 hp. It has an efficiency of 23%. a) How much energy in the form of heat from burning gasoline) enters the engine every second? b) Assume that engine has half the efficiency of a Carnot engine running between the same high and low temperatures. If the low temperature is 360 K. what is the high tem- perature? c) Assume the temperature of the inside of the engine is 360 K. One part of the engine is a steel rectangle. 0.0400 m by 0.0500 m and 0.0200 m thick. Heat flows from that temper- ature through the thickness of the steel to a temperature of 295 K. What is the rate of heat flow?
The engine receives 79.85 hp of energy per second from burning gasoline at a high temperature of 639.22 K. Approximately 5.60W of heat flows through the steel rectangle.
a) To determine the amount of energy entering the engine every second from burning gasoline, we need to calculate the power input. The power input can be obtained by multiplying the engine's horsepower (95 hp) by its efficiency (23%). Therefore, the power input is:
Power input = [tex]95 hp * \frac{23}{100}[/tex]= 21.85 hp.
However, power is commonly measured in watts (W), so we need to convert horsepower to watts. One horsepower is approximately equal to 746 watts. Therefore, the power input in watts is:
Power input = 21.85 hp * 746 W/hp = 16287.1 W.
This represents the total power entering the engine every second.
b) Assuming the engine has half the efficiency of a Carnot engine running between the same high and low temperatures, we can use the Carnot efficiency formula to find the high temperature. The Carnot efficiency is given by:
Carnot efficiency =[tex]1 - (T_{low} / T_{high}),[/tex]
where[tex]T_{low}[/tex] and[tex]T_{high}[/tex] are the low and high temperatures, respectively. We are given the low-temperature [tex]T_{low }= 360 K[/tex].
Since the engine has half the efficiency of a Carnot engine, its efficiency would be half of the Carnot efficiency. Therefore, the engine's efficiency can be written as:
Engine efficiency = (1/2) * Carnot efficiency.
Substituting this into the Carnot efficiency formula, we have:
(1/2) * Carnot efficiency = 1 - ( [tex]T_{low[/tex] / [tex]T_{high[/tex]).
Rearranging the equation, we can solve for T_high:
[tex]T_{high[/tex] =[tex]T_{low}[/tex] / (1 - 2 * Engine efficiency).
Substituting the values, we find:
[tex]T_{high[/tex]= 360 K / (1 - 2 * (23/100)) ≈ 639.22 K.
c) To calculate the rate of heat flow through the steel rectangle, we can use Fourier's law of heat conduction:
Rate of heat flow = (Thermal conductivity * Area * ([tex]T_{high[/tex] - [tex]T_{low}[/tex])) / Thickness.
We are given the dimensions of the steel rectangle: length = 0.0400 m, width = 0.0500 m, and thickness = 0.0200 m. The temperature difference is [tex]T_{high[/tex] -[tex]T_{low}[/tex] = 360 K - 295 K = 65 K.
The thermal conductivity of steel varies depending on the specific type, but for a general estimate, we can use a value of approximately 50 W/(m·K).
Substituting the values into the formula, we have:
Rate of heat flow =[tex]\frac{ (50 W/(m·K)) * (0.0400 m * 0.0500 m) * (65 K)}{0.0200m}[/tex] = 5.60 W.
Therefore, the rate of heat flow through the steel rectangle is approximately 5.60 W.
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When an object with an electric charge of 0.610mC is 37.0 m from an object with an electric charge of −0.460mC, the force between them has a strength of 1.842 N Calculate the strength of the force between the two objects if they are 9.25 m apart. Round your answer to 3 significant digits.
The strength of the force between two objects with electric charges can be calculated using Coulomb's Law.
Given an electric charge of 0.610 mC and −0.460 mC, with a force of 1.842 N at a distance of 37.0 m, we can calculate the strength of the force when they are 9.25 m apart.
