The angle at which light emerges from the opposite face of the equilateral glass prism is 51.2°.
What is Refraction?Refraction is the bending of light when it passes from one medium to another. The angle of refraction is determined by the angle of incidence and the refractive index of the medium through which it is travelling.
The angle at which light emerges from the opposite face of the equilateral glass prism can be calculated using the concept of refraction.
In this case, the angle of incidence is 45° and the refractive index of the medium is 1.52.
Using Snell's law of refraction, the angle of refraction can be calculated as follows:
n₁ sinθ₁ = n₂ sinθ₂
Where n₁ is the refractive index of the incident medium, θ₁ is the angle of incidence, n₂ is the refractive index of the emergent medium, and θ₂ is the angle of refraction.
Therefore, substituting the values for n₁, θ₁ and n₂, the angle of refraction can be calculated as follows:
1.52 sin 45° = n₂ sinθ₂
n₂ sinθ₂ = 1.52 sin 45°
n₂ sinθ₂ = 1.08
θ₂ = sin-¹ (1.08/n₂)
θ₂ = sin-¹ (1.08/1.52)
θ₂ = 51.2°
Therefore, the angle at which light emerges from the opposite face of the equilateral glass prism is 51.2°.
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a wire of 1.0 mm diameter and 2.0 m length and 50 mω is melted and redrawn a 0.2 mm diameter wire. find new resistance of wire.
The resistance of the 0.2 mm diameter wire redrawn from a melted wire of 1.0 mm diameter is approximately 3125 mΩ.
To find the new resistance of the wire, we will first calculate the volume of the original wire and then find the new length using the given diameter. Lastly, we will apply the formula for resistance.
1. Volume of the original wire:
V = πr²h
V = π(0.5mm)²(2000mm) = 1570.8 mm³ (approx)
2. Find the new length (L₂):
V = πr²h
1570.8 = π(0.1mm)²(L₂)
L₂ = 50000 mm (50 m)
3. Calculate new resistance using the formula:
R = ρ(L/A)
Where ρ (rho) is the resistivity, L is the length, and A is the cross-sectional area.
First, find the resistivity (ρ) using the original resistance:
50 mΩ = ρ(2000mm) / (π(0.5mm)²)
ρ = 19.63 µΩm (approx)
Now, calculate the new resistance (R₂):
R₂ = 19.63 µΩm * (50000 mm) / (π(0.1mm)²)
R₂ ≈ 3125 mΩ (approx)
The new resistance of the 0.2 mm diameter wire is approximately 3125 mΩ.
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The resistance of the 0.2 mm diameter wire redrawn from a melted wire of 1.0 mm diameter is approximately 3125 mΩ.
To find the new resistance of the wire, we will first calculate the volume of the original wire and then find the new length using the given diameter. Lastly, we will apply the formula for resistance.
1. Volume of the original wire:
V = πr²h
V = π(0.5mm)²(2000mm) = 1570.8 mm³ (approx)
2. Find the new length (L₂):
V = πr²h
1570.8 = π(0.1mm)²(L₂)
L₂ = 50000 mm (50 m)
3. Calculate new resistance using the formula:
R = ρ(L/A)
Where ρ (rho) is the resistivity, L is the length, and A is the cross-sectional area.
First, find the resistivity (ρ) using the original resistance:
50 mΩ = ρ(2000mm) / (π(0.5mm)²)
ρ = 19.63 µΩm (approx)
Now, calculate the new resistance (R₂):
R₂ = 19.63 µΩm * (50000 mm) / (π(0.1mm)²)
R₂ ≈ 3125 mΩ (approx)
The new resistance of the 0.2 mm diameter wire is approximately 3125 mΩ.
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approximately what is the smallest detail observable with a microscope that uses red light of frequency 4.32×1014 hz ?
The smallest observable detail with a microscope using red light of frequency [tex]4.32×1014 Hz[/tex] is approximately 347 nm. This is due to the diffraction limit of light, which depends on the wavelength and numerical aperture of the lens.
The diffraction limit of a microscope is the smallest resolvable feature based on the wavelength and numerical aperture of the lens. The resolution limit (d) is given by[tex]d ≈ λ/2NA,[/tex] where λ is the wavelength of light and NA is the numerical aperture of the lens. For red light with a wavelength of 690 nm and a numerical aperture of 1.4, the resolution limit is approximately 347 nm. Therefore, this is the smallest observable detail with a microscope using red light of frequency [tex]4.32×1014 Hz.[/tex]
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Solved Least Mass Problem A regular hexagon pivoted
The equilateral triangle will come to rest in a stable equilibrium position when released from the initial position. The square has the smallest moment of inertia about the pivot point.
