Answer:
0.0667 m
Explanation:
λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m
D = screen distance = 2.5 m
d = slit width = 15 x 10⁻⁶ m
n = order = 1
θ = angle = ?
Using the equation
d Sinθ = n λ
(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)
Sinθ = 26.67 x 10⁻³
y = position of first minimum
Using the equation for small angles
tanθ = Sinθ = y/D
26.67 x 10⁻³ = y/2.5
y = 0.0667 m
The first minimum in a single-slit diffraction pattern of light with a wavelength of 400 nm incident on a single slit of width 15 microns, 2.5 m from the screen, is approximately 66.67 mm from the central maximum.
The question asks for the distance of the first minimum from the central maximum in a single-slit diffraction pattern.
The distance to the first minimum from the central maximum in a single-slit diffraction pattern is calculated using the formula: [tex]y = \frac{\lambda X D}{d}[/tex]
where y - is the distance from the central maximum to the first minimum on the screen (meters), λ (lambda) - is the wavelength of light (meters), D - is the distance between the slit and the screen (meters), and d - the width of the slit (meters)
Given λ (lambda) = 400 nm = [tex]400 X 10^{-9} m[/tex], D = 2.5 m, d = 15 microns = [tex]15 X 10^{-6} m[/tex]
Calculation:
[tex]y = \frac{(400 X 10^{-9} m) X (2.5 m)}{(15 X 10^{-6} m)}\\y = 0.0667 \ meters (or 66.7 \ millimeters)[/tex]
The first minimum is 66.7 millimeters from the central maximum.
A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the objects have the same radius. They are all released at the same time and allowed to roll down the plane. Which object reaches the bottom first?
Answer:
Solid sphere will reach first
Explanation:
When an object is released from the top of inclined plane
Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane
initial gravitational potential energy = final total kinetic energy
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]
now we have
[tex]I = mk^2[/tex]
here k = radius of gyration of object
also for pure rolling we have
[tex]v = R\omega[/tex]
so now we will have
[tex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})[/tex]
[tex]mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})[/tex]
[tex]v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}[/tex]
so we will say that more the value of radius of gyration then less velocity of the object at the bottom
So it has less acceleration while moving on inclined plane for object which has more value of k
So it will take more time for the object to reach the bottom which will have more radius of gyration
Now we know that for hoop
[tex]mk^2 = mR^2[/tex]
k = R
For spherical shell
[tex]mk^2 = \frac{2}{3}mR^2[/tex]
[tex]k = \sqrt{\frac{2}{3}} R[/tex]
For solid sphere
[tex]mk^2 = \frac{2}{5}mR^2[/tex]
[tex]k = \sqrt{\frac{2}{5}} R[/tex]
So maximum value of radius of gyration is for hoop and minimum value is for solid sphere
so solid sphere will reach the bottom at first
Relationship between elastic shear stress and shear strain is a) Young's modulus b) Poissons ratio c) Saint-Venant's principle d) Modulus of rigidity e) None of these
Answer:
Option D is the correct answer.
Explanation:
Young's modulus is the ratio of tensile stress and tensile strain.
Bulk modulus is the ratio of pressure and volume strain.
Rigidity modulus is the ratio of shear stress and shear strain.
Here we are asked about relationship between elastic shear stress and shear strain. We have rigidity modulus is the ratio of shear stress and shear strain.
Option D is the correct answer.
The relationship between elastic shear stress and shear strain is characterized by the modulus of rigidity or shear modulus, which is a key property in understanding a material's deformation under shear forces.
The relationship between elastic shear stress and shear strain is represented by the modulus of rigidity, also known as the shear modulus. This modulus is a type of elastic modulus specific to shear stress and is the proportionality constant that relates shear stress to shear strain within the linear elastic region of a material's response to stress, as described by Hooke's Law.
There are different types of elastic modulus for different types of stress and strain; for tensile stress, the modulus is known as Young's modulus, for bulk stress, it is the bulk modulus, and for shear stress - which is the focus of this question - it is the shear or rigidity modulus. The modulus of rigidity is crucial for determining how a material will deform under shear forces and is a fundamental property used in engineering and construction to ensure materials behave as expected under load.
The position vector of a particle of mass 1.70 kg as a function of time is given by r with arrow = (6.00 î + 5.70 t ĵ), where r with arrow is in meters and t is in seconds. Determine the angular momentum of the particle about the origin as a function of time.
Final answer:
The angular momentum of a particle about the origin is given by the cross product of its position vector and linear momentum. In this case, the position vector of the particle is (6.00 î + 5.70 t ĵ) and the mass of the particle is 1.70 kg. To find the angular momentum, calculate the linear momentum and take the cross product with the position vector.
Explanation:
The angular momentum of a particle about the origin is given by the cross product of its position vector and linear momentum. In this case, the position vector of the particle is represented by r = (6.00 î + 5.70 t ĵ) and the mass of the particle is 1.70 kg. To find the angular momentum, we need to calculate the linear momentum first.
The linear momentum of a particle is given by the product of its mass and velocity. The velocity vector is given by the derivative of the position vector with respect to time, which in this case is v = (0 î + 5.70 ĵ) m/s. Substituting the values, we can find the linear momentum, which is p = (1.70 kg)(0 î + 5.70 ĵ) m/s.
To find the angular momentum, we take the cross product of the position vector and linear momentum:
ŕ × p = (6.00 î + 5.70 t ĵ) × (1.70 kg)(0 î + 5.70 ĵ) m/s = (0 î + 34.65 î t + 9.69 ĵ) kg·m²/s
Therefore, the angular momentum of the particle about the origin as a function of time is (0 î + 34.65 î t + 9.69 ĵ) kg·m²/s.
