To solve this problem we will apply the concepts related to angular resolution based on wavelength and the slit width at this case of the cat's pupils. This relationship is given as,
[tex]\theta = \frac{\lambda}{a}[/tex]
Here,
[tex]\lambda[/tex] = Wavelength
a = Slit width
[tex]\theta = \frac{519*10^{-9}m}{0.55*10^{-3}m}[/tex]
[tex]\theta = 9.43*10^{-4} Rad[/tex]
Therefore the angular resolution is [tex]9.43*10^{-4}rad[/tex]
Final answer:
The angular resolution for a cat observing birds can be estimated using the formula θ = 1.22 λ / D, with a pupil width of 0.550 mm and a wavelength of 519 nm, yields an angular resolution of approximately 1.152 × 10⁻⁶ radians.
Explanation:
The question refers to angular resolution in optics, specifically related to the diffraction limit of a cat's eyes observing birds. In physics, the angular resolution for a circular aperture, like the pupil of a cat's eye, can be estimated using the formula θ = 1.22 λ / D, where θ is the angular resolution in radians, λ is the wavelength of the light, and D is the diameter of the aperture (the pupil in this case). Given that the cat's pupil has a slit width of a = 0.550 mm (which we will use in place of the diameter for this rough estimate) and the light has an average wavelength of λ = 519 nm, the angular resolution θ can be calculated as follows:
θ = 1.22 × 519 x 10⁻⁹ m / 0.550 x 10^-3 m
θ = 1.22 × 519 / 550 × 10⁻⁶
θ = 1.22 × 0.944 × 10⁻⁶
θ = 1.152 × 10⁻⁶ radians
Given the specified conditions, this calculation estimates the angular resolution for the cat's eyes. Note, however, that this is a simplification and doesn't take into account the complexities of a vertical slit pupil versus a circular aperture.
106 grams of liquid water are in a cylinder with a piston maintaining 1 atm (101325 Pa) of pressure. It is exactly at the boiling point of water, 373.15 K. We then add heat to boil the water, converting it all to vapor. The molecular weight of water is 18 g/mol and the latent heat of vaporization is 2260 J/g.
The missing part of the question is;
1) How much heat is required to boil the water?
2) Assume that the liquid water takes up approximately zero volume, and the water vapor takes up some final volume Vf. You may also assume that the vapor is an ideal gas. How much work did the vapor do pushing on the piston?
3) How much did the water internal energy change?
Answer:
A) Q = 239.55 KJ
B) W = 18.238 KJ
C) ΔU = 221.31 J
Explanation:
We are giving;
Mass; m = 106 g
Latent heat of vaporization; L = 2260 J/g.
Molecular weight of water; M = 18 g/mol
Pressure; P = 101325 Pa
Temperature; T = 373.15 K
A) Formula for amount of heat required is;
Q = mL
Q = 106 x 2260
Q = 239560J = 239.55 KJ
B) number of moles; n = m/M
n = 106/18
n = 5.889 moles
Now, we know from ideal gas equation that;
PV = nRT
Thus making V the subject, we have; V = nRT/P
However, here we are told V is V_f. Thus, V_f = nRT/P
R is gas constant = 8.314 J/mol·K
Plugging in the relevant values ;
V_f = (5.889 x 8.314 x 373.15)/101325
V_f = 0.18 m³
Now, at constant pressure, work done is;
W = P(ΔV)
W = P(V_f - V_i)
W = 101325(0.18 - 0)
W = 101325 x 0.18
W = 18238.5J = 18.238 KJ
C) Change in water internal energy is gotten from;
ΔU = Q - W
Thus, ΔU = 239.55 KJ - 18.238 KJ
ΔU = 221.31 J
A cleaver physics professor wants to create a situation where a block starts from rest at the top of a 31.0° inclined plane and encounters a spring at the bottom of the incline. The spring has a constant 3.4 kN/m and the block's mass is 33.0 kg. How far does the block travel before hitting the spring, if the spring was compressed 37 cm in it's initial collision?
Answer:
Explanation:
Let the length of inclined plane be L .
work done by gravity on the block
= force x length of path
= mg sinθ x L , m is mass of the block , θ is inclination of path
This in converted into potential energy of compressed spring
1/2 k x² = mgL sin31 , k is force constant . x is compression
.5 x 3400 x .37² = 33 x9.8 x sin31 L
L = 1.4
Length of incline = 1.4 m .
Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a spring whose spring constant is 61.6 N/m. An observer is traveling at a speed of 2.79 × 108 m/s relative to the fixed end of the spring. What does this observer measure for the period of oscillation?
Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer, [tex]v=2.79\times 10^8\ m/s[/tex]
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
[tex]t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s[/tex]
Time period of oscillation measured by the observer is :
[tex]t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s[/tex]
So, the time period of oscillation measured by the observer is 5.79 seconds.
What direction does a S wave move
Answer:
As they travel through rock, the waves move tiny rock particles back and forth -- pushing them apart and then back together -- in line with the direction the wave is traveling. These waves typically arrive at the surface as an abrupt thud. Secondary waves (also called shear waves, or S waves) are another type of body wave
Explanation:
Please mark me as brainliest!! Hope I helped
The maximum current output of a 60 Ω circuit is 11 A. What is the rms voltage of the circuit?
Answer:
660V
Explanation:
V=IR
V=?,I=11A,R=60w
V=60×11
=660V
Current and Resistivity Concepts1.Electric charge is conserved. As a consequence, when current arrives at a junction of wires, the charges can take either of two paths out of the junction and the numerical sum of the currents in the two paths equals the current that entered the junction. Thus, current is which of the following?a)a scalarb)a vector c)neither a vector nor a scalar2.A cylindrical wire has a radius r and length l. If both r and l are doubled, the resistance of the wire does what?a)decreasesb)increases c)remains the same
Answer:
1. A). Scalar
2.C) remains de same
Explanation:
Particles (mass of each = 0.40 kg) are placed at the 60-cm and 100-cm marks of a meter stick of negligible mass. This rigid body is free to rotate about a frictionless pivot at the 0-cm end. The body is released from rest in the horizontal position. What is the magnitude of the initial linear acceleration of the end of the body opposite the pivot?
Answer:
12 m/s ∧2
Explanation:
The picture attached explains it all and i hope it helps. Thank you
To find the initial linear acceleration, calculate the initial torque due to gravity, the moment of inertia, and then apply Newton’s second law for rotation to obtain the initial angular acceleration. The initial linear acceleration can then be found by multiplying the initial angular acceleration by the length of the meter stick.
Explanation:The question is asking for the initial linear acceleration of the end of the rigid body (meter stick) at the moment it is released.
The initial torque τ is equal to the gravitational forces acting on each particle times their respective distances from the pivot, summed up.
τ_initial = m*g*d1 + m*g*d2 = 0.4 kg * 9.8 m/s^2 * 0.6 m + 0.4 kg * 9.8 m/s^2 * 1.0 m
The moment of inertia I for the two-particle system can be calculated with the formula I = ∑mr^2 for each particle:
I = m*d1^2 + m*d2^2 = 0.4 kg * (0.6 m)^2 + 0.4 kg * (1.0 m)^2
According to Newton’s second law for rotation, the initial angular acceleration α is equal to the initial torque divided by the moment of inertia:
α = τ_initial / I
The initial linear acceleration a of the end of the body at the point opposite the pivot is equal to the product of the initial angular acceleration and the total length of the meter stick (1.0 m):
a = α * 1.0 m
Learn more about Rigid Body Dynamics here:https://brainly.com/question/12903545
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A space transportation vehicle releases a 470-kg communications satellite while in a circular orbit 350 km above the surface of the Earth. A rocket engine on the satellite boosts it into an orbit 2350 km above the surface of the Earth. How much energy does the engine have to provide for this boost?
Answer:
E = 3.194 x 10⁹ J = 3.194 GJ
Explanation:
The formula for the absolute potential energy is:
U = - GMm/2r
where,
G = Gravitational Constant = 6.67 x 10⁺¹¹ N m²/kg²
M = mass of Earth = 5.972 x 10²⁴ kg
m = mass of satellite = 470 kg
r = distance between the center of Earth and satellite
Thus, the energy required from engine will be difference between the potential energies.
E = U₂ - U₁
E = - GMm/2r₂ - (- GMm/2r₁)
E = (GMm/2)(1/r₁ - 1/r₂)
where,
r₁ = Radius of Earth + 350 km = 6371 km + 350 km = 6721 km = 6.721 x 10⁶ m
r₂=Radius of Earth + 2350 km=6371 km + 2350 km= 8721 km = 8.721 x 10⁶ m
therefore,
E = [(6.67 x 10⁺¹¹ N m²/kg²)(5.972 x 10²⁴ kg)(470 kg)/2](1/6.721 x 10⁶ m - 1/8.721 x 10⁶ m)
E = 3.194 x 10⁹ J = 3.194 GJ
) So we are in a Universe with no center and no edge, but it is expanding... and it might be infinite. And it is all space and time and mass and energy, but was born out of nothingness. What do you think about this? Does it seem sensible and natural or do you find it odd and confusing? Can you think of any better explanation for what we see? I would like your answer to be at least five (5) sentences long.