This is because the center of mass of the equilateral triangle is directly above the pivot point, while for the square and hexagon, the center of mass is off to one side. To determine which shape has the smallest moment of inertia about the pivot point, we can use the formula for the moment of inertia of a polygon about its centroid:
I = (n/12) * s² * h² * (1 + cos(2*pi/n))
where n is the number of sides, s is the length of each side, and h is the distance from the centroid to a side. For a regular polygon, the centroid is also the center of mass, so we can use the side length as the distance from the pivot point.
Plugging in the values for each shape,
Square: I = (4/12) * 1² * (1/2)² * (1 + cos(pi/2)) = 1/12
Hexagon: I = (6/12) * 2² * (√(3)/2)² * (1 + cos(pi/3)) = 2√(3)/3
Equilateral triangle: I = (3/12) * 3² * (√(3)/3)² * (1 + cos(2*pi/3)) = 9sqrt(3)/4
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--The complete question is, A square, a regular hexagon, and an equilateral triangle are pivoted about one of their vertices such that they can freely rotate in the plane. The sides of the shapes have lengths 1, 2, and 3 units, respectively. Which shape will come to rest in a stable equilibrium position when released from the initial position? Which shape has the smallest moment of inertia about the pivot point?--
an image formed when the light rays do not actually pass through the image location, and wouldnot appear on paper or film placed at that location is referred to as a
An image formed when light rays do not actually pass through the image location and would not appear on paper or film placed at that location is referred to as a "virtual image".
Virtual images are formed by the apparent intersection of light rays, which cannot be projected onto a surface, unlike real images. A virtual image is formed when the light rays appear to be coming from a particular location, but they do not actually converge at that location. Instead, they diverge or appear to be coming from a different location, creating the illusion of an image that would not be visible on paper or film placed at the location where the virtual image appears.
This phenomenon can be observed in mirrors, lenses, and other optical devices that reflect or refract light.
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A large plate subjected to pure shear σxy=S at remote boundaries contains a rigid circular inclusion in the region r
A rigid circular inclusion in a large plate subjected to pure shear at remote boundaries creates stress concentrations around the inclusion, which increase with decreasing shear modulus of the inclusion.
When a plate is subjected to pure shear stress, the stress distribution is uniform throughout the plate, except near the boundaries. However, the presence of a rigid circular inclusion within the plate creates stress concentrations around the inclusion due to the mismatch in stiffness between the inclusion and the surrounding matrix.
These stress concentrations increase with decreasing shear modulus of the inclusion. This phenomenon is important in understanding the behavior of composite materials, where the presence of inclusions of different materials affects the overall mechanical properties of the composite.
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--The complete question is, What is the effect of a rigid circular inclusion on the stress distribution of a large plate subjected to pure shear at remote boundaries? How does the value of the shear modulus of the inclusion affect the stress concentration around the inclusion?--
a mixture of 10.0 g of ne and 10.0 g ar have a total pressure of 1.60 atm. what is the partial pressure of ar? 1.07 atm 0.400 atm 0.537 atm 0.800 atm 1.32 atm
The partial pressure of Ar in the mixture is 0.537 atm.
To find the partial pressure of Ar in a mixture of 10.0 g of Ne and 10.0 g Ar with a total pressure of 1.60 atm, you can use Dalton's Law of Partial Pressures.
Here are the steps:
1. Calculate the moles of each gas using their molar masses (Ne: 20.18 g/mol, Ar: 39.95 g/mol):
Moles of Ne = 10.0 g / 20.18 g/mol = 0.495 moles
Moles of Ar = 10.0 g / 39.95 g/mol = 0.250 moles
2. Calculate the mole fractions of each gas:
Mole fraction of Ne = moles of Ne / (moles of Ne + moles of Ar) = 0.495 / (0.495 + 0.250) = 0.664
Mole fraction of Ar = moles of Ar / (moles of Ne + moles of Ar) = 0.250 / (0.495 + 0.250) = 0.336
3. Use Dalton's Law to find the partial pressure of Ar:
Partial pressure of Ar = mole fraction of Ar * total pressure = 0.336 * 1.60 atm = 0.537 atm
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A hobbyist builds a circuit in which an AC power supply with an rms voltage of 115 V is connected to a 1.86 kΩ resistor. (a) What is the maximum potential difference across the resistor (in V)? ____ V
(b) What is the maximum current through the resistor (in A)? ___ A (c) What is the rms current through the resistor (in A)? ___ A (d) What is the average power dissipated by the resistor (in W)? ____ W
If the RMS voltage across the 1.86 kΩ resistor is 115 V then:
(a) The potential difference across the resistor is 162.6 V.
(b) The maximum current through the resistor is 0.087 A.
(c) The RMS current through the resistor is 0.062 A.
(d) The average power dissipated by the resistor is 7.13 W.
(a) The maximum potential difference across the resistor can be found using the formula Vmax = Vrms x √2, where Vrms is the RMS voltage.