The angular momentum of the particle about the origin as a function of time is L(t) = 58.14 t k kg·m²/s. This result was obtained by calculating the cross product of the position and linear momentum vectors.
To determine the angular momentum of a particle about the origin as a function of time, we start by using the given position vector r(t) = (6.00 î + 5.70t ĵ) in meters.
The linear momentum p of the particle is given by p = m v, where m is the mass and v is the velocity. Since mass m = 1.70 kg, we first need the velocity. The velocity v(t) is the derivative of the position vector:
v(t) = d(r(t))/dt = 0 î + 5.70 ĵ = 5.70 ĵ m/s
Now, the linear momentum p(t) is:
p(t) = m v(t) = 1.70 kg * 5.70 ĵ m/s = 9.69 ĵ kg·m/s
The angular momentum L about the origin is given by L = r × p. Performing the cross product calculation:
r = 6.00 î + 5.70t ĵ
p = 9.69 ĵ
r × p = (6.00 î + 5.70t ĵ) × (9.69 ĵ)
Calculating the cross product, we get:
i-component: 0j-component: 0k-component: 6.00 * 9.69 - 0 = 58.14 t kThus, the angular momentum as a function of time is:
L(t) = 58.14 t k kg·m²/s
Three blocks of metal at the same temperature are placed on a hot stove. Their specific heat capacities are listed below. Rank them from greatest to least in how quickly each warms up. Steel, 450 J/(kg⋅∘C) Aluminum, 910 J/(kg⋅∘C) Copper, 390 J/(kg⋅∘C)
Answer:
Greatest amount of time to warm up: Aluminum
Then steel,
then copper
Explanation:
The higher the heat capacity, the longer it takes to warm the metal up.
Final answer:
To determine which metal heats up the fastest, you must consider their specific heat capacities. Copper, with the lowest specific heat capacity of 390 J/(kg°C), will heat up the fastest, followed by steel (450 J/(kg°C)) and aluminum (910 J/(kg°C)).
Explanation:
The concept of specific heat capacity is critical in understanding the rate at which different materials will heat up or cool down. Specific heat capacity, denoted by Cmetal, refers to the amount of heat needed to raise the temperature of one kilogram of a substance by one degree Celsius.
The metals in question are steel at 450 J/(kg°C), aluminum at 910 J/(kg°C) and copper at 390 J/(kg°C). The lower the specific heat capacity, the faster a material will reach a higher temperature when exposed to a constant heat source. Hence, according to their specific heat capacities, the metals rank in the following order in terms of heating up quickly: copper (390 J/(kg°C)), steel (450 J/(kg°C)), and aluminum (910 J/(kg°C)). This is because copper has the lowest specific heat capacity and therefore will require less energy to increase in temperature compared to steel and aluminum.
The average distance from Earth to the Moon is 384,000 km. In the late 1960s, astronauts reached the Moon in about 3 days. How fast (on average) must they have been traveling (in km/h) to cover this distance in this time? Compare this speed to the speed of a jet aircraft (800 km/h).
Answer:
They must have been traveling at 5333.33 km/h to cover that distance in 3 days.
That speed are 6,66 times higher than the speed of an aircraft jet.
Explanation:
d= 384000 km
t= 3 days = 3*24hr = 72hr
V= 384000km/72hr
V= 5333.33 km/h
comparison:
V1/V2= 5333.33/800
V1/V2= 6.66
When a 13-cm pipeline is filled with air at 30'C a. what is the maximum speed (in m/s) the air could be traveling if flow were laminar? b. what would be the maximum speed (in m/s)'to maintain laminar flow if the pipeline were filled with water at the same temperature instead of air?
Answer:[tex]V_{air}=0.259m/s[/tex]
[tex]V_{water}=0.01293m/s[/tex]
Explanation:
Given data
Length of pipe[tex]\left ( L\right )[/tex]=13cm=0.13m
From tables at [tex]T=30^{\circ}[/tex]
Kinematic viscosity of air[tex]\left ( \mu\right )=1.6036\times 10{-5} m^{2}/s[/tex]
and reynolds number is given by
Re=[tex]\frac{V\times \characteristic\ length}{Kinematic visocity}[/tex]
Flow is laminar up to Re.no. 2100
Re=[tex]\frac{V\times L}{1.6036\times 10{-5}}[/tex]
2100=[tex]\frac{V\times 0.13}{1.6036\times 10{-5}}[/tex]
V=0.259 m/s
For water
Kinematic viscosity of water[tex]\left ( \mu\right )=0.801\times 10{-6}m^{2}/s[/tex]
2100=[tex]\frac{V\times 0.13}{0.801\times 10{-6}}[/tex]
V=0.01293 m/s
A robot car drives off a cliff that is 11 meters above the water below. The car leaves the cliff horizontally with some initial speed and travels down to the water. The car hits the water a distance of 15 meters from the base of the cliff (it has no air resistance as it falls). What was the driving speed of the car? g =< 0,-9.8,0> N/kg.
Answer:
Driving speed of the car = 10 m/s
Explanation:
The car hits the water a distance of 15 meters from the base of the cliff.
Horizontal displacement = 11 m
A robot car drives off a cliff that is 11 meters above the water below.
Vertical displacement = 11 m
We have
s = ut + 0.5 at²
11 = 0 x t + 0.5 x 9.81 x t²
t = 1.50 s
So the car moves 15 meters in 1.50 seconds.