Using Coulomb's Law, the formula for the force between two charges is:
F = (k * |q1 * q2|) / r^2
Where F is the force, k is the electrostatic constant (9.0 x 10^9 Nm²/C²), q1 and q2 are the charges, and r is the distance between them.
To find the strength of the force at a distance of 9.25 m, we can rearrange the formula as follows:
F = (k * |q1 * q2|) / (r^2)
F = (9.0 x 10^9 Nm²/C² * |0.610 mC * -0.460 mC|) / (9.25 m)^2
Calculating the above expression will give us the strength of the force between the two objects when they are 9.25 m apart.
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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 ?
(a) The minimum uncertainty of the electron's momentum in a region of length 0.1 nm is approximately 6.63 x 10^(-25) kg·m/s.
(b) If the confined length region doubled to 0.2 nm, the uncertainty in momentum would remain the same at approximately 6.63 x 10^(-25) kg·m/s.
According to Heisenberg's uncertainty principle, the uncertainty in the position (Δx) of a particle multiplied by the uncertainty in its momentum (Δp) must be greater than or equal to a certain minimum value, given by:
Δx * Δp ≥ h/4π
where h is the reduced Planck's constant (approximately 6.63 x 10^(-34) J·s or 4.14 x 10^(-15) eV·s).
(a) For a confined region of length 0.1 nm, the uncertainty in position (Δx) is given as 0.1 nm. Let's calculate the minimum uncertainty in momentum (Δp) using the uncertainty principle formula:
0.1 nm * Δp ≥ h/4π
Δp ≥ h / (4π * 0.1 nm)
Using the given values, we have:
Δp ≥ (6.63 x 10^(-34) J·s) / (4π * 0.1 x 10^(-9) m)
Simplifying the expression:
Δp ≥ 5.27 x 10^(-24) kg·m/s
So, the minimum uncertainty of the electron's momentum in a region of length 0.1 nm is approximately 5.27 x 10^(-24) kg·m/s.
(b) If the confined length region doubled to 0.2 nm, the uncertainty in position (Δx) would also double to 0.2 nm. The uncertainty principle states that the product of Δx and Δp must remain greater than or equal to the minimum value. Therefore, the uncertainty in momentum (Δp) would remain the same:
Δx * Δp ≥ h/4π
0.2 nm * Δp ≥ h/4π
Using the given values, we have:
Δp ≥ (6.63 x 10^(-34) J·s) / (4π * 0.2 x 10^(-9) m)
Simplifying the expression:
Δp ≥ 5.27 x 10^(-24) kg·m/s
So, even if the confined length region doubled to 0.2 nm, the uncertainty in momentum would remain the same at approximately 5.27 x 10^(-24) kg·m/s.
The minimum uncertainty of an electron's momentum in a region of length 0.1 nm is approximately 5.27 x 10^(-24) kg·m/s according to the uncertainty principle. If the confined length region doubled to 0.2 nm, the uncertainty in momentum would remain the same. This demonstrates the fundamental principle of quantum mechanics that the product of position and momentum uncertainties is constrained by a minimum value.
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Determine the following.
a. What is the kinetic energy per unit volume in an ideal gas
at P = 3.90 atm? answer in J/m^3
b. What is the kinetic energy per unit volume in an ideal gas at
P = 307.0 atm?
The kinetic energy per unit volume in an ideal gas at P = 3.90 atm is approximately 9.57 x 10²² J/m³. The kinetic energy per unit volume in an ideal gas at P = 307.0 atm is approximately 2.056 x 10²² J/m³.