Final answer:
The multiverse concept along with the universe's infinite nature and lack of a center challenges our understanding, yet represents current scientific explanations supported by observational evidence and theoretical models.
Explanation:
The concept of the multiverse, suggesting that our universe is just one among countless others, is a challenging yet intriguing idea. The nature of cosmic expansion, driven by mass and dark energy, adds complexity to our universe's fate, with models predicting expansion forever or eventual contraction. While it may seem sensible to accept that the Big Bang led to a universe with no center or edge, and infinite potential, it's natural to find this overwhelming due to its departure from human intuition. These concepts push the boundaries of our understanding and stimulate philosophical and metaphysical discussions, demonstrating the dynamic involvement of spacetime. As opinions on what constitutes the 'center' or 'boundary' of an infinite universe vary, the scientific community continues to explore why the universe is structured as it is, why certain constants have their values, and what 'accidents' may have been necessary for existence as we perceive it.
The neurons of giant squids, for example, consist of axons with very large radii, which allows the squid to react very quickly when confronted with a predator. Assuming no change in the resistivities or membrane thickness of the axon, by what factor must the radius of the axon increase such that the speed of the pulse increases by a factor of 10
Answer:
100
Explanation:
[tex]\rho_m[/tex] = Resistivity of axon
r = Radius of axon
t = Thickness of the membrane
[tex]\rho_a[/tex] = Resistivity of the axoplasm
Speed of pulse is given by
[tex]v=\sqrt{\dfrac{\rho_mrt}{2\rho_a}}[/tex]
So, radius is given by
[tex]r=\dfrac{2\rho_a}{\rho_mt}v^2[/tex]
If radius is increased by a factor of 10 new radius will be
[tex]r_2=\dfrac{2\rho_a}{\rho_mt}(10v)^2\\\Rightarrow r_2=100\dfrac{2\rho_a}{\rho_mt}v^2\\\Rightarrow r_2=100r[/tex]
So, The radius will increase by a factor of 100.
A plane monochromatic radio wave (? = 0.3 m) travels in vacuum along the positive x-axis, with a time-averaged intensity I = 45.0 W/m2. Suppose at time t = 0, the electric field at the origin is measured to be directed along the positive y-axis with a magnitude equal to its maximum value. What is Bz, the magnetic field at the origin, at time t = 1.5 ns? Bz = I got .04800 but that answer didnt work.
Answer:
The magnetic field [tex]B_Z[/tex] [tex]= - 6.14*10^{-7} T[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 0.3m[/tex]
The intensity is [tex]I = 45.0W/m^2[/tex]
The time is [tex]t = 1.5ns = 1.5 *10^{-9}s[/tex]
Generally radiation intensity is mathematically represented as
[tex]I = \frac{1}{2} c \epsilon_o E_o^2[/tex]
Where c is the speed of light with a constant value of [tex]3.0 *10^8 m/s[/tex]
[tex]E_i[/tex] is the electric field
[tex]\epsilon_o[/tex] is the permittivity of free space with a constant value of [tex]8.85*10^{-12} C^2 /N \cdot m^2[/tex]
Making [tex]E_o[/tex] the subject of the formula we have
[tex]E_i = \sqrt{\frac{2I}{c \epsilon_0} }[/tex]
Substituting values
[tex]E_i = \sqrt{\frac{2* 45 }{(3*10^8 * (8.85*10^{-12}) )} }[/tex]
[tex]= 184.12 \ V/m[/tex]
Generally electric and magnetic field are related by the mathematical equation as follows
[tex]\frac{E_i}{B_i} = c[/tex]
Where [tex]B_O[/tex] is the magnetic field
making [tex]B_O[/tex] the subject
[tex]B_i = \frac{E_i}{c}[/tex]
Substituting values
[tex]B_i = \frac{184.12}{3*10^8}[/tex]
[tex]= 6.14 *10^{-7}T[/tex]
Next is to obtain the wave number
Generally the wave number is mathematically represented as
[tex]n = \frac{2 \pi }{\lambda }[/tex]
Substituting values
[tex]n = \frac{2 \pi}{0.3}[/tex]
[tex]= 20.93 \ rad/m[/tex]
Next is to obtain the frequency
Generally the frequency f is mathematically represented as
[tex]f = \frac{c}{\lambda}[/tex]
Substituting values
[tex]f = \frac{3 *10^8}{0.3}[/tex]
[tex]= 1*10^{9} s^{-1}[/tex]
Next is to obtain the angular velocity
Generally the angular velocity [tex]w[/tex] is mathematically represented as
[tex]w = 2 \pi f[/tex]
[tex]w = 2 \pi (1* 10^9)[/tex]
[tex]= 2 \pi * 10^9 rad/s[/tex]
Generally the sinusoidal electromagnetic waves for the magnetic field B moving in the positive z direction is expressed as