Velocity
[tex]v=\frac{15}{1.5}=10m/s[/tex]
Driving speed of the car = 10 m/s
A young man and woman are sitting on opposite sides of a park bench (1m). If the young man has a mass of 70kg and the woman has a mass of 60kg, what is the force between them?
Answer:
130N
Explanation:
F=(M1+M2)V
F= (70+60)*1
F=130*1
F=130N//
A 0.2 cm diameter wire must carry a 20-A current. If the maximum power dissipation along the wire is 4W/m, what is the minimum allowable conductivity of the wire in Ohm-m? (a) 3.18x10 (b) 3.18x10 () 3.18x10 (d) 3.18x10
Answer:
The conductivity of the wire is [tex]3.18\times10^{7}\ ohm-m[/tex].
Explanation:
Given that,
Diameter = 0.2 cm
Current = 20 A
Power = 4 W/m
We need to calculate the conductivity
We know that,
[tex]\sigma = \dfrac{1}{\rho}[/tex]
Using formula of resistance
[tex]R = \dfrac{\rho l}{A}[/tex]....(I)
Where,
[tex]\rho[/tex] = resistivity
A = area
l = length
Using formula of power
[tex]P = i^2 R[/tex]
[tex]R = \dfrac{P}{i^2}[/tex]
Put the value of R in equation (I)
[tex]\dfrac{P}{i^2}=\dfrac{\rho l}{\pi r^2}[/tex]
[tex]\rho=\dfrac{P\pi r^2}{l\timesi^2}[/tex]
[tex]\sigma=\dfrac{l\times i^2}{P\pi r^2}[/tex]
Put the all values into the formula
[tex]\sigma=\dfrac{1\times(20)^2}{4\times3.14\times(0.1\times10^{-2})^2}[/tex]
[tex]\sigma=3.18\times10^{7}\ ohm-m[/tex]
Hence, The conductivity of the wire is [tex]3.18\times10^{7}\ ohm-m[/tex].
To solve for the minimum allowable conductivity of the wire, we calculate the resistance based on the maximum power dissipation and then rearrange the formula for resistance to solve for conductivity. The minimum allowable conductivity for the wire is 3.18
* 10^4 (
Ω
m)^-1.
Minimum Allowable Conductivity of the Wire
To find the minimum allowable conductivity of a 0.2 cm diameter wire carrying a 20-A current with a maximum power dissipation of 4W/m, we first need to calculate the resistance of a 1-meter length of the wire based on the power dissipation.
Since power dissipation (
P) along the wire is given by P = I^2
* R, where I is the current and R is the instantaneous resistance of the wire, we can rearrange to find R = P / I^2. Substituting the given values, we get R = 4W / (20A)^2 = 0.01
Ω/m.
The resistance R of a conductor is also given by R =
L / (
σ
* A), where L is the length, σ (sigma) is the conductivity, and A is the cross-sectional area. We can rearrange and solve for the conductivity σ = L / (R
* A).
For a wire with a diameter of 0.2 cm, the cross-sectional area A is
π
* (0.1 cm)^2. Converting to meters, A = 3.14 * (0.001 m)^2 = 3.14
* 10^-6 m^2.
Now, substitute L = 1 m, R = 0.01 Ω/m, and A = 3.14
* 10^-6 m^2 into the formula to find the conductivity σ. The minimum allowable conductivity can then be calculated as σ = 1m / (0.01 Ω/m
* 3.14
* 10^-6 m^2) which yields σ = 3.18
* 10^4 (
Ω
*m)^-1.
A golf club strikes a 0.031-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 6240 N, and is in contact with the ball for a distance of 0.011 m. With what speed does the ball leave the club
Answer:
66.5 m/s
Explanation:
m = mass of the golf ball = 0.031 kg
F = magnitude of force applied to the ball = 6240 N
Acceleration experienced by the ball is given as
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{6240}{0.031}[/tex]
a = 201290.32 m/s²
d = distance for which the ball is in contact with the golf club = 0.011 m
v₀ = initial speed of the ball = 0 m/s
v = final speed of the ball = 0 m/s
Using the kinematics equation
v² = v₀² + 2 a d
v² = 0² + 2 (201290.32) (0.011)
v = 66.5 m/s
What is the angular momentum of a 3-kg uni- form cylindrical grinding wheel of radius 0.2 m when rotating at 1500 rpm (Rotational Inertia of a cylinder is mR^2/2).
Answer:
[tex]L = 9.42 kg m^2/s[/tex]
Explanation:
Angular speed of the cylinder is given as
[tex]f = 1500 rpm[/tex]
[tex]f = 1500 round/60 s[/tex]
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi(25) = 50 \pi[/tex]
now moment of inertia of the cylinder is given as
[tex]I = \frac{1}{2}mR^2[/tex]
[tex]I = \frac{1}{2}(3)(0.2)^2[/tex]
[tex]I = 0.06 kg m^2[/tex]
now we have
[tex]L = I\omega[/tex]
[tex]L = (0.06)(50\pi)[/tex]
[tex]L = 9.42 kg m^2/s[/tex]
A solenoid 28.0 cm long with a cross sectional area 0.475 cm^2 contains 645 turns and carries a current 85.0 A. a. Find the inductance of this solenoid. b. Find the total energy contained in the coil's magnetic field (assuming that the field is uniform throughout).