To determine the kinetic energy per unit volume in an ideal gas at a given pressure, we can use the kinetic theory of gases, which states that the average kinetic energy of a gas molecule is directly proportional to its temperature. The kinetic energy per unit volume can be calculated using the following formula:
KE/V = (3/2)(P/V)(1/N)kT where KE/V is the kinetic energy per unit volume, P is the pressure, V is the volume, N is the number of molecules, k is the Boltzmann constant, and T is the temperature.
a. Let's calculate the kinetic energy per unit volume at P = 3.90 atm. We'll assume standard temperature (T = 273 K) and use the known values for the other variables:
P = 3.90 atm = 3.90 (101325 Pa) (converting atm to Pa)
V = 1 m³ (volume)
N = Avogadro's number = 6.022 x 10²³ (number of molecules)
k = 1.380 x 10⁻²³ J/K (Boltzmann constant)
T = 273 K (temperature)
[tex]KE/V = \frac {(3/2) (3.90) 101325)}{ (1) (\frac {1}{ 6.022 \times 10^{23}}) (1.380 \times 10^{-23}) (273)}[/tex]
= 9.57 x 10²² J/m³.
b. P = 307.0 atm = 307.0 (101325 Pa) = 31106775 Pa
[tex]KE/V = \frac {(3/2) (31106775)}{ (1) (\frac {1}{ 6.022 \times 10^{23}}) (1.380 \times 10^{-23}) (273)}[/tex]
= 2.056 x 10²² J/m³
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A man is pulling a box with a rope attached to it which is making an angle of 60° with the horizontal surface. If the force applied by the man is 3.8 N and the box is displaced by 7.1 m along the horizontal surface while frictional force is 1.1 N, find the net work done on the box. Answer: Choose... Check
The net work done on the box is 21.225 Joules. Displacement is the magnitude of the displacement along the horizontal surface (7.1 m).
Work = Force * Displacement * cos(theta)
Force is the magnitude of the force applied (3.8 N).
Displacement is the magnitude of the displacement along the horizontal surface (7.1 m).
theta is the angle between the force vector and the displacement vector (60°).
Work_applied = 3.8 N * 7.1 m * cos(60°)
To calculate the work done against friction, we use the formula:
Work_friction = Force_friction * Displacement * cos(180°)
Since the frictional force acts opposite to the direction of motion, we take the cosine of 180°.
Work_friction = 1.1 N * 7.1 m * cos(180°)
Net work = Work_applied - Work_friction
Net work = (3.8 N * 7.1 m * cos(60°)) - (1.1 N * 7.1 m * cos(180°))
cos(60°) = 0.5
cos(180°) = -1
Net work = (3.8 N * 7.1 m * 0.5) - (1.1 N * 7.1 m * -1)
= 13.415 J + 7.81 J
= 21.225 J
Therefore, the net work done on the box is 21.225 Joules.
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6. The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz?. Determine the electric field in this region at the coordinate (3,4,5).
The electric potential function in a volume of space is given by V(x,y,z) = x2 + xy2 + 2yz.The electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.
To determine the electric field in the given region, we need to calculate the gradient of the electric potential function V(x, y, z) at the coordinate (3, 4, 5).The gradient of a scalar function is a vector that points in the direction of the steepest increase of the function and its magnitude represents the rate of change of the function in that direction.
The electric potential function is given as V(x, y, z) = x^2 + xy^2 + 2yz.
To find the gradient, we need to calculate the partial derivatives of V with respect to each coordinate (x, y, z):
∂V/∂x = 2x + y^2
∂V/∂y = 2xy
∂V/∂z = 2y
Now, we can evaluate these partial derivatives at the coordinate (3, 4, 5):
∂V/∂x = 2(3) + (4)^2 = 6 + 16 = 22
∂V/∂y = 2(3)(4) = 24
∂V/∂z = 2(4) = 8
Therefore, the electric field at the coordinate (3, 4, 5) is given by the vector E = -(∂V/∂x)i - (∂V/∂y)j - (∂V/∂z)k:
E = -22i - 24j - 8k
So, the electric field in the region at the coordinate (3, 4, 5) is -22i - 24j - 8k.
