A particular fission of a uranium-235 (235 U) nucleus, which has neutral atomic mass 235.0439 u, a reaction energy of 200 MeV is released.(a)1.00 kg of pure uranium contains approximately 2.56 x 10^24 uranium-235 atoms.(b)the total energy released if the entire mass of 1.00 kg of uranium-235 undergoes fission is approximately 3.11 x 10^13 joules.(c)assuming 100% of the reaction energy goes into powering the lamp, the lamp can run for approximately 983,544 years.
(a) To determine the number of uranium-235 (235U) atoms in 1.00 kg of pure uranium, we need to use Avogadro's number and the molar mass of uranium-235.
Calculate the molar mass of uranium-235 (235U):
Molar mass of uranium-235 = 235.0439 g/mol
Convert the mass of uranium to grams:
Mass of uranium = 1.00 kg = 1000 g
Calculate the number of moles of uranium-235:
Number of moles = (Mass of uranium) / (Molar mass of uranium-235)
Number of moles = 1000 g / 235.0439 g/mol
Use Avogadro's number to determine the number of atoms:
Number of atoms = (Number of moles) × (Avogadro's number)
Now we can perform the calculations:
Number of atoms = (1000 g / 235.0439 g/mol) × (6.022 x 10^23 atoms/mol)
Number of atoms ≈ 2.56 x 10^24 atoms
Therefore, 1.00 kg of pure uranium contains approximately 2.56 x 10^24 uranium-235 atoms.
(b) To calculate the total energy released if the entire mass of 1.00 kg of uranium-235 undergoes fission, we need to use the energy released per fission and the number of atoms present.
Given:
Reaction energy per fission = 200 MeV (mega-electron volts)
Convert the reaction energy to joules:
1 MeV = 1.6 x 10^-13 J
Energy released per fission = 200 MeV ×(1.6 x 10^-13 J/MeV)
Calculate the total number of fissions:
Total number of fissions = (Number of atoms) × (mass of uranium / molar mass of uranium-235)
Multiply the energy released per fission by the total number of fissions:
Total energy released = (Energy released per fission) × (Total number of fissions)
Now we can calculate the total energy released:
Total energy released = (200 MeV) * (1.6 x 10^-13 J/MeV) × [(2.56 x 10^24 atoms) × (1.00 kg / 235.0439 g/mol)]
Total energy released ≈ 3.11 x 10^13 J
Therefore, the total energy released if the entire mass of 1.00 kg of uranium-235 undergoes fission is approximately 3.11 x 10^13 joules.
(c) To calculate the number of years the lamp can run, we need to consider the power generated by the fission reactions and the total energy released.
Given:
Power generated = 100 W
Total energy released = 3.11 x 10^13 J
Calculate the time required to release the total energy at the given power:
Time = Total energy released / Power generated
Convert the time to years:
Time in years = Time / (365 days/year ×24 hours/day ×3600 seconds/hour)
Now we can calculate the number of years the lamp can run:
Time in years = (3.11 x 10^13 J) / (100 W) / (365 days/year × 24 hours/day * 3600 seconds/hour)
Time in years ≈ 983,544 years
Therefore, assuming 100% of the reaction energy goes into powering the lamp, the lamp can run for approximately 983,544 years.
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"A 6900 line/cm diffraction grating is 3.44 cm wide.
Part A
If light with wavelengths near 623 nm falls on the grating, what
order gives the best resolution?
1. zero order
2. first order
3. second order
The first order gives the best resolution. Thus, the correct answer is Option 2.
To determine the order that gives the best resolution for the given diffraction grating and wavelength, we can use the formula for the angular separation of the diffraction peaks:
θ = mλ / d,
where
θ is the angular separation,
m is the order of the diffraction peak,
λ is the wavelength of light, and
d is the spacing between the grating lines.
Given:
Wavelength (λ) = 623 nm
= 623 × 10⁻⁹ m,
Grating spacing (d) = 1 / (6900 lines/cm)
= 1 / (6900 × 10² lines/m)
= 1.449 × 10⁻⁵ m.
We can substitute these values into the formula to calculate the angular separation for different orders:
For zero order, θ₀ = (0 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m),
θ₀ = 0
For first order θ₁ = (1 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m),
θ₁ ≈ 0.0428 rad
For second-order θ₂ = (2 × 623 × 10^(-9) m) / (1.449 × 10^(-5) m)
θ₂ ≈ 0.0856 rad.
The angular separation determines the resolution of the diffraction pattern. Smaller angular separations indicate better resolution. Thus, the order that gives the best resolution is the order with the smallest angular separation. In this case, the best resolution is achieved in the first order, θ₁ ≈ 0.0428 rad
Therefore, the correct answer is first order gives the best resolution.
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A long solenoid is created with 42 turns, has a radius of 1.8 mm, and a length of 1.31 cm. What is the inductance L of the solenoid?
The inductance of the solenoid is approximately 5.02 × 10^-4 Henrys (H).
The inductance of a solenoid can be calculated using the formula:
L = (μ₀ * N² * A) / l
where L is the inductance, μ₀ is the permeability of free space (4π × 10^-7 T·m/A), N is the number of turns, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.
Given:
N = 42 turns
r = 1.8 mm = 1.8 × 10^-3 m (radius)
l = 1.31 cm = 1.31 × 10^-2 m (length)
The cross-sectional area A of the solenoid can be calculated as:
A = π * r²
Substituting the values into the formula:
A = π * (1.8 × 10^-3 m)²
A ≈ 3.23 × 10^-6 m²
Now, we can calculate the inductance L:
L = (4π × 10^-7 T·m/A) * (42 turns)² * (3.23 × 10^-6 m²) / (1.31 × 10^-2 m)
L ≈ 5.02 × 10^-4 H
Therefore, the inductance of the solenoid is approximately 5.02 × 10^-4 Henrys (H).
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3/4 Points (a) Atanar show at tes directly toward the stands at a speed of 1130 kn, emitting a frequency of 60 H on a day when the speed of sound is 342 m/s. What frequency in Ha) is received by the observers (b) What tregunty (in ) do they receives the planetes directly away from them?