Explanation:
It is given that,
Length of solenoid, l = 28 cm = 0.28 m
Area of cross section, A = 0.475 cm² = 4.75 × 10⁻⁵ m²
Current, I = 85 A
(a) The inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_oN^2A}{l}[/tex]
[tex]L=\dfrac{4\pi\times 10^{-7}\times (645)^2\times 4.75\times 10^{-5}}{0.28}[/tex]
L = 0.0000886 H
[tex]L=8.86\times 10^{-5}\ H[/tex]
(b) Energy contained in the coil's magnetic field is given by :
[tex]U=\dfrac{1}{2}LI^2[/tex]
[tex]U=\dfrac{1}{2}\times 8.86\times 10^{-5}\ H\times (85\ A)^2[/tex]
U = 0.3200675 Joules
or
U = 0.321 Joules
Hence, this is the required solution.
A container is filled with water to a depth of 26.2 cm. On top of the water floats a 16.1 cm thick layer of oil with a density of 0.85 × 103 kg/m3. The acceleration of gravity is 9.81 m/s2 . a) What is the pressure at the surface of the water? Answer in units of Pa.
Answer:
1.34 x 10^3 Pa
Explanation:
density of oil = 0.85 x 10^3 kg/m^3
g = 9.81 m/s^2
height of oil column = 16.1 cm = 0.161 m
Pressure on the surface of water = height of oil column x density of oil x g
= 0.161 x 0.85 x 10^3 x 9.81 = 1.34 x 10^3 Pa
Thus, the pressure on the surface of water is 1.34 x 10^3 Pa.
A proton moves through a region of space where there is a magnetic field B⃗ =(0.64i+0.40j)T and an electric field E⃗ =(3.3i−4.5j)×103V/m. At a given instant, the proton's velocity is v⃗ =(6.6i+2.8j−4.8k)×103m/s.
Determine the components of the total force on the proton.
Express your answers using two significant figures. Enter your answers numerically separated by commas.
Answer:
[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]
Explanation:
When a charge is moving in constant magnetic field and electric field both then the net force on moving charge is vector sum of force due to magnetic field and electric field both
so first the force on the moving charge due to electric field is given by
[tex]\vec F_e = q\vec E[/tex]
[tex]\vec F_e = (1.6 \times 10^{-19})(3.3 \hat i - 4.5 \hat j) \times 10^3[/tex]
[tex]\vec F_e = (5.28 \times 10^{-16}) \hat i - (7.2 \times 10^{-16}) \hat j[/tex]
Now force on moving charge due to magnetic field is given as
[tex]\vec F_b = q(\vec v \times \vec B)[/tex]
[tex]\vec F_b = (1.6 \times 10^{-19})((6.6 \hat i+2.8 \hat j−4.8 \hat k) \times 10^3 \times (0.64 \hat i + 0.40 \hat j) )[/tex]
[tex]\vec F_b = (4.22 \times 10^{-16})\hat k - (2.87 \times 10^{-16})\hat k - (4.92 \times 10^{-16})\hat j + (3.07 \times 10^{-16}) \hat i[/tex]
[tex]\vec F_b = (3.07\times 10^{-16})\hat i - (4.92 \times 10^{-16})\hat j + (1.35 \times 10^{-16})\hat k[/tex]
Now net force due to both
[tex]F = F_e + F_b[/tex]
[tex]F = (8.35 \times 10^{-16})\hat i - (12.12 \times 10^{-16})\hat j +(1.35 \times 10^{-16})\hat k[/tex]
An electric and magnetic field exerts force on a proton moving with velocity in that field. The total force can be calculated from the Lorentz Force equation, which requires knowledge of the charge of the proton, its velocity, and the electric and magnetic fields it is experiencing.
Explanation:In Physics, the total force acting on a charged particle moving through an electric field E and a magnetic field B is given by the Lorentz Force equation: F = q(E + v × B), where q is the charge of the particle, and v is its velocity.
By given that the proton's charge is q = 1.6×10^-19 C and proton's velocity v = (6.6i+2.8j-4.8k)x10^3 m/s, electric field E = (3.3i-4.5j)x10^3 V/m, and magnetic field B = (0.64i+0.40j)T, we can plug these values into the equation.
To find the cross product of v and B, we use the determinant of a 3x3 matrix. The value for the F vector can be calculated as follows: F = q [E + v × B] and the cross product 'v × B' is calculated as the determinant of a 3x3 matrix. This yields force values that are expected to be in i, j, and k components. These calculations need to be done carefully to ensure accuracy, but are straightforward with the use of any standard physics formula sheet.
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What is the common trade name of the polymer polytetrafluoroethylene? For what is it commonly used?
Answer:
Common trade name of polytetrafluoroethylene is Teflon
Many uses are there some of them are given in explanation.
Explanation:
Common trade name of polytetrafluoroethylene is Teflon.
Main uses of polytetrafluoroethylene:
1) To coat non stick pans
2) Used as ski bindings
3) Used as fabric protector to repel stains on formal school-wear
4) Used to make to make a waterproof, breathable fabric in outdoor apparel.
All neutron stars are things that produce intense gravity. All neutron stars are extremely dense objects. Therefore, all extremely dense objects are things that produce intense gravity. Identify the major, minor, and middle terms.
Answer:
Major term is 'things that provide intense gravity'
Minor term is 'extremely dense objects'
Middle term is 'neutron stars'
Explanation:
Major term is given by the predicate part of the conclusionMinor term is given by the subject part of sentence in conclusionMiddle term is given by the subject part and not the conclusionA Canadian driving from Quebec to Montreal finds he has traveled 271 km. How many miles is this? There are 1.61 km in 1 mi
Answer:
168.32 mile
Explanation:
1 mile = 1.61 km
1.61 km = 1 mile
1 km = 1 / 1.61 mile
So, 271 km = 271 / 1.61 = 168.32 mile
Two 20 ohm resistors are connected in parallel and two 10 ohm resistors are connected in parallel. If these two combinations are connected in series the equivalent resistance of the combination is: a) 20 ohm b) 10 ohm c) 15 Ohm d) 30 Ohm
The equivalent resistance of the combination is 15 ohm. Option C is correct.