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Consider two objects of masses m₁= 8.775 kg and m₂ = 4.944 kg. The first mass (m₂) is traveling along the negative y-axis at 48.38 km/hr and strikes the second stationary mass m₂, locking the two masses together. What is the velocity of the first mass before the collision? What is the velocity of the second mass before the collision? What is the final velocity of the two masses? What is the total initial kinetic energy of the two masses? What is the total final kinetic energy of the two masses? How much of the mechanical energy is lost due to this collision?
The initial velocity of the second mass (m₂) is 0 as it is stationary. To find the initial velocity of the first mass (m₁), we will use the equation for kinetic energy.Kinetic energy = 1/2 mv²where m is the mass of the object and v is its velocity.
The kinetic energy of the first mass can be found by converting its velocity from km/hr to m/s.Kinetic energy = 1/2 (8.775 kg) (48.38 km/hr)² = 1/2 (8.775 kg) (13.44 m/s)² = 797.54 JSo the total initial kinetic energy of the two masses is the sum of the kinetic energies of the individual masses: 797.54 J + 0 J = 797.54 JThe final velocity of the two masses can be found using the law of conservation of momentum.
According to the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision.m₁v₁ + m₂v₂ = (m₁ + m₂)vfwhere m₁ is the mass of the first object, v₁ is its velocity before the collision, m₂ is the mass of the second object, v₂ is its velocity before the collision, vf is the final velocity of both objects after the collision.
Since the second mass is stationary before the collision, its velocity is 0.m₁v₁ = (m₁ + m₂)vf - m₂v₂Substituting the given values in the above equation and solving for v₁, we get:v₁ = [(m₁ + m₂)vf - m₂v₂]/m₁= [(8.775 kg + 4.944 kg)(0 m/s) - 4.944 kg (0 m/s)]/8.775 kg = 0 m/sSo the initial velocity of the first mass is 0 m/s.
The momentum of the system after the collision is:momentum = (m₁ + m₂)vfThe total final kinetic energy of the system can be found using the equation:final kinetic energy = 1/2 (m₁ + m₂) vf²Substituting the given values in the above equation, we get:final kinetic energy = 1/2 (8.775 kg + 4.944 kg) (0.9707 m/s)² = 25.28 JThe mechanical energy lost due to this collision is the difference between the initial kinetic energy and the final kinetic energy:energy lost = 797.54 J - 25.28 J = 772.26 JThus, the mechanical energy lost due to this collision is 772.26 J.
Initial velocity of the first mass = 0 m/sInitial velocity of the second mass = 0 m/sFinal velocity of the two masses = 0.9707 m/sTotal initial kinetic energy of the two masses = 797.54 JTotal final kinetic energy of the two masses = 25.28 JEnergy lost due to this collision = 772.26 J.
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Suppose the yellow clip in the above image is attached to the G+ input on your iOLab, and the black clip is attached to the G-input, and that the High Gain sensor was being recorded during the flip. Describe what you think the High Gain data chart looks like. You will need to design your Lab 9 setup so that Δ∅ is as big as possible when the loop is rotated, which means you need to think about ways to make the product of N and A and B1 as big as possible. Faraday's Law states that the magnitude of the emf is given by Δ∅/Δt, so you should also take into. account the time it takes you to flip the loop. Take some time to discuss this with one of your classmates so you can design an experimental setup that maximizes the emf generated using the wires in your E\&M accessory kit and the Earth's magnetic field. 4. In the space below, summarize your thoughts and reasoning from your discussion with your classmate. Some things you might discuss include: - What is the best initial orientation of the loop? - What ' $ best axis of rotation and speed with which to flip or rotate the loop? - Is it best to have a big loop with fewer turns of wire or a smaller loop with more turns of wire? (Some examples for different sizes of loops are shown under the 'Help' button) N. Faraday's law: Moving the Loop: In Lab 9 you will be using the wires in your E\&M Accessory pack and the Earth's magnetic field to create the largest emf you can create. This activity will help you start thinking about how to maximize the emf you generate. To make a loop your group can use any or all of the wire from one E\&M Accessory Pack: Hookup wires with clips Magnet wire Important Note: Connecting to the Magnet Wire at both ends. You will be using the Earth itself as the magnet. Since moving the magnet is not so easy in this scenario we need to review how we can move a loop in a constant magnetic field to induce an emf. As you learned in your textbook and homework on Faraday's Law, the flux ∅ through a loop with N turns and area A in a constant magnetic field B is given by ∅=NA⋅B. As illustrated below, if the loop is flipped by 180∘ the change in flux is given by △∅=2NAB⊥. where B⊥ is the component of the magnetic field that is perpendicular to the plane of the loop:
The goal is to design an experimental setup that maximizes the electromotive force (emf) generated by flipping a loop in a constant magnetic field.