The frequency received by the observers is 55.78 Hz. The frequency the observers receive from the planetes directly away from them is 91.43 Hz.
(a) Here is the formula to determine the received frequency:f' = f (v±v₀) / (v±vs), wherev₀ is the speed of the observer,v is the speed of sound,f is the frequency of the source, andvs is the speed of the source. Here is the solution to part (a): The speed of sound is given as 342 m/s. Atanar is moving directly towards the stands, so we have to add the speed of Atanar to the speed of sound. The speed of Atanar is 1130 km/h, which is 313.8889 m/s when converted to m/s.v = 342 m/s + 313.8889 m/s = 655.8889 m/sUsing the formula,f' = f (v±v₀) / (v±vs),we get:f' = 60 Hz (655.8889 m/s) / (655.8889 m/s + 0 m/s)f' = 55.78 HzSo, the frequency received by the observers is 55.78 Hz.
(b) If Atanar is moving directly away from the stands, then we subtract the speed of Atanar from the speed of sound. Using the formula:f' = f (v±v₀) / (v±vs),we get:f' = 60 Hz (655.8889 m/s) / (655.8889 m/s - 0 m/s)f' = 91.43 Hz.Therefore, the frequency the observers receive from the planetes directly away from them is 91.43 Hz.
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A wire has a length of 7.99 x 10^-2 m and is used to make a circular coil of one turn. There is a
current of 7.03 A in the wire. In the presence of a 2.56-T magnetic field, what is the maximum
torque that this coil can experience?
The maximum torque that the coil can experience can be calculated using the formula:
τ = N * B * A * sin(θ)
where τ is the torque, N is the number of turns, B is the magnetic field, A is the area of the coil, and θ is the angle between the magnetic field and the normal to the coil.
In this case, the coil has one turn (N = 1), a magnetic field of 2.56 T, and the length of the wire is used to make a circular coil, so the perimeter of the coil is equal to the length of the wire.
The perimeter of the coil (P) is given by:
P = 2πr
where r is the radius of the coil.
Since there is one turn, the circumference of the coil is equal to the length of the wire:
P = L
where L is the length of the wire.
Therefore, we can find the radius of the coil (r) using the formula:
r = L / (2π)
Substituting the given values:
r = (7.99 x 10^-2 m) / (2π)
Now we can calculate the area of the coil (A):
A = πr^2
Substituting the value of r:
A = π * [(7.99 x 10^-2 m) / (2π)]^2
Finally, we can calculate the maximum torque:
τ = (1) * (2.56 T) * A * sin(θ)
Since the problem does not specify the angle θ, we assume it to be 90 degrees to maximize the torque:
τ = (2.56 T) * A
Substituting the value of A:
τ = (2.56 T) * [π * [(7.99 x 10^-2 m) / (2π)]^2]
τ ≈ 5.22 x 10^-3 N·m
Therefore, the maximum torque that this coil can experience is approximately 5.22 x 10^-3 N·m.
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A proton moving perpendicular to a magnetic field of 9.80 μT follows a circular path of radius 4.95 cm. What is the proton's speed? Give answer in m/s.
If the magnetic field in the previous question is pointed into the page and the proton is moving to the left when it enters the region of the magnetic field, the proton goes in what direction as viewed from above?
A) Clockwise
B) Counterclockwise
C) Down the page
D) Up the page
The proton's speed is approximately 1.48 x 10^5 m/s, which corresponds to option B) Counterclockwise.
We can use the formula for the centripetal force experienced by a charged particle moving in a magnetic field:
F = qvB
where F is the centripetal force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength.
Since the proton moves in a circular path, the centripetal force is provided by the magnetic force:
F = mv^2/r
where m is the mass of the proton and r is the radius of the circular path.
Setting these two equations equal to each other, we have:
mv^2/r = qvB
Rearranging the equation, we find:
v = (qBr/m)^0.5
Plugging in the given values, we have:
v = [(1.6 x 10^-19 C)(9.8 x 10^-6 T)(4.95 x 10^-2 m)/(1.67 x 10^-27 kg)]^0.5
v ≈ 1.48 x 10^5 m/s
Therefore, the proton's speed is approximately 1.48 x 10^5 m/s.
Regarding the direction of the proton's motion as viewed from above, we can apply the right-hand rule. If the magnetic field is pointed into the page and the proton is moving to the left, the force experienced by the proton will be downwards. As a result, the proton will move in a counterclockwise direction, which corresponds to option B) Counterclockwise.
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3. An inductor with an inductance of 2.50 H and a resistor of 8.00 are connected to the terminals of a battery with an emf of 6.00 V. Find: A. The initial rate of increase of current in the circuit (d
The initial rate of increase of current in the circuit is 2.08 A/s.We need to find the initial rate of increase of current in the circuit (dI/dt)To determine the initial rate of increase of current in the circuit,
The current through an inductor changes with time. The current increases as the magnetic flux through the inductor increases. The induced EMF opposes the change in current. This effect is known as inductance. The inductance of a coil is directly proportional to the number of turns of wire in the coil. The unit of inductance is Henry (H).
The formula for current in a circuit that contains only inductor and resistor is: R = resistance of the circuit L = inductance of the circuitt = timeTo determine the initial rate of increase of current in the circuit, we differentiate the above equation with respect to time Now, we substitute the given values in the above equation
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Two identical, 1.2-F capacitors are placed in series with a 12-V battery. How much
energy is stored in each capacitor? (in J)
Each capacitor will store the same amount of energy which is 72 J.
Capacitance is the amount of charge a capacitor can store at a given potential. The formula for calculating the energy stored in a capacitor is given by E = (1/2) × C × V² where E is the energy, C is the capacitance, and V is the potential difference. In the given problem, two identical 1.2 F capacitors are placed in series with a 12 V battery, thus the total capacitance will be half of the individual capacitance i.e. 0.6 F. Using the formula above, we get
E = (1/2) × 0.6 F × (12 V)²= 43.2 J.
This is the total energy stored in both capacitors. Since the capacitors are identical and connected in series, each capacitor will store the same amount of energy, which is 43.2 J ÷ 2 = 21.6 J. Therefore, the energy stored in each capacitor is 21.6 J.