What is resistance?Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
Case 1: Two 20 ohm resistors are connected in parallel;
[tex]\rm \frac{1}{R_{eq}_1} =\frac{1}{R_1} +\frac{1}{R_2} \\\\ \rm \frac{1}{R_{eq}_1} =\frac{1}{20} +\frac{1}{20} \\\\ R_{eq}_1}= 10 \ ohm[/tex]
Case 2: Two 10-ohm resistors are connected in parallel.
[tex]\rm \frac{1}{R_{eq}_2} =\frac{1}{R_1} +\frac{1}{R_2} \\\\ \rm \frac{1}{R_{eq}_2} =\frac{1}{10} +\frac{1}{10} \\\\ R_{eq}_2}= 5 \ ohm[/tex]
Case 3; Two combinations are connected in series.
[tex]\rm R= R_{eq_1}+R_{eq_2}\\\\ \rm R= 10+ 5 \\\\ R=15 \ ohm[/tex]
The equivalent resistance of the combination is 15 ohm.
Hence, option C is correct.
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Final answer:
Two 20 ohm resistors in parallel have an equivalent resistance of 10 ohms, two 10 ohm resistors in parallel have an equivalent resistance of 5 ohms. When these two combinations are connected in series, the total equivalent resistance is the sum of both, which is c) 15 ohms.
Explanation:
To find the equivalent resistance of the given combinations of resistors, we first need to understand how resistors combine in parallel and in series.
When resistors are in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances:
Req = 1 / (1/R1 + 1/R2)
For two 20 ohm resistors in parallel:
Req = 1 / (1/20 + 1/20) = 1 / (2/20) = 10 ohms
For two 10 ohm resistors in parallel:
Req = 1 / (1/10 + 1/10) = 1 / (2/10) = 5 ohms
When we have resistors in series, their resistances simply add up:
Rtotal = Req1 + Req2
So, for the 10 ohm equivalent and the 5 ohm equivalent resistors in series:
Rtotal = 10 ohms + 5 ohms = 15 ohms
Therefore, the correct answer to the student's question is (c) 15 Ohm.
In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 10-5 m. (c = 3.00 × 108 m/s) (a) At what angle away from the central bright spot does the third bright fringe past the central bright spot occur? (b) At what angle does the second dark fringe occur?
Explanation:
It is given that,
Frequency of monochromatic light, [tex]f=5\times 10^{14}\ Hz[/tex]
Separation between slits, [tex]d=2.2\times 10^{-5}\ m[/tex]
(a) The condition for maxima is given by :
[tex]d\ sin\theta=n\lambda[/tex]
For third maxima,
[tex]\theta=sin^{-1}(\dfrac{n\lambda}{d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{n\lambda}{d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{nc}{fd})[/tex]
[tex]\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})[/tex]
[tex]\theta=4.69^{\circ}[/tex]
(b) For second dark fringe, n = 2
[tex]d\ sin\theta=(n+1/2)\lambda[/tex]
[tex]\theta=sin^{-1}(\dfrac{5\lambda}{2d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{5c}{2df})[/tex]
[tex]\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})[/tex]
[tex]\theta=3.90^{\circ}[/tex]
Hence, this is the required solution.
(a) The angle of the third bright fringe (θ₃) is approximately 4.69 degrees (b) The angle of the second dark fringe (θ₂) is approximately 3.90 degrees.
To solve this problem, we can use the formula for the angles of the maxima and minima in a Double-Slit Experiment diffraction pattern. For the third bright fringe, the condition for constructive interference is given by:
[tex]\[ d \sin(\theta_m) = m \lambda \][/tex]
where:
- d is the slit separation,
- [tex]\( \theta_m \)[/tex] is the angle of the mth bright fringe,
- m is the order of the fringe
- λ is the wavelength of the light.
For the second dark fringe, the condition for destructive interference is given by:
[tex]\[ d \sin(\theta_n) = \left( n + \frac{1}{2} \right) \lambda \][/tex]
where:
- [tex]\( \theta_n \)[/tex] is the angle of the nth dark fringe,
- n is the order of the dark fringe
Given that the frequency f is related to the wavelength λ by the speed of light (c) as c = fλ, we can express λ in terms of f.
Part (a): Third Bright Fringe
[tex]\[ d \sin(\theta_m) = m \lambda \][/tex]
[tex]\[ \sin(\theta_3) = \frac{3 \lambda}{d} \][/tex]
[tex]\[ \sin(\theta_3) = \frac{3 c}{d f} \][/tex]
[tex]\[ \theta_3 = \sin^{-1}\left(\frac{3 c}{d f}\right) \][/tex]
Part (b): Second Dark Fringe
[tex]\[ d \sin(\theta_n) = \left( n + \frac{1}{2} \right) \lambda \][/tex]
[tex]\[ \sin(\theta_2) = \frac{(2 + 0.5) \lambda}{d} \][/tex]
[tex]\[ \sin(\theta_2) = \frac{2.5 c}{d f} \][/tex]
[tex]\[ \theta_2 = \sin^{-1}\left(\frac{2.5 c}{d f}\right) \][/tex]
Now, plug in the values:
Given:
- [tex]\( d = 2.20 \times 10^{-5} \)[/tex] m,
- [tex]\( f = 5.00 \times 10^{14} \)[/tex] Hz,
- [tex]\( c = 3.00 \times 10^8 \)[/tex] m/s,
- m = 3,
- n = 2.