[tex]B_z = B_i cos (nx -wt)[/tex]
Since the magnetic field is induced at the origin then the equation above is reduced to
[tex]B_z = B_i cos (n(0) -wt) = B_i cos ( -wt)[/tex]
x =0 because it is the origin we are considering
Substituting values
[tex]B_z = (6.14*10^{-7}) cos (- (2 \pi * 10^{9})(1.5 *10^{-9}))[/tex]
[tex]= - 6.14*10^{-7} T[/tex]
o study torque experimentally, you apply a force to a beam. One end of the beam is attached to a pivot that allows the beam to rotate freely. The pivot is taken as the origin or your coordinate system. You apply a force of F = Fx i + Fy j + Fz k at a point r = rx i + ry j + rz k on the beam. show answer No Attempt 33% Part (a) Enter a vector expression for the resulting torque, in terms of the unit vectors i, j, k and the components of F and r. τ = | γ θ i j k d Fx Fy Fz g m n rx ry rz ( ) 7 8 9 HOME ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback. No Attempt No Attempt 33% Part (b) Calculate the magnitude of the torque, in newton meters, when the components of the position and force vectors have the values rx = 0.76 m, ry = 0.035 m, rz = 0.015 m, Fx = 3.6 N, Fy = -2.8 N, Fz = 4.4 N. No Attempt No Attempt 33% Part (c) If the moment of inertia of the beam with respect to the pivot is I = 442 kg˙m2, calculate the magnitude of the angular acceleration of the beam about the pivot, in radians per second squared. All content © 2020 Expert TA, LLC
Answer:
(a) Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
(b) Magnitude of resulting torque = = 3.99 Nm
(c) angular acceleration = = 0.009027 rad/s²
Explanation:
Given Data;
I = 442 kg˙m2
rx = 0.76 m,
ry = 0.035 m,
rz = 0.015 m,
Fx = 3.6 N,
Fy = -2.8 N,
Fz = 4.4 N
F = Fx i + Fy j + Fz ------------------------------1
r = rx i + ry j + rz k ------------------------------2
(a) Torgue is given by the formula;
T = r * F ------------------------------------3
Putting equation 1 and 2 into equation 3, we have;
Torque= r x F
= (rx i +ry j +rz k) x (Fx i + Fy j +Fz k )
= (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
Therefore,
Resulting torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
b)
Putting given values into the above expression, we have
torque = (ryFz - rzFy) i +(rzFx - rxFz) j+(rxFy - ryFx) k
=(0.035*4.4 - (0.015*-2.8))i +(0.015*3.6 - 0.76*4.4)j+(0.76* -2.8 - 0.035*3.6)k
= (0.154 +0.041) i + (0.054 - 3.344) j + (-2.128 -0.126) k
= (0.196) i - (3.29) j + (-2.254) k
Magnitude of resulting torque = √(0.196² + 3.29² +2.254²
=√15.943031
= 3.99 Nm
c) Angular acceleration is given by the formula;
angular acceleration = torque/moment of inertia
= 3.99/ 442
= 0.009027 rad/s²
A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60-T magnetic field. What is the radius of the resulting path
Answer:
1.2cm
Explanation:
V=(2ev/m)^1/2
=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2
=6.2x10^5m/s
Radius of resulting path= MV/qB
= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6
=0.012m
=1.2cm
Answer:
0.012m
Explanation:
To find the radius of the path you can use the formula for the radius od the trajectory of a charge that moves in a constant magnetic field:
[tex]r=\frac{mv}{qB}[/tex]
m: mass of the proton 9.1*10^{-31}kg
q: charge of the proton 1.6*10^{-19}C
B: magnitude of the magnetic field 0.60T
v: velocity of the proton
In order to use the formula you need to calculate the velocity of the proton. This can be made by using the potential difference and charge, that equals the kinetic energy of the proton:
[tex]qV=E_k=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.5*10^{3}V)}{1.67*10^{-27}kg}}=6.92*10^{5}\frac{m}{s}[/tex]
Then, by replacing in the formula for the radius you obtain:
[tex]r=\frac{(1.67*10^{-27}kg)(6.92*10^5\frac{m}{s})}{(1.6*10^{-19}C)(0.60T)}=0.012m[/tex]
hence, the radius is 0.012m
Young's experiment is performed with light of wavelength 502 nmnm from excited helium atoms. Fringes are measured carefully on a screen 1.40 mm away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.4 mmmm from the center of the central bright fringe. Part A What is the separation of the two slits? dd = nothing mmmm SubmitRequest Answer Provide Feedback
Answer:
Explanation:
wavelength of light λ = 502 x 10⁻⁹ m /s
screen distance D = 1.4 m
Slit separation d = ?