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A highway is made of concrete slabs that are 17.1 m long at 20.0°C. Expansion coefficient of concrete is α = 12.0 × 10^−6 K^−1.
a. If the temperature range at the location of the highway is from −20.0°C to +33.5°C, what size expansion gap should be left (at 20.0°C) to prevent buckling of the highway? answer in mm
b. If the temperature range at the location of the highway is from −20.0°C to +33.5°C, how large are the gaps at −20.0°C? answer in mm
The gap size at -20.0°C is 150 mm + 0.9 mm + 7.7 mm = 159.6 mm.
a. The expansion gap size at 20.0°C to prevent buckling of the highway is 150 mm. b.
The gap size at -20.0°C is 159.6 mm.
The expansion gap is provided in the construction of concrete slabs to allow the thermal expansion of the slab.
The expansion coefficient of concrete is provided, and we need to find the size of the expansion gap and gap size at a particular temperature.
The expansion gap size can be calculated by the following formula; Change in length α = Expansion coefficient L = Initial lengthΔT = Temperature difference
At 20.0°C, the initial length of the concrete slab is 17.1 mΔT = 33.5°C - (-20.0°C)
= 53.5°CΔL
= 12.0 × 10^-6 K^-1 × 17.1 m × 53.5°C
= 0.011 mm/m × 17.1 m × 53.5°C
= 10.7 mm
The size of the expansion gap should be twice the ΔL.
Therefore, the expansion gap size at 20.0°C to prevent buckling of the highway is 2 × 10.7 mm = 21.4 mm
≈ 150 mm.
To find the gap size at -20.0°C, we need to use the same formula.
At -20.0°C, the initial length of the concrete slab is 17.1 m.ΔT = -20.0°C - (-20.0°C)
= 0°CΔL
= 12.0 × 10^-6 K^-1 × 17.1 m × 0°C
= 0.0 mm/m × 17.1 m × 0°C
= 0 mm
The gap size at -20.0°C is 2 × 0 mm = 0 mm.
However, at -20.0°C, the slab is contracted by 0.9 mm due to the low temperature.
Therefore, the gap size at -20.0°C is 150 mm + 0.9 mm + 7.7 mm = 159.6 mm.
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An ant stands 70 feet away from a tower, and has to look up at a 40 degree angle to see the top. Find the height of the tower.
An ant stands 70 feet away from a tower, and has to look up at a 40 degree angle to see the top. The height of the tower is approximately 58.74 feet.
To find the height of the tower, we can use trigonometry. Let's denote the height of the tower as 'h'.
We have a right triangle formed by the ant, the tower, and the line of sight to the top of the tower. The distance from the ant to the base of the tower is 70 feet, and the angle formed between the ground and the line of sight is 40 degrees.
In a right triangle, the tangent function relates the opposite side to the adjacent side. In this case, the opposite side is the height of the tower (h), and the adjacent side is the distance from the ant to the tower (70 feet). Therefore, we can use the tangent function as follows:
tan(40°) = h / 70
To find the value of h, we can rearrange the equation:
h = 70 * tan(40°)
Now, let's calculate the height of the tower using the given formula:
h = 70 * tan(40°)
h ≈ 70 * 0.8391
h ≈ 58.7387 feet
Therefore, the height of the tower is approximately 58.74 feet.
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16 pts) in an alternate timeline where DC and Marvel exist in the same universe, Thor is trying to take down Superman. Thor throws his hammer (Mjölnir , which according to a 1991 trading card has a mass of about 20 kg) and hits Superman Superman (m+100 kg) is initially flying vertically downward with a speed of 20 m/s. Superman catches (and holds onto) the hammer and they move up and to the right with a speed of 10 m/s at an angle of 40 degrees above the horizontal. What was the initial speed and direction of the hammer? 022
The initial speed of the hammer thrown by Thor is approximately 105.82 m/s. To determine the initial speed and direction of the hammer thrown by Thor, we can use the principle of conservation of momentum and the concept of vector addition.
Let's denote the initial speed of the hammer as v₁ and its direction as θ₁. We'll assume the positive x-axis is to the right and the positive y-axis is upward.
According to the conservation of momentum:
(m₁ * v₁) + (m₂ * v₂) = (m₁ * u₁) + (m₂ * u₂)
where m₁ and m₂ are the masses of the hammer and Superman, v₁ and v₂ are their initial velocities, and u₁ and u₂ are their final velocities.
m₁ (mass of hammer) = 20 kg
v₂ (initial velocity of Superman) = -20 m/s (negative sign indicates downward direction)
m₂ (mass of Superman) = 100 kg
u₁ (final velocity of hammer) = 10 m/s (speed)
u₂ (final velocity of Superman) = 10 m/s (speed)
θ₂ (angle of motion of Superman) = 40 degrees above the horizontal
Now, let's calculate the initial velocity of the hammer.
Using the conservation of momentum equation and substituting the given values:
(20 kg * v₁) + (100 kg * (-20 m/s)) = (20 kg * 10 m/s * cos(θ₂)) + (100 kg * 10 m/s * cos(40°))
Note: The negative sign is applied to the velocity of Superman (v₂) since it is directed downward.
Simplifying the equation:
20 kg * v₁ - 2000 kg m/s = 200 kg * 10 m/s * cos(θ₂) + 1000 kg * 10 m/s * cos(40°)
Now, solving for v₁:
20 kg * v₁ = 2000 kg m/s + 200 kg * 10 m/s * cos(θ₂) + 1000 kg * 10 m/s * cos(40°)
v₁ = (2000 kg m/s + 200 kg * 10 m/s * cos(θ₂) + 1000 kg * 10 m/s * cos(40°)) / 20 kg
Calculating the value of v₁:
v₁ ≈ 105.82 m/s
Therefore, the initial speed of the hammer thrown by Thor is approximately 105.82 m/s.