[tex]\[ \theta_3 = \sin^{-1}\left(\frac{3 \times 3.00 \times 10^8}{2.20 \times 10^{-5} \times 5.00 \times 10^{14}}\right) \][/tex]
[tex]\[ \theta_3 \approx 4.69^\circ \][/tex] (rounded to two decimal places, as given)
[tex]\[ \theta_2 = \sin^{-1}\left(\frac{2.5 \times 3.00 \times 10^8}{2.20 \times 10^{-5} \times 5.00 \times 10^{14}}\right) \][/tex]
[tex]\[ \theta_2 \approx 3.90^\circ \][/tex] (rounded to two decimal places, as given)
So, the correct answers are indeed [tex]\( \theta_3 \approx 4.69^\circ \)[/tex] for part (a) and [tex]\( \theta_2 \approx 3.90^\circ \)[/tex] for part (b).
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The work function for metallic caesium is 2.24 eV. Calculate the kinetic energy and speed of electrons ejected if a light source of a) 250 nm and b) 600 nm is used.
Answer:
a) [tex]4.362580048\times 10^{-19}\ Joule[/tex]
b)[tex]0.27566898\times 10^{-19}\ Joule[/tex]
Explanation:
a) When, wavelength=λ=250 nm
[tex]\text{Work function of metallic caesium}=2.24\ eV\\=2.24\times 1.6021\times 10^{-19}\\=3.588704\times 10^{-19}\ Joule \ \text{(converting to SI units)}\\\lambda =\text {Wavelength of light}=250\ nm\\Energy\\E=\frac{hc}{\lambda}\\\text{Where h=Plancks constant}=6.62607004\times 10^{-34} m^2kg / s\\\text{c=speed of light}=3\times 10^8\ m/s\\\Rightarrow E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{250\times 10^{-9}}\\\Rightarrow E=0.07951284048\times 10^{-17} Joule\\[/tex]
[tex]\text{Kinetic energy}=\text{E - Work function}\\K.E.=(7.951284048\times 10^{-19})-(3.588704\times 10^{-19})\\K.E.=4.362580048\times 10^{-19}\ Joule\\[/tex]
b) When, λ=600 nm
[tex]E=\frac{hc}{\lambda}\\E=\frac{6.62607004\times 10^{-34}\times 3\times 10^8}{600\times 10^{-9}}\\\Rightarrow E=3.313030502\times 10^{-19}\\\text{Kinetic energy}=\text{E - Work function}\\K.E.=(3.31303502\times 10^{-19})-(3.588704\times 10^{-19})\quad \text{Work function remains constant}\\K.E.=-0.27566898\times 10^{-19}\ Joule[/tex]
A commuter train blows its 200-Hz horn as it approaches a crossing. The speed of sound is 335 m/s. An observer waiting at the crossing receives a frequency of 207 Hz. What is the speed of the train?
Answer:
346m/s
Explanation:
v1=335m/s
f1=200hz
f2=207hz
v2=?
v1/f1=v2/f2
335/200=v2/207
v2=335*207/200
v2=346m/s
The motion of a particle is defined by the relation x = t3 – 12t2 +36t +32, where x and t are expressed in feet and seconds, respectively. Determine the time, position, and acceleration of the particle when v = 0 ft/s.
1) Time: 2 s, 6 s
The position of the particle is given by:
[tex]x=t^3 -12t^2 +36t+32[/tex]
where t is the time in seconds and x is the position in feet.
The velocity of the particle can be found by differentiating the position:
[tex]v(t)=x'(t)=3t^2 -24t+36[/tex]
and it is expressed in ft/s.
In order to find the time at which the velocity is v=0 ft/s, we substitute v=0 into the previous equation:
[tex]0=3t^2-24t+36\\0=t^2 -8t+12\\0=(t-2)(t-6)[/tex]
So the two solutions are
t = 2 s
t = 6 s
2) Position: x = 64 ft and x = 32 ft
The position at which the velocity of the particle is v = 0 can be found by susbtituting t = 2 and t = 6 into the equation for the position.
For t = 2 s, we have
[tex]x=(2)^3-12(2)^2 +36(2)+32=64[/tex]
For t = 6 s, we have
[tex]x=(6)^3-12(6)^2 +36(6)+32=32[/tex]
So the two positions are
x = 64 ft
x = 32 ft
3) Acceleration: [tex]-12 ft/s^2[/tex] and [tex]+12 ft/s^2[/tex]
The acceleration of the particle can be found by differentiating the velocity. We find:
[tex]a(t)=v'(t)=6t-24[/tex]
And substituting t = 2 and t = 6, we find the acceleration when the velocity of the particle is zero:
[tex]a(2)=6(2)-24=-12[/tex]
[tex]a(6)=6(6)-24=12[/tex]
So the two accelerations are
[tex]a=-12 ft/s^2[/tex]
[tex]a=12 ft/s^2[/tex]
The time, position, and acceleration of the particle when v = 0 ft/s are t = 2 s and t = 6 s, x = 24 ft and x = 184 ft, and a = -12 ft/s² and a = 12 ft/s².
Explanation:To find the time, position, and acceleration of the particle when the velocity is 0 ft/s, we need to determine the values of t, x, and a when v = 0.