position of n the separation is given by the formula
x = n Dλ / d , n is order of fringe , x = distance of n th fringe
10.4 x 10⁻³ = 20 x 1.4 x 502 x 10⁻⁹ / d
d = 20 x 1.4 x 502 x 10⁻⁹ / 10.4 x 10⁻³
= 1351.54 x 10⁻⁶
= 1.35 x 10⁻³ m
1.35 mm.
1.) A metal sphere is held in a fluid flowing with a temperature of 30°C and velocity 2.5 m/s. The sphere has a radius of 10 mm and a constant surface temperature of 60°C. Find the drag force on the sphere and the rate of heat transfer from the sphere for: a.)Water b.)Air at 1 bar pressure Discuss what factors influence the answers.
Answer:
Explanation:
The pictures attached herewith shows the explanation and i hope it all helps. Thank you
Preparing an Accounts Payable Schedule Pilsner Inc. purchases raw materials on account for use in production. The direct materials ected purchases April May June Pilsner typically pays 25% on account in the month of billing and 75% the next month Required: 1. How much cash is required for payments on account in May? $374,400 411,200 416,000 2. How much cash is expected for payments on account in June?
Answer:
How much cash is expected for payments on account in June? = $412,400.
Explanation:
Cash Required for expected payments on account in June
Pilsner Inc. pays 25% on account in the month of billing and the remaining 75% in the next month, therefore, for the month of June the total payment will include 75% of Purchases made in May and 25% of purchases made in June.
Purchases in May = $411,200
Purchases in June = $416,000
Total Payment for May = (75% of 411,200)+(25% of 416,000)
= $308,400 + $104,000
= $412,400.
A small mirror is attached to a vertical wall, and it hangs a distance of 1.70 m above the floor. The mirror is facing due east, and a ray of sunlight strikes the mirror early in the morning and then again later in the morning. The incident and reflected rays both lie in a plane that is perpendicular to both the wall and the floor. Early in the morning, the ray is observed to strike the floor at a distance of 3.75 m from the base of the wall. Later on in the morning, the ray strikes the floor at a distance of 1.18 m from the wall. The earth rotates at 15.0 degrees per hour. How much time (in hours) has elapsed between the two observations
Answer:
Around 2 hours
Explanation:
The answer is given in the pictures below. All the page numbers are circled on the top left corner of each page.
Final answer:
The question deals with calculating time based on the change in position of a reflected sunlight ray due to Earth's rotation, with the Earth rotating at a rate of 15 degrees per hour.
Explanation:
The question concerns the reflection of light from a mirror and the calculation of time based on the Earth's rotation rate. The sunlight's path changes throughout the morning due to the Earth's rotation, which moves at a consistent rate of 15 degrees per hour. Given the different positions where the ray of sunlight strikes the floor at different times, we can calculate the angle change and then the time elapsed.
To find the time elapsed, we first determine the angular change between the two observations. The change in position on the floor from 3.75 m to 1.18 m corresponds to a difference in angle when considering the mirror as the vertex of a right triangle with the wall. Next, we apply the Earth's rotational rate to find the corresponding time.
In an experiment of a simple pendulum, measurements show that the pendulum has length し 0.397 ± 0.006 m, mass M-0.3172 ± 0.0002 kg, and period Tem-1.274 ± 0.005 s. Take g = 9.8 m/s of error (use the error propagation method you learned in Lab 1). and the predicted value Ttheo-
i. Use the measured length L to predict the theoretical pendulum period Ttheo with a range
ii. Compute the percentage difference (as defined in Lab 1) between the measured value Texp
Answer:
1.)1.265+or minus 0.0006m
2).0.71%
Explanation:
See attached file
One possible explanation for a galaxy's type invokes the angular momentum of the protogalactic cloud from which it formed. Suppose a galaxy forms from a protogalactic cloud with a lot of angular momentum. Assuming its type has not changed as a result of other interactions, we'd expect this galaxy to be ______.
Answer:
A spiral galaxy
Explanation:
Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecules is reduced because each other molecule occupies volume v. Instead of pV=NkT, we get: p(V-Nb)=NkT. Let b=1.2 × 10-28 m3. Let's look at 3moles of this gas at T=300K starting in 0.001 m3 volume. 1) What's the initial value of the pressure?