Plugging in the values, we get Vmax = 115 x √2 = 162.6 V.
Therefore, the maximum potential difference across the resistor is 162.6 V.
(b) The maximum current through the resistor can be found using Ohm's Law, which states that I = V/R, where V is the potential difference and R is the resistance.
Plugging in the values, we get I = 162.6/1860 = 0.087 A.
Therefore, the maximum current through the resistor is 0.087 A.
(c) The RMS current through the resistor is equal to the maximum current divided by √2.
Therefore, Irms = 0.087/√2 = 0.062 A.
(d) The average power dissipated by the resistor can be found using the formula Pavg = Vrms x Irms x cos(θ), where cos(θ) is the power factor.
For a purely resistive circuit like this one, the power factor is 1.
Plugging in the values, we get Pavg = 115 x 0.062 x 1 = 7.13 W.
Therefore, the average power dissipated by the resistor is 7.13 W.
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Johnny, Susie, and Joe are working together to push a 65 kg filing cabinet across the floor to the right Johnny pushes with a force of 85 N, Susie pushes with a force of 85 N, and Joe pushes with a force of 50 N. The cabinet moves at a constant speed. what is closest to the weight of the filing cabinet
The weight of the filing cabinet is closest to 638 N.
How to solve for the weightSince the cabinet is moving at a constant speed, we know that the net force acting on it is zero (otherwise it would be accelerating).
Therefore, the force of friction acting on the cabinet must be equal in magnitude and opposite in direction to the total force applied by Johnny, Susie, and Joe.
The total force applied is:
85 N + 85 N + 50 N = 220 N
Therefore, the force of friction is also 220 N.
The weight of the cabinet can be calculated using the formula:
weight = mass x gravity
where mass is given as 65 kg and gravity is approximately 9.81 m/s^2.
weight = 65 kg x 9.81 m/s^2 = 637.65 N
So the weight of the filing cabinet is closest to 638 N.
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A series of small machine components being moved by a conveyor belt pass over a 120-mm-radius idler pulley. At the instant shown, the velocity of point A is 300 mm/s to the left and its acceleration is 180 mm/s^2 to the right. Determine (a) the angular velocity and angular acceleration of the idler pulley, (b) the total acceleration of the machine component at B.
(a) The angular velocity is 2.5 rad/s and angular acceleration is 1.5 rad/s²
(b) The total acceleration of the machine component at B is 360 mm/s² to the right.
(a) To find the angular velocity of the idler pulley, we can use the formula:
ω = v / r
where v is the velocity of point A and r is the radius of the pulley. Thus, we have:
ω = 300 mm/s / 120 mm = 2.5 rad/s
To find the angular acceleration of the pulley, we can use the formula:
α = a / r
where a is the acceleration of point A and r is the radius of the pulley. Thus, we have:
α = 180 mm/s² / 120 mm = 1.5 rad/s²
(b) To find the total acceleration of the machine component at B, we can use the formula:
aB = aA + α × r
where aA is the acceleration of point A and α is the angular acceleration of the pulley. We know that aA = 180 mm/s² to the right, and from part (a) we found that α = 1.5 rad/s². The radius of the pulley is 120 mm. Thus, we have:
aB = 180 mm/s² + (1.5 rad/s²) × 120 mm
aB = 360 mm/s² to the right
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A sound wave with wavelength λ0 and frequency f0 moves into a new medium in which the speed of sound is v1=2v0.
What is the new wavelength λ1?
Express your answer in terms of λ0.
What is the new frequency f1?
Express your answer in terms of f0.
The new frequency f1 is half of the original frequency f0.
When a sound wave moves into a new medium, its wavelength changes while its frequency remains constant. The new wavelength can be found using the equation:
λ1 = λ0 * (v0 / v1)
where λ0 is the wavelength in the original medium, v0 is the speed of sound in the original medium, and v1 is the speed of sound in the new medium.
Substituting the given values, we get:
λ1 = λ0 * (v0 / 2v0)
λ1 = λ0 / 2
Therefore, the new wavelength λ1 is half of the original wavelength λ0.
The new frequency f1 can be found using the equation:
f1 = f0 * (v0 / v1)
where f0 is the frequency in the original medium.
Substituting the given values, we get:
f1 = f0 * (v0 / 2v0)
f1 = f0 / 2
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A 0.72 mm diameter hole is illuminated by infrared light of wavelength 2.5 micrometers. What is the angle of the first dark fringe?
The angle of the first dark fringe is 0.398°, when the diameter hole is illuminated.
To find the angle of the first dark fringe when a 0.72 mm diameter hole is illuminated by infrared light with a wavelength of 2.5 micrometers, you can use the formula for the angular width of the central maximum in a single-slit diffraction pattern:
θ = (2 * λ) / d
where θ is the angle of the first dark fringe, λ is the wavelength of the light, and d is the diameter of the hole.