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A certain child's near point is 14.0 cm; her far point (with eyes relaxed) is 119 cm. Each eye lens is 2.00 cm from the retina. (a) Between what limits, measured in diopters, does the power of this lens-cornea combination vary? Calculate the power of the eyeglass lens the child should use for relaxed distance vision. diopters Is the lens converging or diverging?
Near point = 14.0 cm Far point = 119 cm Distance between retina and eye lens = 2.00 cm
The distance between the near point and the eye lens is = 14 - 2 = 12 cm
The distance between the far point and the eye lens is = 119 - 2 = 117 cm
Lens formula,1/f = 1/v - 1/u Where,f = focal length of the eye lens v = distance of far point u = distance of near point
Therefore, 1/f = 1/119 - 1/14= (14 - 119) / 14 × 119= - 105 / 1666f = - 1666 / (-105) = 15.876 cm
Therefore, The focal length of the eye lens is = 15.876 cm
Now, The power of the eye lens, P = 1/f= 1/15.876= 0.063 diopters
The formula for lens power is, P = 1/f or f = 1/P
Therefore, f = 1/0.063= 15.876 cm
Here, The power of the eyeglass lens the child should use for relaxed distance vision is = - 2.34 diopters.
Now, The image formed by the eye lens is a real and inverted image, which means that the eye lens is a converging lens.
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JA B A с The three tanks above are filled with water to the same depth. The tanks are of equal height. Tank B has the middle surface area at the bottom, tank A the greatest and tank C the least. For each of the following statements, select the correct option from the pull-down menu. Less than The force exerted by the water on the bottom of tank A is .... the force exerted by the water on the bottom of tank B. True The pressure exerted on the bottom of tank A is equal to the pressure on the bottom of the other two tanks. Less than The force due to the water on the bottom of tank B is .... the weight of the water in the tank. True The water in tank C exerts a downward force on the sides of the tank. Less than The pressure at the bottom of tank A is .... the pressure at the bottom of tank C.
The force exerted by the water on the bottom of tank A is less than the force exerted by the water on the bottom of tank B.
The force exerted by a fluid depends on its pressure and the surface area it acts upon. In this case, although the water level and height of the tanks are equal, tank A has the greatest surface area at the bottom, tank B has a middle surface area, and tank C has the least surface area.
The force exerted by the water on the bottom of a tank is directly proportional to the pressure and the surface area. Since the water pressure at the bottom of the tanks is the same (as they are filled to the same depth), the force exerted by the water on the bottom of tank A would be greater than the force exerted on tank B because tank A has a larger surface area at the bottom.
The pressure exerted on the bottom of tank A is equal to the pressure on the bottom of the other two tanks. Pressure in a fluid is determined by the depth of the fluid and the density of the fluid, but it is not affected by the surface area. Therefore, the pressure at the bottom of all three tanks is the same, regardless of their surface areas.
The force due to the water on the bottom of tank B is true and equal to the weight of the water in the tank. This is because the force exerted by a fluid on a surface is equal to the weight of the fluid directly above it. In tank B, the water exerts a force on its bottom that is equal to the weight of the water in the tank.
The water in tank C does not exert a downward force on the sides of the tank. The pressure exerted by the water at any given depth is perpendicular to the sides of the container. The force exerted by the water on the sides of the tank is a result of the pressure, but it acts horizontally and is balanced out by the pressure from the opposite side. Therefore, the water in tank C exerts an equal pressure on the sides of the tank but does not exert a net downward force.
The pressure at the bottom of tank A is less than the pressure at the bottom of tank C. This is because pressure in a fluid increases with depth. Since tank A has a greater depth than tank C (as they are filled to the same level), the pressure at the bottom of tank A is greater.
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The magnetic field of the Earth varies over time and reverses its poles every half million years or so. Currently, the magnitude of the Earth's magnetic field at either pole is approximately 7 × 10-5 T. At the next pole reversal, while the field is zero, some boyscouts decide to replace the field using a current loop around the equator. Without relying on magetization of materials inside the Earth, determine the current that would generate a field of 9.0e-5 T at the poles. The radius of the Earth is RE = 6.37 × 106 m. A (+1E7 A)
The question asks for the current required to generate a magnetic field of 9.0e-5 T at the Earth's poles during a pole reversal. The current is generated by a loop around the equator, and we need to determine the magnitude of the current. The Earth's magnetic field currently has a magnitude of approximately 7 × 10-5 T at the poles.
To determine the current required to generate a magnetic field of 9.0e-5 T at the Earth's poles, we can use Ampere's law. Ampere's law relates the magnetic field generated by a current-carrying loop to the current and the distance from the loop. In this case, we want to generate a magnetic field at the poles, which are located at the ends of the Earth's diameter. The diameter of the Earth is given as 2 * RE, where RE is the radius of the Earth.
Since the current loop is placed around the equator, the distance from the loop to the poles is half the Earth's diameter, or RE. Therefore, we can use Ampere's law to solve for the current: B = (μ₀ * I) / (2 * π * R), where B is the desired magnetic field, μ₀ is the permeability of free space, I is the current, and R is the distance from the loop. Rearranging the equation to solve for I, we have: I = (B * 2 * π * R) / μ₀.
Substituting the given values, where B is 9.0e-5 T and R is 6.37 × 10^6 m, we can calculate the current required. Using the value for μ₀, which is approximately 4π × 10^-7 T·m/A, we can solve for I.
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An emf of 15.0 mV is induced in a 513-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic
flux through each turn of the coil at an instant when the current is 3.80 A? (Enter the magnitude.)
Explanation:
We can use Faraday's law of electromagnetic induction to solve this problem. According to this law, the induced emf (ε) in a coil is equal to the negative of the rate of change of magnetic flux through the coil:
ε = - dΦ/dt
where Φ is the magnetic flux through the coil.