Answer:
P = 7482600 Pa = 7.482 MPa
Explanation:
We have the equation:
P(V - Nb) = NKT
here,
P = Initial Pressure = ?
V = Initial Volume = 0.001 m³
N = No. of moles = 3
b = constant = 1.2 x 10⁻²⁸ m³
T = Temperature = 300 k
k = Gas Constant = 8.314 J/mol . k
Therefore,
P[0.001 m³ - (3)(1.2 x 10⁻²⁸ m³)] = (3 mol)(8.314 J/mol. k)(300 k)
P = (7482.6 J)/(0.001 m³)
P = 7482600 Pa = 7.482 MPa
A mass m = 1.1 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k = 115 N/m and negligible mass. At time t = 0 the mass is released from rest at a distance d = 0.35 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by y(t) = A cos(ωt – φ). The positive y-axis points upward. show answer Correct Answer 17% Part (a) Find the angular frequency of oscillation, in radians per second. ω = 10.22 ✔ Correct! show answer Incorrect Answer 17% Part (b) Determine the value of the coefficient A, in meters.
Answer:
a) = 10.22 rad/s
b) = 0.35 m
Explanation:
Given
Mass of the particle, m = 1.1 kg
Force constant of the spring, k = 115 N/m
Distance at which the mass is released, d = 0.35 m
According to the differential equation of s Simple Harmonic Motion,
ω² = k / m, where
ω = angular frequency in rad/s
k = force constant in N/m
m = mass in kg
So,
ω² = 115 / 1.1
ω² = 104.55
ω = √104.55
ω = 10.22 rad/s
If y(0) = -0.35 m and we want our A to be positive, then suffice to say,
The value of coefficient A in meters is 0.35 m
In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move a bit farther across the ice and other times players want the stones to stop a bit sooner. Suggest a way to increase the kinetic friction between the stone and the ice so that the stone stops more quickly. Next, suggest a way to decrease the kinetic friction between the stone and the ice so that the stone slides farther along the ice before coming to a stop
Answer:
To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.
To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.
Explanation:
In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.
On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.
The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.
Final answer:
To increase kinetic friction in curling, roughen the ice or stone's bottom; to decrease it, smooth the ice or use lubricants, often achieved by sweeping.
Explanation:
To increase the kinetic friction between the curling stone and the ice, thereby causing the stone to stop more quickly, you could roughen the surface of the ice or the bottom of the stone. This creates more irregularities which bump against each other, leading to increased vibrations and energy conversion to heat and sound, thus slowing down the stone. Additionally, players could sweep less or not at all in front of the stone so that the ice surface remains rougher and produces more resistance.
To decrease the kinetic friction and allow the stone to slide farther across the ice before coming to a stop, you could smooth the surface of the ice or apply a substance to the bottom of the stone to make it more slippery. Sweeping the ice in front of the stone is a common technique used to achieve this; it melts and smooths the ice surface, reducing the irregularities that cause friction and allowing the stone to glide more easily.
Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm^ 2 after it passes through the stack.
If the incident intensity is kept constant:
1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?
Answer:
1
When second polarizer is removed the intensity after it passes through the stack is
[tex]I_f_3 = 27.57 W/cm^2[/tex]
2 When third polarizer is removed the intensity after it passes through the stack is
[tex]I_f_2 = 102.24 W/cm^2[/tex]
Explanation:
From the question we are told that
The angle of the second polarizing to the first is [tex]\theta_2 = 21^o[/tex]
The angle of the third polarizing to the first is [tex]\theta_3 = 61^o[/tex]
The unpolarized light after it pass through the polarizing stack [tex]I_u = 60 W/cm^2[/tex]
Let the initial intensity of the beam of light before polarization be [tex]I_p[/tex]
Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as
[tex]I_1 = \frac{I_p}{2}[/tex]
Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as
[tex]I_2 = I_1 cos^2 \theta_1[/tex]
[tex]= \frac{I_p}{2} cos ^2 \theta_1[/tex]
The intensity of light that will emerge from the third filter is mathematically represented as
[tex]I_3 = I_2 cos^2(\theta_2 - \theta_1 )[/tex]
[tex]I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)][/tex]
making [tex]I_p[/tex] the subject of the formula
[tex]I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}[/tex]
Note that [tex]I_u = I_3[/tex] as [tex]I_3[/tex] is the last emerging intensity of light after it has pass through the polarizing stack
Substituting values
[tex]I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}[/tex]
[tex]I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}[/tex]
[tex]=234.622W/cm^2[/tex]
When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as
[tex]I_f_3 = \frac{I_p}{2} cos ^2 \theta_2[/tex]
[tex]I_f_3[/tex] is the intensity of the light emerging from the stack
substituting values
[tex]I_f_3 = \frac{234.622}{2} * cos^2(61)[/tex]
[tex]I_f_3 = 27.57 W/cm^2[/tex]
When the third polarizer is removed the second polarizer becomes the
the final polarizer and the intensity of light emerging from the stack would be
[tex]I_f_2 = \frac{I_p}{2} cos ^2 \theta_1[/tex]
[tex]I_f_2[/tex] is the intensity of the light emerging from the stack
Substituting values
[tex]I_f_2 = \frac{234.622}{2} cos^2 (21)[/tex]
[tex]I_f_2 = 102.24 W/cm^2[/tex]
The intensity of light after the second polarizer is removed is 22.88 w/cm² and after the third polarizer is removed is 26.73 w/cm².