Given the relation x = t³ - 12t² + 36t + 32, we need to solve for t when v = 0. We can use the equation v = dx/dt to find the velocity function and set it equal to 0.
By differentiating x with respect to t, we get v = 3t² - 24t + 36. Setting v = 0, we can solve the quadratic equation 3t² - 24t + 36 = 0 to find the values of t. The solutions are t = 2 and t = 6.
Therefore, when v = 0, the time is t = 2 s and t = 6 s. We can substitute these values into the position function to find the corresponding positions. When t = 2 s, x = (2)³ - 12(2)² + 36(2) + 32 = 24 ft. When t = 6 s, x = (6)³ - 12(6)² + 36(6) + 32 = 184 ft.
To find the acceleration, we can differentiate the velocity function with respect to t. By differentiating v = 3t² - 24t + 36, we get a = 6t - 24. Substituting t = 2 s and t = 6 s into this equation, we get a = 6(2) - 24 = -12 ft/s² and a = 6(6) - 24 = 12 ft/s².
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A 20.0 kg box slides down a 12.0 m long incline at an angle of 30.0 degrees with the horizontal. A force of 50.0 N is applied to the box to try to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is: 300 J.
372 J.
410 J.
455 J.
525 J.
Answer:
[tex]KE_f = 372 J[/tex]
Explanation:
The forces on the box while it is sliding down are
1). Component of the weight along the inclined plane
2). Friction force opposite to the motion of the box
3). Applied force on the box
now we know that component of the weight along the inclined plane is given as
[tex]F_g = mgsin\theta[/tex]
[tex]F_g = (20.0)(9.8)sin30 = 98 N[/tex]
Now we know that other component of the weight of object is counterbalanced by the normal force due to inclined plane
[tex]F_n = mgcos\theta[/tex]
[tex]F_n = (20.0)(9.8)cos30 = 170 N[/tex]
now the kinetic friction force on the box is given as
[tex]F_k = \mu F_n[/tex]
[tex]F_k = 0.100(170) = 17 N[/tex]
now the Net force on the box which is sliding down is given as
[tex]F_{net} = F_g - F_k - F_{applied}[/tex]
[tex]F_{net} = 98 - 17 - 50 = 31 N[/tex]
now the work done by net force = change in kinetic energy of the box
[tex]F.d = KE_f - KE_i[/tex]
[tex]31(12) = KE_f - 0[/tex]
[tex]KE_f = 372 J[/tex]
Final answer:
The increase in the kinetic energy of the box is calculated by finding the net work done on the box as it slides down the incline, which is found to be 373.2 J. The closest answer from the options provided is 372 J.
Explanation:
We need to calculate the increase in the kinetic energy of the box as it slides down the incline. First, let's determine the forces acting on the box:
Gravitational force component along the incline: Fg = m*g*sin(θ) = 20.0 kg * 9.81 m/s2 * sin(30°) = 98.1 N
Kinetic friction force: [tex]F_{k}[/tex] = μk * N = μk * m*g*cos(θ) = 0.100 * 20.0 kg * 9.81 m/s2 * cos(30°) = 17.0 N
Applied force up the incline: Fa = 50.0 N
Next, calculate the net force on the box:
Fnet = Fg - [tex]F_{k}[/tex] - Fa = 98.1 N - 17.0 N - 50.0 N = 31.1 N
Now, calculate the work done by the net force, which equals the increase in kinetic energy:
Work = Fnet * d = 31.1 N * 12.0 m = 373.2 J
The closest answer to this calculated value is 372 J.
A large bottle contains 150 L of water, and is open to the atmosphere. If the bottle has a flat bottom with an area of 2 ft, calculate the absolute pressure at the bottom of the bottle. Give your answer in Pa. Assume the water has a density of 1000 kg/m, and assume the system is at sea level on Earth. Answer: over 100,000 Pa
Answer:
Total pressure exerted at bottom = 119785.71 N/m^2
Explanation:
given data:
volume of water in bottle = 150 L = 0.35 m^3
Area of bottle = 2 ft^2
density of water = 1000 kg/m
Absolute pressure on bottom of bottle will be sum of atmospheric pressure and pressure due to water
Pressure due to water P = F/A
F, force exerted by water = mg
m, mass of water = density * volume
= 1000*0.350 = 350 kg
F = 350*9.8 = 3430 N
A = 2 ft^2 = 0.1858 m^2
so, pressure P = 3430/ 0.1858 = 18460.71 N/m^2
Atmospheric pressure
At sea level atmospheric pressure is 101325 Pa
Total pressure exerted at bottom = 18460.71 + 101325 = 119785.71 N/m^2
Total pressure exerted at bottom = 119785.71 N/m^2
An automobile starter motor draws a current of 1.7 A from a 17.2 V battery when operating at normal speed. A broken pulley locks the armature in position, and the current increases to 12.3 A. What was the back emf of the motor when operating normally? Answer in units of V.
Answer:
14.82 Volts
Explanation:
[tex]V_{back}[/tex] = Back emf of the motor
[tex]V_{battery}[/tex] = battery voltage = 17.2 Volts
[tex]i_{locked}[/tex] = Current in locked condition = 12.3 A
R = resistance
In locked condition, using ohm's law
[tex]R = \frac{V_{battery}}{i_{locked}}[/tex]
[tex]R = \frac{17.2}{12.3}[/tex]
R = 1.4 Ω
[tex]i_{normal}[/tex] = Current in normal condition = 1.7 A
Back emf of the motor is given as
[tex]V_{back}[/tex] = [tex]V_{battery}[/tex] - [tex]i_{normal} R[/tex]
[tex]V_{back}[/tex] = 17.2 - (1.7 x 1.4)
[tex]V_{back}[/tex] = 14.82 Volts
A far sighted person can not see clearly objects that are closer to his eyes than 60.0 cm. Which one of the following combinations represents the correct focal length and the refractive power of the contact lenses that will enable him to see the objects at a distance of 25.0 cm from his eyes?