First, convert the diameter to micrometers: 0.72 mm = 720 micrometers.
Now, plug in the values:
θ = (2 * 2.5) / 720
θ = 5 / 720
To find the angle in radians, calculate:
θ = 0.00694 radians
To convert radians to degrees, multiply by (180 / π):
θ = 0.00694 * (180 / π)
θ ≈ 0.398°
The angle of the first dark fringe is approximately 0.398°.
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Find the tension in an elevator cable if the 1,000kg elevator is descending with an acceleration of 1.8 m/s2, downward.a. 16,000 Nb. 8,000 Nc. 12,000 Nd. 20,000 N
The tension in the elevator cable if the 1,000kg elevator is descending with an acceleration of 1.8 m/s2, downward is 8,000 N. The correct option is b.
To find the tension in the elevator cable, we need to use the formula: Tension (T) = m(g - a), where m is the mass of the elevator (1,000 kg), g is the acceleration due to gravity (9.8 m/s²), and a is the acceleration of the descending elevator (1.8 m/s²).
T = 1,000 kg * (9.8 m/s² - 1.8 m/s²)
T = 1,000 kg * 8 m/s²
T = 8,000 N
Therefore, the tension in the elevator cable is 8,000 N (option b).
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In Eq. (5), show that vpi = Ro √g/2h Why was R0 used to calculate this velocity?
this is the equation (5) its refering to in question number 3
po = mp vpi
po = mp (Ro/t)
R0 is used to calculate this velocity because it represents a distance related to the particle's motion. By incorporating it into the equation, we can determine the initial velocity (vpi) based on the distance (Ro) and the height (h) under the influence of gravity (g).
In the given equation, po represents the force exerted on the fluid by the piston, mp represents the mass of the piston and vpi represents the velocity of the piston.
To calculate vpi, we can rearrange the equation as:
vpi = po/mp
We know that the force exerted on the fluid by the piston is equal to the weight of the column of fluid above it. Therefore, we can substitute po with ρghA, where ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid column above the piston and A is the cross-sectional area of the piston.
vpi = (ρghA)/mp
We can simplify this equation further by substituting A with πR0^2, where R0 is the radius of the piston.
vpi = (ρghπR0^2)/mp
Now, we can substitute mp with ρπR0^2t, where t is the thickness of the piston.
vpi = (ρghπR0^2)/(ρπR0^2t)
Simplifying further, we get:
vpi = gh/2t
Finally, we can substitute t with R0/2 to get:
vpi = Ro√g/2h
Therefore, R0 was used in the calculation of vpi because it is the radius of the piston, which determines the cross-sectional area of the piston and hence the force exerted on the fluid.
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in a sad turn of events, the same sports car traveling at 35 m/s. plows into a rock wall and comes to rest in 0.25 seconds. determine the size of the force to stop the car
a. 75,000 N
b. -5,000 N
c. -10,000 N
d. -140,000 N
Each pulse produced by an argon-fluoride excimer laser used in PRK and LASIK ophthalmic surgery lasts only 10.0 ns but delivers an energy of 2.50 mJ.
part a: What is the power produced during each pulse?
part b: If the beam has a diameter of 0.850 mm, what is the average intensity of the beam during each pulse?
part c: If the laser emits 55 pulses per second, what is the average power it generates?
Part a: The power produced during each pulse can be calculated by dividing the energy (2.50 mJ) by the duration of the pulse (10.0 ns). This yields a power of 250 kW.
Part b: The average intensity of the beam during each pulse can be calculated by dividing the power (250 kW) by the area of the beam (0.850 mm). This yields an intensity of 294.12 MW/m^2.
Part c: The average power generated by the laser can be calculated by multiplying the number of pulses (55) per second by the power of each pulse (250 kW). This yields an average power of 13.75 MW.
Therefore, the argon-fluoride excimer laser produces a pulse of energy 2.50 mJ, lasting 10.0 ns, with a power of 250 kW. The beam has an average intensity of 294.12 MW/m^2 and an average power of 13.75 MW when it emits 55 pulses per second. This laser is commonly used in PRK and LASIK ophthalmic surgery due to its ability to produce a very short pulse of high intensity and power.
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A force of 100N acts on a body of mass 20kg. The force accelerates the body from rest until it attains a velocity of 20ms-1 . Through what distance the force acts?
Okay, let's break this down step-by-step:
* There is a force of 100N acting on the body.
* The mass of the body is 20kg.
* The body accelerates from rest to a velocity of 20ms^-1.
To calculate the distance over which this force acts:
1) Calculate the acceleration: Force / Mass = 100N / 20kg = 5ms^-2
2) Calculate the displacement (distance traveled) using: displacement = 1/2 * acceleration * time^2
Since the acceleration is constant, we can set the initial velocity to 0 and final velocity to 20ms^-1.
