The moment of inertia of the given cube, which is the rotational equivalent of mass in linear motion, is 0.000075 kg.m^2. This is calculated by applying the formula I = (M*s^2)/6 with given parameters. The parallel axis theorem helps in finding the moment of inertia about any axis parallel to and a distance away from an object's center of mass.
Explanation:The moment of inertia of a cube about an axis normal to and through the center of one of its faces is calculated using the formula I = (M*s^2)/6, where M is mass, and s is the side length of the cube. In this case, M = 0.500 kg and s = 0.030 m. Therefore, the moment of inertia I = (0.500 kg * (0.030 m)^2)/6 = 0.000075 kg.m^2. Moment of inertia can be thought of as the rotational equivalent of mass for linear motion.
The moment of inertia of a cube, or any object, about any axis through its center of mass can be calculated using the parallel axis theorem. The theorem links the moment of inertia about an axis passing through the object's center of mass to the moment of inertia about any other parallel axis. It states that the moment of inertia about any axis parallel to and a distance d away from the axis through the center of mass is equal to the moment of inertia of the object about the axis through the center of mass plus the mass of the object times the square of the distance between the axes (I=I_cm + md^2). Therefore, the moment of inertia is also related to the distribution of mass within an object.
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An airplane is initially flying horizontally (not gaining or losing altitude), and heading exactly North. Suppose that the earth's magnetic field at this point is also exactly horizontal, and points from South to North (it really does!). The airplane now starts a different motion. It maintains the same speed, but gains altitude at a constant rate, still heading North. The magnitude of the electric potential difference between the wingtips has... [Remember that the airplane is made of metal, and is a conductor] Group of answer choices
Note: The answer choices are :
a) Increased
b) Decreased
c) stayed the same
Answer:
The correct option is Increased
The magnitude of the electric field potential difference between the wingtips increases.
Explanation:
The magnitude of the electric potential difference is the induced emf and is given by the equation:
[tex]emf = l (v \times B)[/tex]
where l = length
v = velocity
B = magnetic field
As the altitude of the airplane increases, the magnetic flux becomes stronger, the speed of the airplane becomes perpendicular to the magnetic field, i.e. [tex]v \times B = vB sin90 = vB\\[/tex] ,
the induced emf = vlB, and thus increases.
The magnitude of the electric field potential difference between the wingtips increases
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
Consider two pulses (a wave with only a single peak) traveling towards each other on a string. When the instant that the peaks of these two pulses cross, the resultant disturbance has a maximum displacement of 3.80 A where A is the amplitude of the first pulse. What must be the amplitude of the second pulse be
Answer:
2.8A
Explanation:
To calculate the total amplitude (when both pulses meet), we need to add up the amplitudes of each pulse. Since A is the amplitude of the first pulse, and we can call B the amplitude of the second pulse and C the total amplitude, we have that A+B=C=3.8A, which means that B=3.8A-A=2.8A, which we have already said is the amplitude of the second pulse.
2. An electrical heater 200mm long and 15mm in diameter is inserted into a drilled hole normal to the surface of a large block of material having a thermal conductivity of 5W/m·K. Estimate the temperature reached by the heater when dissipating 25 W with the surface of the block at a temperature of 35 °C.
Answer:
The final temperature is 50.8degrees celcius
Explanation:
Pls refer to attached handwritten document
Answer: 50.63° C
Explanation:
Given
Length of heater, L = 200 mm = 0.2 m
Diameter of heater, D = 15 mm = 0.015 m
Thermal conductivity, k = 5 W/m.K
Power of the heater, q = 25 W
Temperature of the block, = 35° C
T1 = T2 + (q/kS)
S can be gotten from the relationship
S = 2πL/In(4L/D)
On substituting we have
S = (2 * 3.142 * 0.2) / In (4 * 0.2 / 0.015)
S = 1.2568 / In 53.33
S = 1.2568 / 3.98
S = 0.32 m
Proceeding to substitute into the main equation, we have
T1 = T2 + (q/kS)
T1 = 35 + (25 / 5 * 0.32)
T1 = 35 + (25 / 1.6)
T1 = 35 + 15.625
T1 = 50.63° C
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.