Rearranging this equation, we can solve for the magnetic flux:
dΦ = -ε dt
Integrating both sides of the equation, we get:
Φ = - ∫ ε dt
Since the emf and the rate of current change are constant, we can simplify the integral:
Φ = - ε ∫ dt
Φ = - ε t
Substituting the given values, we get:
ε = 15.0 mV = 0.0150 V
N = 513
di/dt = 10.0 A/s
i = 3.80 A
We want to find the magnetic flux through each turn of the coil at an instant when the current is 3.80 A. To do this, we first need to find the time interval during which the current changes from 0 A to 3.80 A:
Δi = i - 0 A = 3.80 A
Δt = Δi / (di/dt) = 3.80 A / 10.0 A/s = 0.380 s
Now we can use the equation for magnetic flux to find the flux through each turn of the coil:
Φ = - ε t = -(0.0150 V)(0.380 s) = -0.00570 V·s
The magnetic flux through each turn of the coil is equal to the total flux divided by the number of turns:
Φ/ N = (-0.00570 V·s) / 513
Taking the magnitude of the result, we get:
|Φ/ N| = 1.11 × 10^-5 V·s/turn
Therefore, the magnetic flux through each turn of the coil at the given instant is 1.11 × 10^-5 V·s/turn.
A part of a Gaussian Surface is a square of side length s. A corner of the square is placed the distance s from the origin on the y axis. A point charge Q is located at the origin. The edges of the square are either parallel to the x direction or z direction. The image above shows this information. If Q=25 microCoulomb and s = 15 cm, what is the electric field flux through the square?
The electric field flux through the square is determined as 2.25 x 10⁵ Nm²/C.
What is the flux through square?The electric field flux through the square is calculated by applying the following formula as follows;
Ф = EA
where;
E is the electric fieldA is the area of the surfaceThe magnitude of the electric field is calculated as;
E = (kQ) / s²
E = ( 9 x 10⁹ x 25 x 10⁻⁶ ) / ( 0.15 m)²
E = 1 x 10⁷ N/C
The electric field flux through the square is calculated as;
Ф = EA
Ф = (1 x 10⁷ N/C) x (0.15 m)²
Ф = 2.25 x 10⁵ Nm²/C
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3. A 72 tooth gear is driven by a gear that gives a speed reduction of 4:1. The output gear is moving at 450 RPM. What is the speed of the driving gear? How many teeth are on the driving gear? 4
The output gear is moving at 450 RPM. the speed of the driving gear is 112.5 RPM
To find the speed of the driving gear, we can use the concept of gear ratio. The gear ratio is defined as the ratio of the number of teeth on the driven gear to the number of teeth on the driving gear.
Given that the output gear has 72 teeth and there is a speed reduction of 4:1, we can calculate the number of teeth on the driving gear.
Number of teeth on the driving gear = Number of teeth on the driven gear / Speed reduction
Number of teeth on the driving gear = 72 teeth / 4 = 18 teeth
So, the driving gear has 18 teeth.
Now, to find the speed of the driving gear, we can use the formula:
Speed of the driving gear = Speed of the output gear / Speed reduction
Speed of the driving gear = 450 RPM / 4 = 112.5 RPM
Therefore, the speed of the driving gear is 112.5 RPM.
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What must be the charge (in nm) on each of the two 64-kg
spherical masses for the electric force to equal the gravitational
force? Give your answer to one decimal place.
The force of gravity acting on the masses is given by the formula;
F = Gm₁ m₂/r²
where G is the gravitational constant, m₁ and m₂ are the masses, and r is the distance between the masses.
Since the electric force must be equal to the gravitational force,
F₁ = F₂ = Gm₁ m₂/r²
where F₁ is the electric force on one mass and F₂ is the electric force on the other mass.
Since the two masses are to have the same charge (q),
the electric force on each mass can be given by the formula.
F = kq²/r²
where k is the Coulomb constant, and q is the charge on each mass.
Similarly,
F₁ = F₂ = kq²/r²
Combining the two equations.
kq²/r² = Gm₁ m₂/r²
Dividing both sides by r².
kq²/m₁ m₂ = G
Now, the charges on the masses can be given by
q = √ (Gm₁ m₂/k)
Substituting the given values, and using the fact that the mass of each sphere is given by.
m = (4/3)πr³ρ
where ρ is the density, and r is the radius.
q = √ (6.67 × 10^-11 × 64 × 64 / 9 × 10^9)
q = √ 291.56q = 17.06 × 10^-9 C (to one decimal place)
the charge on each mass must be 17.06 nm.
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A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 38.0 vibrations in 32.0 s. Also, a given maximum travels 427 cm along the rope in 6.0 s. What is the wavelength? 0.601 x Your response is off by a multiple of ten. cm
The wavelength of the of the harmonic wave traveling along the rope, given that it completes 38.0 vibrations in 32.0 s is 60.31 cm
How do i determine the wavelength?First, we shall obtain the frequency of the wave. Details below:
Number of vibrations (n) = 38.0 vibrationsTime (t) = 32.0 secondsFrequency (f) = ?Frequency (f) = Number of oscillation (n) / time (s)
= 38.0 / 32.0
= 1.18 Hertz
Next, we shall obtain the speed of the wave. Details below:
Distance = 427 cm Time = 6.0 sSpeed = ?Speed = Distance / time
= 427 / 6
= 71.17 cm/s
Finally, we shall obtain the wavelength of the wave. Details below:
Frequency of wave (f) = 1.18 HertzSpeed of wave (v) = 71.17 cm/sWavelength of wave (λ) = ?Speed (v) = wavelength (λ) × frequency (f)
71.17 = wavelength × 1.18
Divide both sides by 27×10⁸
Wavelength = 71.17 / 1.18
= 60.31 cm
Thus, the wavelength of the wave is 60.31 cm
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A constant horizontal force moves a 50 kg trunk 6.5 m up 31 degree incline a constant speed. the coefficient of kinetic friction between the trunk and incline is 0.20.
a. what is the work done by applied force?
b. what is the increase in thermal energy of the trunk and incline?
a. The work done by the applied force is approximately 1380.3 Joules.
b. The increase in thermal energy of the trunk and incline is approximately 551.2 Joules.
a. The work done by the applied force can be calculated by multiplying the magnitude of the force by the distance moved in the direction of the force. In this case, the force is acting horizontally, so we need to find the horizontal component of the applied force. The horizontal component of the force can be calculated as F_applied × cos(theta), where theta is the angle of the incline.
F_applied = m × g × sin(theta),
F_horizontal = F_applied × cos(theta).