A) -42.9 cm, +2.33 diopters
B) -42.9 cm, -2.33 diopters
C) +60 cm, +42.9 diopters
D) +42.9 cm, +2.33 diopters
E) +42.9 cm, -2.33 diopters
Answer:
[tex]f = +42.9 cm[/tex]
[tex]P =+2.33Dioptre[/tex]
Explanation:
As we know that Far sighted person has near point shifted to 60 cm distance
so he is able to see the object 60 cm
and the person want to see the objects at distance 25 cm
so here the image distance from lens is 60 cm and the object distance from lens is 25 cm
now from lens formula we have
[tex]\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}[/tex]
[tex]-\frac{1}{60} + \frac{1}{25} = \frac{1}{f}[/tex]
[tex]f = +42.9 cm[/tex]
Now we know that power of lens is given as
[tex]P = \frac{1}{f}[/tex]
[tex]P = \frac{1}{0.429} = +2.33Dioptre[/tex]
A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of the water? b) What is the buoyancy force on the rock? c) What is the mass of the water that the rock displaces? d) What is the weight of the rock under water?
Answer:
a) Weight of the rock out of the water = 16.37 N
b) Buoyancy force = 4.61 N
c) Mass of the water displaced = 0.47 kg
d) Weight of rock under water = 11.76 N
Explanation:
a) Mass of the rock out of the water = Volume x Density
Volume = 470 cm³
Density = 3.55 g/cm³
Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg
Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N
b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.
Volume = 470 cm³
Density of liquid = 1 g/cm³
[tex]\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N[/tex]
c) Mass of the water displaced = Volume of body x Density of liquid
Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg
d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force
Weight of rock under water = 16.37 - 4.61 =11.76 N
The deepest part of the Atlantic Ocean is called Milwaukee Deep with a depth of 8648 m. Determine the pressure in the ocean water at the base of Milwaukee Deep.
Answer:
8.475 x 10^7 Pa
Explanation:
h = 8648 m, g = 9.8 m/s^2, density of water, d = 1000 kg/m^3
Pressure at a depth is defined as the product of depth of water , acceleration due to gravity and density of water.
P = h x d x g
P = 8648 x 1000 x 9.8 = 8.475 x 10^7 Pa
Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship travels at 22 miles per hour and the second ship travels at 34 miles per hour, how far apart are the two ships after 1.2 hours?
Answer:
49.07 miles
Explanation:
Angle between two ships = 110° = θ
First ship speed = 22 mph
Second ship speed = 34 mph
Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b
Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c
Here the angle between the two sides of a triangle is 110° so from the law of cosines we get
a² = b²+c²-2bc cosθ
⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110
⇒a² = 2408.4
⇒a = 49.07 miles
A solid ball with mass M and radius R is rolling (without slipping) on a flat surface at 6 m/s. It then gets to a small smooth hill and rolls up 80 cm to the top and continues rolling on a flat surface. (a) Explain why we are allowed to apply the conservation of energy to the ball although rolling requires friction. (b) Use conservation of energy to find the velocity of the ball after it has passed the hill. (Moment of inertia of a solid ball around its center of mass is 2 5 MR 2
Answer:
Part B)
v = 4.98 m/s
Explanation:
Part a)
As the ball is rolling on the inclined the the friction force will be static friction and the contact point of the ball with the plane is at instantaneous rest
The point of contact is not slipping on the ground so we can say that the friction force work done would be zero.
So here in this case of pure rolling we can use the energy conservation
Part b)
By energy conservation principle we know that
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
so we will have
[tex]\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega'^2 + mgh[/tex]
here in pure rolling we know that
[tex]v = R\omega[/tex]
now from above equation we have
[tex]\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v}{R})^2 = \frac{1}{2}mv'^2 + \frac{1}{2}(\frac{2}{5}mR^2)(\frac{v'}{R})^2 + mgh[/tex]
now we have
[tex]\frac{1}{2}mv^2(1 + \frac{2}{5}) = \frac{1}{2}mv'^2(1 + \frac{2}{5}) + mgh[/tex]
now plug in all values in it
[tex]\frac{1}{2}m(6^2)(\frac{7}{5}) = \frac{1}{2}mv'^2(\frac{7}{5}) + m(9.81)(0.80)[/tex]
[tex]25.2 = 0.7v^2 + 7.848[/tex]
[tex]v = 4.98 m/s[/tex]
Final answer:
The conservation of energy can be applied to the rolling ball because rolling motion involves both kinetic energy and potential energy. By equating the initial total energy of the ball to the final kinetic energy of the ball, we can find its velocity after it has passed the hill.
Explanation:
(a) The conservation of energy can be applied to the rolling ball because rolling motion involves both kinetic energy and potential energy. Although rolling requires friction, the work done by friction is accounted for in the system's total energy. Therefore, the law of conservation of energy can still be applied.
(b) To find the velocity of the ball after it has passed the hill, we can apply the conservation of energy. At the top of the hill, the ball has both potential energy and kinetic energy. At the bottom of the hill, the ball only has kinetic energy. By equating the initial total energy of the ball to the final kinetic energy of the ball, we can solve for its velocity.