Explanation:The intensity of light passing through a polarizing filter can be determined by Malus's Law, which states that the intensity of the transmitted light is equal to the incident light multiplied by the square of the cosine of the angle between the filters. If unpolarized light is incident on the stack, the intensity is halved after passing through the first filter.
1) When the second polarizer is removed, the angle difference will be 40 degrees (61 degrees - 21 degrees). Hence, the intensity would be (60/2) * cos²(40 degree) = 22.88 w/cm².
2) When the third polarizer is removed, the angle difference is only 21 degrees, hence the intensity would be (60/2) * cos²(21 degree) = 26.73 w/cm².
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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm
2
after it passes through the stack.
If the incident intensity is kept constant:
1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?
Answer:
A) I_f3 = 27.58 W/cm²
B) I_f2 = 102.26 W/cm²
Explanation:
We are given;
-The angle of the second polarizing to the first; θ_2 = 21°
-The angle of the third polarizing to the first; θ_1 = 61°
- The unpolarized light after it pass through the polarizing stack; I_u = I_3 = 60 W/cm²
A) Let the initial intensity of the beam of light before polarization be I_p
Thus, when the unpolarized light passes through the first polarizing filter, the intensity of light that emerges would be given as;
I_1 = (I_p)/2
According to Malus’s law,
I = I_max(cos²Φ)
Thus, we can say that;
the intensity of light that would emerge from the second polarizing filter would be given as;
I_2 = I_1(cos²Φ1) = ((I_p)/2)(cos²Φ1)
Similarly, the intensity of light that will emerge from the third filter would be given as;
I_3 = I_2(cos²Φ1) = ((I_p)/2)(cos²Φ1)(cos²(Φ2 - Φ1)
Thus, making I_p the subject of the formula, we have ;
I_p = (2I_3)/[(cos²Φ1)(cos²(Φ2 - Φ1)]
Plugging in the relevant values, we have;
I_p = (2*60)/[(cos²21)(cos²(61 - 21)]
I_p = 234.65 W/cm²
Now, when the second polarizer is removed, the third polarizer becomes the second and final polarizer so the intensity of light emerging from the stack would be given as;
I_f3 = (I_p/2)(cos²Φ2)
I_f3 = (234.65/2)(cos²61)
I_f3 = 27.58 W/cm²
B) Similarly, when the third polarizer is removed, the second polarizer becomes the final polarizer and the intensity of light emerging from the stack would be given as;
I_f2 = (I_p/2)(cos²Φ1)
I_f2 = (234.65/2)(cos²21)
I_f2 = 102.26 W/cm²
A 6.85-m radius air balloon loaded with passengers and ballast is floating at a fixed altitude. Determine how much weight (ballast) must be dropped overboard to make the balloon rise 114 m in 17.0 s. Assume a constant value of 1.2 kg/m3 for the density of air. Ballast is weight of negligible volume that can be dropped overboard to make the balloon rise.
Answer: 120 kg
Explanation:
Given
Radius of balloon, r = 6.85 m
Distance moved by the balloon, d = 114 m
Time spent in moving, t = 17 s
Density of air, ρ = 1.2 kg/m³
Volume of the balloon = 4/3πr³
Volume = 4/3 * 3.142 * 6.85³
Volume = 4/3 * 3.142 * 321.42
Volume = 4/3 * 1009.90
Volume = 1346.20 m³
Density = mass / volume ->
Mass = Density * volume
Mass = 1.2 * 1346.2
Mass = 1615.44 kg
Velocity = distance / time
Velocity = 114 / 17
Velocity = 6.71 m/s
If it starts from rest, 0 m/s, then the final velocity is 13.4 m/s
acceleration = velocity / time
acceleration = 13.4 / 17 m/s²
The mass dropped from the balloon decreases Mb and increases buoyancy
F = ma
mg = (Mb - m) * a
9.8 * m = (1615.44 - m) * 13.4/17
9.8m * 17/13.4 = 1615.44 - m
12.43m = 1615.44 - m
12.43m + m = 1615.44
13.43m = 1615.44
m = 1615.44 / 13.43
m = 120.29 kg
A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angular speed ωi. What is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other? Note: the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to the disk is I = (1/2)MR .