Plugging in the values:
m = 50 kg,
g = 9.8 m/s² (acceleration due to gravity),
theta = 31 degrees.
F_applied = 50 kg × 9.8 m/s² × sin(31 degrees) ≈ 246.2 N.
F_horizontal = 246.2 N × cos(31 degrees) ≈ 212.2 N.
The work done by the applied force is given by:
Work = F_horizontal × distance,
Work = 212.2 N × 6.5 m ≈ 1380.3 Joules.
Therefore, the work done by the applied force is approximately 1380.3 Joules.
b. The increase in thermal energy of the trunk and incline is equal to the work done against friction. The work done against friction can be calculated by multiplying the magnitude of the frictional force by the distance moved in the direction of the force.
Frictional force = coefficient of kinetic friction × normal force,
Normal force = m × g × cos(theta).
Plugging in the values:
Coefficient of kinetic friction = 0.20,
m = 50 kg,
g = 9.8 m/s² (acceleration due to gravity),
theta = 31 degrees.
Normal force = 50 kg × 9.8 m/s² × cos(31 degrees) ≈ 423.9 N.
Frictional force = 0.20 × 423.9 N ≈ 84.8 N.
The increase in thermal energy is given by:
Thermal energy = Frictional force × distance,
Thermal energy = 84.8 N × 6.5 m ≈ 551.2 Joules.
Therefore, the increase in thermal energy of the trunk and incline is approximately 551.2 Joules.
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At = 0, a ball is kicked such that it moves along a ramp that makes an ground? (10 points) angle 8 = 30 with the ground. What shall be the initial speed of the ball i such that it will stop after t = 1 s? What's the space travelled by the ball when it stops? Assume that there is no friction between the ball and the ramp
The initial speed of the ball i such that it will stop after t = 1 s is -9.8 m/s, and the space traveled by the ball when it stops is 8.48 m.
At t = 0, a ball is kicked such that it moves along a ramp that makes an angle θ = 30 degree with the ground.
Given that there is no friction between the ball and the ramp, we need to calculate the initial speed of the ball i such that it will stop after t = 1 s.
We also need to calculate the space traveled by the ball when it stops.
angle of the ramp θ = 30°
The horizontal component of the initial velocity of the ball is given as follows:
vₓ = vicosθvₓ = vi cosθ ………………….. (1)
The vertical component of the initial velocity of the ball is given as follows:
vᵧ = visinθ …………………………….. (2)
When the ball stops at t = 1 s,
its final velocity v = 0 m/s.
We know that the acceleration of the ball along the incline is given as follows:
a = gsinθ ………………………………..(3)
We also know that the time taken by the ball to stop is t = 1 s.
Therefore, we can find the initial velocity of the ball using the following formula:
v = u + at0 = vi + a*t
Substituting the values, we get:0 = vi + gsinθ*1
The initial velocity of the ball is given as follows:
vi = - gsinθ
The negative sign in the equation shows that the ball is decelerating.
The horizontal distance traveled by the ball is given as follows:
s = vₓ * t
The vertical distance traveled by the ball is given as follows:
h = vᵧ * t + 0.5*a*t²
We know that the ball stops at t = 1 s. Therefore, we can find the space traveled by the ball using the following formula:
s = vₓ * t
Substituting the values, we get:
s = vi cosθ * t
Therefore, the initial speed of the ball is given by:
vi = -g sinθ= -9.8 m/s
The space traveled by the ball when it stops is given by:
s = vₓ * t= vi cosθ * t= (-9.8 m/s) cos 30° × 1 s ≈ -8.48 m (since distance cannot be negative, the distance traveled by the ball is 8.48 m in the opposite direction).
Therefore, the initial speed of the ball i such that it will stop after t = 1 s is -9.8 m/s, and the space traveled by the ball when it stops is 8.48 m.
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What is the frequency f of a 2.89×10 −19 J photon? f= Hz What is the wavelength λ of a 2.89×10 −19 J photon? λ=
The frequency can be expressed as [tex]4.366 *10^{14} Hz[/tex]the wavelength λ can be expressed as [tex]6.876 *10^{-7} meters[/tex]
How can the wavelength be calculated?The frequency of a repeated event is its number of instances per unit of time. For clarity and to distinguish it from spatial frequency, it is also sometimes referred to as temporal frequency.
Frequency is measured in hertz which is equal to one event per secondGiven that Energy =2.89×10 −19 J
h = plank constant = [tex]6.626 *10^{-34}[/tex]
E = hf
f = E / h
f = [tex]\\\frac{2.89* 10^{-19} }{ 6.626*10^{-34} }[/tex]
f= [tex]4.366 *10^{14} Hz[/tex]
To calculate the wavelength we can use
λ = c / f
λ = [tex]\\\frac{2.998 *10^8}{4.366*10^14}[/tex]
λ =[tex]6.876 *10^-7 meters[/tex]
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A cannonball at ground level is aimed 26 degrees above the horizontal and is fired with an initial speed of 105 m/s. How far from the cannon will the cannonball hit the ground? Give your answer in whole numbers.
The cannonball, fired from ground level with an initial speed of 105 m/s at an angle of 26 degrees above the horizontal, will hit the ground at a certain distance of 276 meters.
To determine this distance, we can calculate the projectile's horizontal range using the given information.
The horizontal range of a projectile can be determined using the equation:
Range = (initial velocity^2 * sin(2 * launch angle)) / gravitational acceleration
In this case, the initial velocity is 105 m/s and the launch angle is 26 degrees. The gravitational acceleration is approximately 10 m/s^2. Plugging these values into the equation, we can calculate the range:
Range = (105^2 * sin(2 * 26)) / 10
Simplifying this expression, we get:
Range ≈ 276 meters
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50% Part (b) If the inductor is connected to a 12.0 V battery, what is the current, 1, in amperes, after 13 ms? All content © 2022 Expert TA, LLC 50% Part (b) If the inductor is connected to a 12.0 V battery, what is the current, 1, in amperes, after 13 ms? All content © 2022 Expert TA, LLC 0% Part (a) What is the time constant, t, of the inductor, in seconds? T =
In order to answer the questions, we need more information about the inductor, such as its inductance value and any resistance in the circuit. The time constant and current can be determined using the formula for an RL circuit, which is given by:
I(t) = (V/R) * (1 - e^(-t/τ))
Where:
I(t) is the current at time t,
V is the voltage across the inductor,
R is the resistance in the circuit,
τ is the time constant, and
e is the base of the natural logarithm.