Then: time = (20ms^-1) / 5ms^-2 = 4s
3) Displacement = 1/2 * 5ms^-2 * 4s^2 = 20m
Therefore, the force of 100N acts on the body over a distance of 20m to accelerate it from rest to 20ms^-1.
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by what percentage is the maximum range of a 0.850 kgkg ball reduced if fwindfwindf_wind == 0.900 nn ? express your answer as a percentage. view available hint(s)
the maximum range of the 0.850 kg ball is reduced by approximately 10.8% due to the 0.900 N wind force.we need to find the percentage by which the maximum range of the 0.850 kg ball is reduced due to the 0.900 N wind force.
Step 1: Calculate the force due to gravity acting on the ball
Force = mass × acceleration due to gravity
F_gravity = 0.850 kg × 9.81 m/s²
F_gravity = 8.3385 N
Step 2: Calculate the net force acting on the ball
Net force = F_gravity - F_wind
Net force = 8.3385 N - 0.900 N
Net force = 7.4385 N
Step 3: Calculate the percentage reduction in force
Percentage reduction = [(F_gravity - Net force) / F_gravity] × 100
Percentage reduction = [(8.3385 N - 7.4385 N) / 8.3385 N] × 100
Percentage reduction ≈ (0.9 N / 8.3385 N) × 100
Percentage reduction ≈ 10.8%
So, the maximum range of the 0.850 kg ball is reduced by approximately 10.8% due to the 0.900 N wind force.
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An airplane passenger has 120 cm3 of air in his stomach just before the plane takes off from a sea-level airport. What volume, in cubic centimeters, will the air have at cruising altitude at the same temperature (body temperature) if cabin pressure drops to 7.50 × 104 Pa and no air leaves their stomach?
The volume will the air have at cruising altitude at the same temperature (body temperature) if cabin pressure drops to 7.50 × 10⁴ Pa and no air leaves their stomach is 162 cm³.
To determine the volume of air in the passenger's stomach at cruising altitude, we need to consider the initial and final pressure. The initial volume is 120 cm³ and the cabin pressure at cruising altitude is 7.50 × 10⁴ Pa. We can use Boyle's law, which states that the product of pressure and volume remains constant when temperature is constant: P₁V₁ = P₂V₂.
However, we first need to determine the initial pressure at sea level. Standard atmospheric pressure at sea level is approximately 1.013 × 10⁵ Pa. Now we can use Boyle's law:
(1.013 × 10⁵ Pa)(120 cm³) = (7.50 × 10⁴ Pa)(V₂)
To solve for V₂, the final volume at cruising altitude:
V₂ = (1.013 × 10⁵ Pa)(120 cm³) / (7.50 × 10⁴ Pa)
≈ 162 cm³
So, the volume of air in the passenger's stomach at cruising altitude at the same temperature will be approximately 162 cm³.
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what is the heat change when a 225 g sample of olive oil (c = 1.79 j/g c) is cooled from 95.8°c to 52.1°c?a. 1.76x104) b. 1.76x10 c. -9,83 x 10 d. 9.83 x 10) e. not enough information given
The solidification of olive oil is a slow and gradual process that takes the consistency from a liquid to a soft state to a solid state. The cooling process begins at 40-50°F. This means that the temperature of this oil will start to solidify.
To find the heat change when a 225g sample of olive oil (c = 1.79 J/g°C) is cooled from 95.8°C to 52.1°C, this formula can be used:
q = mcΔT
where:
q = heat change
m = mass of the sample (225 g)
c = specific heat capacity of olive oil (1.79 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
Step 1: Calculate ΔT
ΔT = 52.1°C - 95.8°C = -43.7°C
Step 2: Plug the values into the formula
q = (225 g)(1.79 J/g°C)(-43.7°C)
Step 3: Calculate q
q = -17637.975 J
Since the answer should be expressed in scientific notation, you can round it to two significant figures:
q ≈ -1.76 x 10^4 J
The heat change when the 225 g sample of olive oil is cooled from 95.8°C to 52.1°C is approximately -1.76 x 10^4 J.
So, the correct answer is (a.)
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A ferry boat is sailing at 12 km 30 degrees W of N with respect to a river that is flowing at 6.0 km/h E. As observed from the shore, what direction is the ferry boat sailing?
The ferry boat is sailing in a direction of 0 degrees (due North) at a speed of 10.39 km/h. To determine the direction of the ferry boat as observed from the shore, we must consider both the ferry boat's velocity and the river's velocity. The ferry boat is sailing at 12 km/h 30 degrees W of N, and the river is flowing at 6.0 km/h E.