Answer:
Explanation:
Moment of inertia of larger disk I₁ = 1/2 MR²
Moment of inertia of smaller disk I₂ = 1/2 m r ²
Initial angular velocity
We shall apply law of conservation of angular momentum .
initial total momentum = final angular momentum
I₁ X ωi = ( I₁ + I₂ )ωf
1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf
ωf = ωi / ( 1 + m r²/MR² )
If the bore of an engine is increased without any other changes except for the change to proper-size replacement pistons. the displacement will ________ and the compression rate will ________.
Answer:
increase, increase
Explanation:
If the bore of an engine is increased without any other changes except for the change to proper-size replacement pistons. the displacement will increase and the compression rate will increase.
Further Explanation:
If the bore is increased, there are several benefits to this. There is more area for the valves this improves flow through the engine. The compression ratio will increase, thereby increasing thermal efficiency. The displacement also increases as the power output increase. Also, higher RPM will further increase power.
However, a decrease in bore will result in corresponding reduction in displacement and power output.
A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is the period of small oscillations of the hoop? Type your answer below, accurate to two decimal places, and assuming it is in seconds. [Recall that the moment of inertia of a hoop around its center is Icm=MR2.]
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds
An unusual lightning strike has a vertical portion with a current of –400 A downwards. The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT. In units of N (Newtons), the force exerted by the Earth’s magnetic field on the 25 m-long current is Group of answer choices 300 A, East. 0.30 A, West. 0 0.012 A, East. 0.012 A, West.
Given that,
Current, I = 400 A (downwards)
The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT.
We need to find the force exerted by the Earth’s magnetic field on the 25 m-long current. We know that the magnetic force is given by :
[tex]F=iLB\sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=400\times 25\times 30\times 10^{-6}\\\\F=0.3\ N[/tex]
The force is acting in a plane perpendicular to the current and the magnetic field i.e out of the plane.
After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. The rockets explode high in the air and the sound travels out uniformly in all directions. If the sound intensity is 1.52 10-6 W/m2 at a distance of 123 m from the explosion, at what distance from the explosion is the sound intensity half this value
Answer:
The distance when the intensity is halved is 173.95 m
Explanation:
Given;
initial intensity of the sound, I₁ = 1.52 x 10⁻⁶ W/m²
initial distance from the explosion, d₁ = 123 m
final intensity of the sound, I₂ = ¹/₂ (1.52 x 10⁻⁶ W/m²) = 0.76 x 10⁻⁶ W/m²
Intensity of sound is inversely proportional to the square of distance between the source and the receiver.
I ∝ ¹/d²
I₁d₁² = I₂d²
(1.52 x 10⁻⁶)(123)² = (0.76 x 10⁻⁶)d₂²
d₂² = (1.52 x 10⁻⁶ x 123²) / (0.76 x 10⁻⁶)
d₂² = 30258
d₂ = √30258
d₂ = 173.95 m
Therefore, the distance when the intensity is halved is 173.95 m
The sound intensity varies inversely as the square of the distance from the
explosion source.
The distance from the explosion at which the sound intensity is half of 1.52 × 10⁻⁶ W/m², is approximately 173.95 meters
Reasons:
The sound intensity at 123 m = 1.52 × 10⁻⁶ W/m²
Required:
The distance at which the sound intensity is half the given value.
Solution:
Sound intensity is given by the formula;
[tex]I \propto \mathbf{ \dfrac{1}{d^2}}[/tex]
Which gives;
I × d² = Constant
I₁ × d₁² = I₂ × d₂²
Where;
I₁ = The sound intensity at d₁
I₂ = The sound intensity at d₂
[tex]d_2 = \mathbf{ \sqrt{\dfrac{I_1 \times d_1^2}{I_2} }}[/tex]
When the sound intensity is half the given value, we have;
I₂ = 0.5 × I₁
I₂ = 0.5 × 1.52 × 10⁻⁶ = 7.6 × 10⁻⁷
Therefore;
[tex]d_2 = \sqrt{\dfrac{1.52 \times 10^{-6} \times 123^2}{7.6 \times 10^{-7}} } \approx 173.95[/tex]
The distance from the explosion at which the sound intensity is half of the
sound intensity at 123 meters from the explosion, d₂ ≈ 173.95 m.
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