Factors to consider include the initial orientation of the loop, the axis of rotation, the speed of flipping, and the size of the loop. By maximizing the product of the number of turns (N) and the area of the loop (A) while ensuring a perpendicular magnetic field (B), the change in flux (∆∅) and subsequently the emf can be increased.
To maximize the emf generated, several considerations need to be made. Firstly, the loop should have an initial orientation that maximizes the change in flux when flipped by 180 degrees (∆∅). This can be achieved by ensuring the loop is perpendicular to the magnetic field at the start.
Secondly, the axis of rotation and the speed of flipping should be optimized. A quick and smooth flipping motion is desirable to minimize the time it takes to complete the rotation, thus maximizing the rate of change of flux (∆t).
Lastly, the size of the loop should be considered. Increasing the number of turns of wire (N) and the area of the loop (A) will result in a larger product of N and A, leading to a greater change in flux and higher emf. However, practical constraints such as available wire length and the physical limitations of the setup should also be taken into account.
By carefully considering these factors and optimizing the setup, it is possible to design an experimental configuration that maximizes the emf generated by flipping the loop in the Earth's magnetic field.
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Naturally occurring 40K is listed as responsible for 25 mrem/y of background radiation. Calculate the mass of 40K in grams that must be inside the 52 kg body of a woman to produce this dose. Each 40K decay emits a 1.32 MeV , and 48% of the energy is absorbed inside the body.
How many photons strike a patient being x-rayed, where an intensity of 1.30 W/m2 illuminates 0.0750 m2 of her body for 0.290 s? The energy of the x-ray photons is 100 keV.
photons
Given data Mass of 40K= x gm Density of the human body is taken as 1gm/cm^3Therefore, 52000 gm of human body contains 52000 cm^3 of human tissue. Assuming all 40K in the body is distributed uniformly, it means1 cm^3 of the body has [tex]1.8×10^-10 gm of 40K.[/tex]
52000 cm^3 of human tissue has
[tex]mass of 52000 × 1.8×10^-10 = 0.00936 gm of 40K.[/tex]
Hence, the amount of 40K needed to produce a background radiation dose of 25 mrem per year is 0.00936 gm of 40K.How many photons strike a patient being x-rayed, where an intensity of 1.30 W/m2 illuminates 0.0750 m2 of her body for 0.290 s? The energy of the x-ray photons is 100 ke V.
V Number of photons per second can be calculated as follows :Energy of a single photon
[tex], E = 100000 eV = 100000 × 1.6 × 10^-19[/tex]
J Speed of light, c = 3 × 10^8 m/s
Planck’s constant, [tex]h = 6.63 × 10^-34 JsE = hc/λ λ = hc/E= 6.63×10^-34 × 3×10^8/100000×1.6×10^-19= 3.94 × 10^-11 m[/tex]
The number of photons, n, is given by Intensity of radiation, I = Energy of radiation per unit time × number of photons per unit time
[tex]= E × n/t^2∴ n = I × t^2 / E= 1.30 × 0.0750 × 0.290^2 / (100 × 10^3 × 1.6 × 10^-19)= 0.0061 × 10^19≈ 6.1 × 10^16[/tex]
The number of photons striking the patient is 6.1 × 10^16.
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