Step 1: Break the ferry boat's velocity into its components:
- Northward component: 12 km/h * cos(30°) = 10.39 km/h
- Westward component: 12 km/h * sin(30°) = 6 km/h
Step 2: Add the river's velocity to the ferry boat's components:
- Northward component: 10.39 km/h (unchanged)
- Eastward component: 6.0 km/h (from the river) - 6 km/h (from ferry boat) = 0 km/h
Step 3: Determine the resultant velocity's magnitude and direction:
- Magnitude: √(10.39^2 + 0^2) = 10.39 km/h
- Direction: tan^-1(0/10.39) = 0° (N).
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you buy a gold crown at a flea market. when you get home, you use your physics knowledge to check whether the crown is pure gold. you hang it form a scale and find its weigh to be 7.84 N. them you weigh the crown when it is completely in water, now the scale reads 6.84 N. The density of gold is 19300 kgm^-3. density of water is 1000kgm^-3. what is the buoyant force on the crown?
To determine the buoyant force on the crown, we can use the principle of buoyancy, which states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object.
First, we need to calculate the volume of the crown. We can use the fact that the density of gold is 19300 kg/m^3, and the weight of the crown is 7.84 N, to find the volume of the crown:
Density = Mass/Volume
Volume = Mass/Density
Volume = 7.84 N / (19300 kg/m^3)
Volume = 4.06 x 10^-4 m^3
Next, we need to determine the weight of the water displaced by the crown when it is submerged in water. We can use the fact that the weight of the crown in water is 6.84 N, to find the weight of the water displaced by the crown:
Weight of water displaced = Weight of crown - Weight of crown in water
Weight of water displaced = 7.84 N - 6.84 N
Weight of water displaced = 1 N
Since the density of water is 1000 kg/m^3, we can use the weight of the water displaced to find the volume of water displaced:
Density = Mass/Volume
Volume = Mass/Density
Volume = 1 N / (1000 kg/m^3)
Volume = 1 x 10^-3 m^3
Since the volume of water displaced is equal to the volume of the crown, we can conclude that the crown is made of pure gold, since its density is the same as that of gold.
Therefore, the buoyant force on the crown is equal to the weight of the water displaced, which is 1 N.
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What is the definition of velocity and acceleration?
The definition of velocity is vector quantity that represents the rate of change of an object's position with respect to time. The definition of acceleration another vector quantity that represents the rate of change of an object's velocity with respect to time
Velocity has both magnitude (speed) and direction, making it different from speed, which is a scalar quantity with only magnitude. In simple terms, velocity indicates how fast an object is moving and in which direction. The formula for velocity is v = Δx/Δt, where v is velocity, Δx is the change in position, and Δt is the change in time.
Acceleration indicates how quickly an object is speeding up or slowing down, as well as changing its direction. The formula for acceleration is a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time. Both velocity and acceleration are crucial concepts in physics, particularly in the study of motion. Understanding these terms helps us analyze and predict the behavior of moving objects in various situations, from everyday experiences to complex scientific phenomena.
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A cart is moving on a horizontal track. A heavy bag falls off the cart and moves straight down relative to the cart. Describe what happens to the speed of the cart. Represent your answer with the impulse-momentum bar chart.
When a heavy bag falls off a moving cart on a horizontal track, the speed of the cart will be affected due to the conservation of linear momentum.
As the bag falls straight down, an equal and opposite impulse acts on the cart, causing it to increase in speed. In an impulse-momentum bar chart, the initial momentum (mass x initial velocity) of the cart-bag system is represented by a bar. After the bag falls, the final momentum is split into two separate bars: one for the cart and one for the bag. The cart's momentum bar will be larger than its initial momentum, indicating an increase in its speed, while the bag's momentum bar will represent its downward momentum. The sum of the final momentum bars will be equal to the initial momentum bar, conserving the total linear momentum of the system.
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When a heavy bag falls off a moving cart on a horizontal track, the speed of the cart will be affected due to the conservation of linear momentum.
As the bag falls straight down, an equal and opposite impulse acts on the cart, causing it to increase in speed. In an impulse-momentum bar chart, the initial momentum (mass x initial velocity) of the cart-bag system is represented by a bar. After the bag falls, the final momentum is split into two separate bars: one for the cart and one for the bag. The cart's momentum bar will be larger than its initial momentum, indicating an increase in its speed, while the bag's momentum bar will represent its downward momentum. The sum of the final momentum bars will be equal to the initial momentum bar, conserving the total linear momentum of the system.
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a solenoid that is 85.0cm long has a radius of 1.50 cm and a winding of 1500 turns; it carries a current of 4.60a. calculate the magnitude of the magnetic field inside the solenoid. (4 pt)
A solenoid that is 85.0cm long has a radius of 1.50 cm and a winding of 1500 turns; it carries a current of 4.60a, then the magnitude of the magnetic field inside the solenoid is approximately [tex]1.65 \times 10^{-2}[/tex] T.