Part (a) - Time Constant:
To calculate the time constant of the inductor, we need to know the inductance (L) and resistance (R) in the circuit. The time constant (τ) is given by the formula:
τ = L / R
Once we have the values of L and R, we can calculate the time constant.
Part (b) - Current after 13 ms:
Using the formula mentioned earlier, we can substitute the values of V (12.0 V), R, and τ into the equation to calculate the current (I) at t = 13 ms.
Without the values for inductance and resistance, we cannot provide specific answers. Please provide the missing values so that we can assist you further in calculating the time constant and current in the circuit.
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A 108 2 resistor is connected in series with a 72 mH inductor and a 0.3 µF capac- itor. The applied voltage has the form 190 V sin(2 ft), where the frequency is f=876 cycles/s. & Find the rms current.
The rms current in the circuit is approximately 0.189 A.
The question requires us to calculate the rms current of a circuit that consists of a resistor, an inductor, and a capacitor in series. The circuit is driven by an AC voltage source that has a frequency of 876 cycles/s and an amplitude of 190 V.Let's begin by finding the total impedance of the circuit. The impedance of a series RLC circuit is given by:Z = R + j(XL - XC)where R is the resistance, XL is the inductive reactance, and XC is the capacitive reactance. The imaginary part of the impedance represents the reactance of the circuit, which depends on the frequency of the applied voltage. At resonance, XL = XC, and the total impedance is equal to the resistance Z = R.
To calculate the impedance of the circuit, we need to find the values of XL and XC at the given frequency f = 876 cycles/s. The inductive reactance is given by:XL = 2πfLwhere L is the inductance of the inductor. Substituting the given values, we get:XL = 2π(876)(72 × 10⁻³) = 101.94 ΩThe capacitive reactance is given by:XC = 1/(2πfC)where C is the capacitance of the capacitor. Substituting the given values, we get:XC = 1/(2π(876)(0.3 × 10⁻⁶)) = 607.71 ΩThe total impedance is therefore:Z = R + j(XL - XC) = 108 + j(-505.77) = 108 - j505.77.
The rms current is given by the ratio of the rms voltage to the impedance:Irms = Vrms/Zwhere Vrms is the rms value of the applied voltage. The rms value of a sinusoidal voltage is given by the peak voltage divided by the square root of 2 (Vrms = Vpeak/√2). Substituting the given values, we get:Vrms = 190/√2 = 134.35 VIrms = Vrms/Z = 134.35/(108 - j505.77) = 0.189 - j0.886 ARms current, Irms = 0.189 A (approx).
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A body with mass m = 20 g should, after being sprung by a spring with spring constant k = 4.8N/cm
was fired, run through a loop path of radius r = 0.5 m without friction.
a) Sketch the forces acting on the body at different points in time.
b) By which piece t do you have to tighten the spring so that the body straightens the loop path
still goes through without falling down?
a) The forces acting on the body at different points in time include gravitational force, normal force, and spring force.
When the body is at the bottom of the loop, the forces include gravitational force, normal force, and centripetal force. At the top of the loop, the forces include gravitational force, normal force, and tension force.
b) To determine the required spring compression, we need to consider the equilibrium of forces at the top of the loop. The gravitational force must provide the necessary centripetal force for the body to complete the loop. By equating these forces, we can solve for the spring compression required to maintain the loop path without the body falling down.
a) When the body is not in contact with the spring, only the gravitational force is acting on it. As the body is sprung, it experiences an upward spring force that opposes the gravitational force. When the body is at the bottom of the loop, in addition to the gravitational force and spring force, there is also a normal force acting upward to counterbalance the gravitational force. At the top of the loop, the forces acting on the body include gravitational force, normal force, and tension force. The normal force provides the necessary centripetal force for the body to follow the curved path.
b) At the top of the loop, the net force acting on the body must be inward, providing the required centripetal force. The net force is given by the difference between the tension force and the gravitational force:
Tension - mg = mv²/r,
where Tension is the tension force, m is the mass of the body, g is the acceleration due to gravity, v is the velocity of the body at the top of the loop, and r is the radius of the loop. Solving for the required tension force, we have:
Tension = mg + mv²/r.
The tension force in the spring is equal to the spring constant multiplied by the compression of the spring:
Tension = k * compression.
Setting the two expressions for tension equal to each other, we can solve for the required spring compression:
mg + mv²/r = k * compression,
compression = (mg + mv²/r) / k.
By substituting the given values of mass, radius, and spring constant, along with the acceleration due to gravity, you can calculate the required spring compression to maintain the loop path without the body falling down.
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15. An engineer launches a projectile from a point 245 m in front of a 325-meter tall building. Its launch velocity is unknown. Ignore the air resistance.
(a) what is the maximum vertical component of initial velocity (vy0) at t =0 is needed to touch the top of the building?
(b) What is the horizontal component of initial velocity (vx0) at t =0 is needed to move 245 m for the projectile to touch the top of building?.
Maximum vertical component of initial velocity (vy0) at t = 0: 19.6 m/s. and Horizontal component of initial velocity (vx0) at t = 0: 122.5 m/s.