Two very small planets orbit a much larger star. Planet A orbits the star with a period TA . Planet B orbits the star at four times the distance of planet A, but planet B is four times as massive as planet A. Assume that the orbits of both planets are approximately circular.
Planet B orbits the star with a period TB equal to
a. TA/4.
b. 16TA.
c. 8TA.
d. 4TA
d. TA
Answer:
Explanation:
Given that,
Period of Planet A around the star
Pa = Ta
Let the semi major axis of planet A from the star be
a(a) = x
Given that, the distance of Planet B from the star (i.e. it semi major axis) is 4 time that of Planet A
a(b) = 4 × a(a) = 4 × x
a(b) = 4x
Also planet B is 4 times massive as planet A
Using Kepler's law
P² ∝a³
P²/a³ = k
Where
P is the period
a is the semi major axis
Then,
Pa² / a(a)³ = Pb² / a(b)³
Pa denotes Period of Planet A and it is Pa = Ta
a(a) denoted semi major axis of planet A and it is a(a) = x
Pb is period of planet B, which is the required question
a(b) is the semimajor axis of planet B and it is a(b) = 4x
So,
Ta² / x³ = Pb² / (4x)³
Ta² / x³ = Pb² / 64x³
Cross multiply
Ta² × 64x³ = Pb² × x³
Divide both sides by x³
Ta² × 64 = Pb²
Then, Pb = √(Ta² × 64)
Pb = 8Ta
Then, the period of Planet B is eight times the period of Planet A.
The correct answer is C
The period of orbit for planet B is 4 times the period of orbit for planet A.
Explanation:To determine the period of orbit for planet B, we can use Kepler's Third Law, which states that the square of the period of an orbit is proportional to the cube of its average distance from the center of attraction. Since planet B is four times the distance of planet A and four times as massive, its period will be 4^3/4^2 times that of planet A. Simplifying, this means that TB = 4 * TA, so the correct answer is d. 4TA.
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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.
) 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]
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
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
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:
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
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
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
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The same pipe is used to carry both air and water. For the same fluid velocity and friction factor for the air and water flows: the pressure drop for the air flow is greater than that for the water flow. the pressure drop for the air flow equals the pressure drop for the water flow. the pressure drop for the water flow is greater than that for the air flow.
Answer:
the pressure drop for the water flow is greater than that for the air flow.
Explanation:
Detailed analysis of the problem is show below.
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
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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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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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
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:
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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.
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
A firm wants to determine the amount of frictional torque in their current line of grindstones, so they can redesign them to be more energy efficient. To do this, they ask you to test the best-selling model, which is basically a diskshaped grindstone of mass 1.1 kg and radius 0.09 m which operates at 73.3 rad/s. When the power is shut off, you time the grindstone and find it takes 42.4 s for it to stop rotating. What is the frictional torque exerted on the grindstone in newton-meters
Explanation:
Mass of the diskshaped grindstone, m = 1.1 kg
Radius of disk, r = 0.09 m
Angular velocity, [tex]\omega=73.3\ rad/s[/tex]
Time, t = 42.4 s
We need to find the frictional torque exerted on the grindstone. Torque in the rotational kinematics is given by :
[tex]\tau=I\alpha[/tex]
I is moment of inertia of disk, [tex]I=\dfrac{mr^2}{2}[/tex]
[tex]\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{1.1\times (0.09)^2\times 73.3 }{2\times 42.4}\\\\\tau=7.7\times 10^{-3}\ N-m[/tex]
So, the frictional torque exerted on the grindstone is [tex]7.7\times 10^{-3}\ N-m[/tex].
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.
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
Space satellites in the inner solar system (such as those in orbit around Earth) are often powered by solar panels. Satellites that travel into the outer solar system, however, are usually powered by the decay of nuclear isotopes – a more expensive and dangerous technology. Why can’t they use solar panels as well?
a. At greater distances, sunlight is more likely to destructively interfere.
b. At greater distances, sunlight is more likely to constructively interfere.
c. At greater distances from the sun, the frequency of sunlight will be too low.
d. At greater distances from the sun, the intensity of sunlight will be too low.
Answer:
D
Explanation:
See attached file