To calculate the magnitude of the magnetic field inside the solenoid, we can use the formula B = μ₀ * n * I, where B is the magnetic field, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current.
Here's a step-by-step explanation:
1. First, we need to find the number of turns per unit length (n). We're given that the solenoid has a length of 85.0 cm and a winding of 1500 turns. Convert the length to meters: 85.0 cm = 0.85 m.
n =( number of turns) /( length) = 1500 turns /[tex]\frac{1500 turns}{ 0.85 m}[/tex] = 1764.71 turns/m.
2. Next, we're given that the solenoid carries a current (I) of 4.60 A.
3. Now, we need to find the permeability of free space (μ₀). This is a constant value: μ₀ = [tex]4\pi \times 10^{-7}[/tex]T·m/A.
4. Finally, we can calculate the magnetic field (B) using the formula: B = μ₀ * n * I.
[tex]B = (4\pi \times 10^{-7} T \frac{m}{A} ) * (1764.71 turns/m) * (4.60 A).[/tex]
5. Solving the equation, we get:
[tex]B \approx 1.65 \times 10^{-2} T.[/tex]
So, the magnitude of the magnetic field inside the solenoid is approximately [tex]1.65 \times 10^{-2}[/tex] T.
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two parallel 3.0-cm-diameter flat aluminum electrodes are spaced 0.50 mm apart. the electrodes are connected to a 50 v battery.A) What is the capacitance?B) What is the magnitude of the charge on each electrode?
The capacitance (A) is 3.54 pF, and the magnitude of the charge (B) on each electrode is 177 pC.
To calculate the capacitance (A), use the formula C = ε₀ * A / d, where ε₀ is the vacuum permittivity (8.85 * 10⁻¹² F/m), A is the area of the electrode, and d is the distance between the electrodes. First, find the area of the electrode by using A = π * r², with r = 1.5 cm. Then, plug the values into the capacitance formula and solve.
To find the magnitude of the charge (B) on each electrode, use Q = C * V, where Q is the charge, C is the capacitance, and V is the voltage. Plug in the calculated capacitance and the given voltage (50 V) and solve for the charge.
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On a standard day at 20,000 ft, if ve is 430 kn, what is the true velocity and mach number?
The true velocity at 20,000 ft is approximately 568.4 knots and the Mach number is approximately 1.93.
The true velocity (Vt) and Mach number (M) can be calculated using the following equations:
Vt = Ve / √(ρ/ρ₀)
M = Vt / a
where:
Ve = indicated airspeed
ρ = density of air at altitude
ρ₀ = density of air at sea level (1.225 kg/m³)
a = speed of sound at altitude
Assuming standard atmospheric conditions, the density of air at 20,000 ft is approximately 0.286 kg/m³ and the speed of sound is approximately 295 m/s. Substituting these values into the equations, we get:
Vt = (430 kn) / √(0.286/1.225) = 568.4 kn
M = (568.4 kn) / (295 m/s) = 1.93
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Antonina throws a coin straight up from a height of 1.2m above the water surface in a fountain. The coin strikes the water 1.3s later. We want to know the vertical velocity of the coin at the moment it strikes the water. We can ignore air resistance.
Which kinematic formula would be most useful to solve for the target unknown?
Answer: v=1,56(m/s)
Explanation: Called: g(Gravity acceleration) =10[tex]m/s^{2} \\[/tex](or 9,8[tex]m/s^{2}[/tex])
h=1,2m ; t=1,3s
Because the coin fell freely, the velocity is calculated by the formula:
v=[tex]\sqrt{2gh}[/tex] (1)
besides, you have time to touch the bottom of the water is 1,3s:
t=[tex]\sqrt\frac{2g}{h} }[/tex] (2)
(1) and (2) => v=1,56 (m/s)
A 62 kg sprinter, starting from rest, runs 51 m in 7.0 s at constant acceleration. What is the magnitude of the horizontal force acting on the sprinter? What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s? Express your answers using two significant figures separated by commas.
The magnitude of the horizontal force acting on the sprinter is 76 N, and the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s is 684 W, 342 W, and 228 W, respectively.