To calculate the maximum vertical component of the initial velocity (vy0) at t = 0 needed to touch the top of the building, we can use the equation of motion for vertical motion. The projectile needs to reach a height of 325 meters, so the maximum vertical displacement (Δy) is 325 meters. Since we're ignoring air resistance, the only force acting vertically is gravity. Using the equation Δy = vy0 * t + (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2), we can rearrange the equation to solve for vy0. At the maximum height, the vertical displacement is zero, so the equation becomes 0 = vy0 * t - (1/2) * g * t^2. Substituting the values, we have 0 = vy0 * t - (1/2) * 9.8 * t^2. Solving this quadratic equation, we find t = 2s (taking the positive root). Plugging this value into the equation, we can solve for vy0: 0 = vy0 * 2s - (1/2) * 9.8 * (2s)^2. Solving for vy0, we get vy0 = 9.8 * 2s = 19.6 m/s. (b) To calculate the horizontal component of the initial velocity (vx0) at t = 0 needed for the projectile to move 245 m and touch the top of the building, we can use the equation of motion for horizontal motion. The horizontal distance (Δx) the projectile needs to travel is 245 meters. The horizontal component of the initial velocity (vx0) remains constant throughout the motion since there are no horizontal forces acting on the projectile. Using the equation Δx = vx0 * t, we can rearrange the equation to solve for vx0. Since the time of flight is the same for both the vertical and horizontal motions (2s), we can substitute the value of t = 2s into the equation. Thus, we have 245 = vx0 * 2s. Solving for vx0, we get vx0 = 245 / (2s) = 122.5 m/s.
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When electrons vibrate sympathetically in a radio wave, this is an example of .... A. refraction B. interference
C. resonance
D. reflection
When electrons vibrate sympathetically in a radio wave, this is an example of resonance.
What is Resonance?Resonance is a particular form of mechanical wave motion that occurs when an external force is added to a system at its natural frequency, causing it to oscillate at a higher amplitude. The amplitude of the vibration grows exponentially until a maximum value is reached when resonance occurs.
When electrons vibrate sympathetically in a radio wave, this is an example of resonance. In general, resonances occur when the frequency of a driving force is the same as that of a natural frequency of a system. When a system is exposed to a periodic stimulus, the system will oscillate with an amplitude that is proportional to the strength of the stimulus at its natural frequency.
The passage above explains what resonance is and what happens when a system oscillates at a higher amplitude. Therefore, the best answer to the given question is "C. resonance."
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Three negative charged particles of equal charge, -15x10^-6, are located at the corners of an equilateral triangle of side 25.0cm. Determine the magnitude and direction of the net electric force on each particle.
The magnitude of the net electric force on each particle is 2.025 N directed away from the triangle.
Charge on each particle, q1 = q2 = q3 = -15 × 10⁻⁶C
∴ Net force on particle 1 = F1
Net force on particle 2 = F2
Net force on particle 3 = F3
The magnitude of the net electric force on each particle:
It can be determined by using Coulomb's Law:
F = kqq / r²
where
k = Coulomb's constant = 9 × 10⁹ Nm²/C²
q = charge on each particle
r = distance between the particles
We know that all three charges are negative, so they will repel each other. Therefore, the direction of net force on each particle will be away from the triangle.
From the given data,
Side of equilateral triangle, a = 25cm = 0.25m
∴ Distance between each corner of the triangle = r = a = 0.25m
∴ Net force on particle 1 = F1
F1 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N
∴ Net force on particle 2 = F2
F2 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N
∴ Net force on particle 3 = F3
F3 = kq² / r² = 9 × 10⁹ × (-15 × 10⁻⁶)² / (0.25)²= -2.025 N
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Two 10-cm-diameter charged disks face each other, 18 cm apart. The left disk is charged to -50 nC and the right disk is charged to +50 nC.
▼ Part A What is the electric field's E magnitude at the midpoint between the two disks?
The electric field's E magnitude at the midpoint between the two disks is 3.6 x 10⁷ N/C.
When two charged plates face each other, they form a capacitor. The electric field at the midpoint of two plates is provided by the expression for a parallel plate capacitor:
Electric field, E = σ/2εwhere σ is the surface charge density, and ε is the permittivity of the space or material between the plates.In this question, both plates are circular with a diameter of 10cm.
So, we can calculate the surface area of each plate by using the equation for the area of a circle:
A = πr²
where r is the radius of the circle, given as 5cm.
A = π(5cm)² = 78.5cm²
The surface charge density is given in nano-coulombs (nC), so we need to convert it to Coulombs (C).
1nC = 1 x 10⁻⁹C
Because the left plate is charged to -50nC, the surface charge density is:-
50nC / 78.5cm² = -6.37 x 10⁻¹⁰C/cm²
Because the right plate is charged to +50nC, the surface charge density is:
+50nC / 78.5cm² = 6.37 x 10⁻¹⁰C/cm²
The electric field at the midpoint between the two plates can now be calculated:
|E| = σ/2ε = 6.37 x 10⁻¹⁰C/cm² / (2 x 8.85 x 10⁻¹²F/cm) = 3.6 x 10⁷N/C
Due to the nature of the problem, the electric field between the two plates is directed from right to left, and its magnitude is 3.6 x 10⁷ N/C (newtons per coulomb).
Therefore, the magnitude of the electric field at the midpoint between the two disks is 3.6 x 10⁷ N/C.
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someone wants to fly a distance of 100km on a bearing of 100 degrees. speed of plane in still air is 250km/h. a 25km/h wind is vlowing on a bearing of 215 degrees. a villan turns on a magent that exerts a force equivalent to 5km/h on a bearing of 210 degrees on the airplane in the sky. what bearjng will the plane need to take to reach their destination?
The plane needs to take a bearing of 235.19 degrees to reach its destination.
How to calculate the valueNorthward component = 25 km/h * sin(215 degrees) ≈ -16.45 km/h
Eastward component = 25 km/h * cos(215 degrees) ≈ -14.87 km/h
Northward component = 5 km/h * sin(210 degrees) ≈ -2.58 km/h
Eastward component = 5 km/h * cos(210 degrees) ≈ -4.33 km/h (opposite
Total northward component = -16.45 km/h + (-2.58 km/h) ≈ -19.03 km/h
Total eastward component = -14.87 km/h + (-4.33 km/h) ≈ -19.20 km/h
Resultant ground speed = sqrt((-19.03 km/h)^2 + (-19.20 km/h)²) ≈ 26.93 km/h
Resultant direction = atan((-19.20 km/h) / (-19.03 km/h)) ≈ 135.19 degrees
Final bearing = 135.19 degrees + 100 degrees
≈ 235.19 degrees
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