The first step is to find the sprinter's acceleration using the equation:
distance = (initial velocity x time) + (1/2 x acceleration x [tex]time^2[/tex])
Rearranging this equation gives us:
acceleration = 2 x (distance - initial velocity x time) / [tex]time^2[/tex])
Plugging in the given values, we get:
acceleration = 2 x (51 m - 0 m/s x 7.0 s) / [tex](7.0 s)^2[/tex] = [tex]1.22 m/s^2[/tex]
Next, we can use Newton's second law of motion, F = ma, to find the magnitude of the horizontal force acting on the sprinter:
force = mass x acceleration = 62 kg x [tex]1.22 m/s^2[/tex] = 76 N
To find the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s, we can use the equation for power:
power = work/time
The work done by the sprinter is equal to the force exerted multiplied by the distance traveled, since the force and displacement are in the same direction. Therefore, we can use the equation:
work = force x distance
Plugging in the values, we get:
work = 76 N x (2 m + 4 m + 6 m) = 1368 J
Using the given times, we can calculate the power output at each time:
power at 2.0 s = work done in 2.0 s / 2.0 s = 684 J/s = 684 W
power at 4.0 s = work done in 4.0 s / 4.0 s = 342 J/s = 342 W
power at 6.0 s = work done in 6.0 s / 6.0 s = 228 J/s = 228 W
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Application 2 Consider an object (A) which moves in a uniform rectilinear motion in the negative direction of the x-axis. The speed of (A) is 15 m/s, and its initial abscissa is XoA 30 m. 1. a. Determine the time equation of the motion of (A). b. Draw the (V-1) graph of (A). 2. Another particle (B), whose time equation is XB = 101-70 (S.1), is moving on the same axis. We start timing for (A) and (B) simultaneously. a. Determine the instant at which (A) and (B) meet as well as the position at where they meet. b. Determine the distance separating (A) and (B) at t = 8 s. B- Acceleration rectilinear motion : ARM
1. For an object (A) moving uniformly in the negative direction of the x-axis with a speed of 15 m/s and initial abscissa of 30 m, the time equation of its motion is X = 30 - 15*t, and its (V-1) graph is a straight line with a slope of -15 and a y-intercept of 30, and 2. When another particle (B) with a time equation of XB = 101-70 (S.1) moves on the same axis, (A) and (B) meet at time t = 2.13 s and position X = -32.7 m. The distance separating (A) and (B) at t = 8 s is 369 m.
1.a. The time equation of the motion of (A) is given by:
X = XoA + Vt
where X is the position of (A) at time t, XoA is the initial position of (A), V is the velocity of (A) and t is the time elapsed since the start of the motion.
Plugging in the given values, we get:
X = 30 - 15t
b. The (V-1) graph of (A) is a straight line with a slope of -15 (since the velocity is constant and negative) and a y-intercept of 30 (since the initial position is 30). The graph looks like this:( below)
2a. To determine the instant at which (A) and (B) meet, we need to find the time t at which their positions are equal. Equating the time equations of (A) and (B), we get:
30 - 15t = 101 - 70t
Solving for t, we get:
t = 2.13 s
To find the position at which they meet, we can plug this value of t into either of the time equations and get:
X = 101 - 70*2.13 = -32.7 m
So (A) and (B) meet at time t = 2.13 s and position X = -32.7 m.
b. To determine the distance separating (A) and (B) at t = 8 s, we need to find their positions at that time. Using the time equation of (A), we get:
Xa = 30 - 158 = -90 m
Using the time equation of (B), we get:
Xb = 101 - 708 = -459 m
The distance separating (A) and (B) at t = 8 s is:
|Xb - Xa| = |-459 - (-90)| = 369 m.
Hence, Two particles moving on the same axis, where one is uniformly moving with an initial abscissa of 30 m and a speed of 15 m/s, and the other is moving with a time equation of XB = 101-70 (S.1), meet at time t = 2.13 s and position X = -32.7 m, while the distance separating them at t = 8 s is 369 m.
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1. To use the measured d, we assume the current flows along the central axes of Rod CD and Rod AB. Because of the repulsive forces, the conduction electrons in each rod however tend to move as far away from the other rod as possible. Considering this effect, should the actual μ0 value be higher or lower than the measured μ0 value? Why?
7*10^-7 = μ0*L/2*pi*d*g
L = 0.296
d = 0.011
g = 9.8
μ0 = 1.76*10^-6
2. If the length of Rod AB is doubled while the length of Rod CD remains the same, will the result change?
The actual μ₀ value should be higher than the measured μ₀ value and doubling the length of the rod will not affect the μ₀ value.
Detailed explanation of the answer is given below:
1. If we consider the effect of repulsive forces causing the conduction electrons in each rod to move as far away from the other rod as possible, the actual μ₀ value should be higher than the measured μ₀ value.
The reason for this is that the effective distance between the centers of the rods would be slightly larger due to the repulsion, causing the denominator in the equation to increase, and thus requiring a higher μ₀ value to maintain the equality.
2. If the length of Rod AB is doubled while the length of Rod CD remains the same, the result will not change.
In the given equation, 7*10^-7 = μ0*L/2*pi*d*g, the length of the rods does not directly affect μ₀. The equation only depends on the distance (d) between the rods and the gravitational constant (g), so doubling the length of Rod AB will not affect the μ0 value.
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