Answer:
a) The final angular velocity of the system is 4 radians per second, b) The amount of mechanical energy lost due to friction is 5.25 joules.
Explanation:
a) The problem is a clear representation of the Principle of the Angular Momentum Conservation, where moment of inertia of the system is increased by the adding of the uniform rod and there are no external forces exerted on the system. This system is represented by the following model:
[tex]I_{d} \cdot \omega_{o} = (I_{d} + I_{r})\cdot \omega_{f}[/tex]
Where:
[tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex] - Initial and final angular velocities, measured in radians per second.
[tex]I_{d}[/tex], [tex]I_{r}[/tex] - Moments of inertia of the uniform solid disk and uniform rod, measured in [tex]kg\cdot m^{2}[/tex].
Now, the final angular speed is cleared:
[tex]\omega_{f} = \frac{I_{d}}{I_{d}+I_{r}}\cdot \omega_{o}[/tex]
The moments of inertia of the uniform solid disk and uniform rod are modelled by these formulas:
Solid Disk
[tex]I_{d} = \frac{1}{2}\cdot m_{d}\cdot r_{d}^{2}[/tex]
Where:
[tex]m_{d}[/tex] - Mass of the disk, measured in kilograms.
[tex]r_{d}[/tex] - Radius of the disk, measured in meters.
Given that [tex]m_{d} = 1\,kg[/tex] and [tex]r_{d} = 1\,m[/tex], the moment of inertia of the disk is:
[tex]I_{d} = \frac{1}{2}\cdot (1\,kg)\cdot (1\,m)^{2}[/tex]
[tex]I_{d} = 0.5\,kg\cdot m^{2}[/tex]
Rod (rotating about its center)
[tex]I_{r} = \frac{1}{12}\cdot m_{r}\cdot l_{r}^{2}[/tex]
Where:
[tex]m_{r}[/tex] - Mass of the rod, measured in kilograms.
[tex]l_{r}[/tex] - Length of the rod, measured in meters.
Given that [tex]m_{r} = 0.5\,kg[/tex] and [tex]l_{r} = 3\,m[/tex], the moment of inertia of the rod is:
[tex]I_{r} = \frac{1}{12}\cdot (0.5\,kg)\cdot (3\,m)^{2}[/tex]
[tex]I_{r} = 0.375\,kg\cdot m^{2}[/tex]
Now, knowing that [tex]\omega_{o} = 7\,\frac{rad}{s}[/tex], the final angular velocity is:
[tex]\omega_{f} = \left(\frac{0.5\,kg\cdot m^{2}}{0.5\,kg\cdot m^{2}+0.375\,kg\cdot m^{2}}\right)\cdot \left(7\,\frac{rad}{s} \right)[/tex]
[tex]\omega_{f} = 4\,\frac{rad}{s}[/tex]
The final angular velocity of the system is 4 radians per second.
b) According to the Principle of Energy Conservation, the inclusion of the uniform rod on the turntable is represented by the following expression:
[tex]K_{1} = K_{2} + \Delta E_{loss}[/tex]
Where:
[tex]K_{1}[/tex] - Rotational kinetic energy of the uniform disk, measured in joules.
[tex]K_{2}[/tex] - Rotational kinetic energy of the system (uniform disk + uniform rod), measured in joules.
[tex]\Delta E_{loss}[/tex] - Mechanical energy lost due to friction, measured in joules.
The mechanical energy lost due to friction is cleared:
[tex]\Delta E_{loss} = K_{1} - K_{2}[/tex]
Now, the expression is expanded and mechanical energy losses is calculated:
[tex]\Delta E_{loss} = \frac{1}{2}\cdot I_{d}\cdot \omega_{o}^{2} - \frac{1}{2}\cdot (I_{d}+I_{r})\cdot \omega_{f}^{2}[/tex]
[tex]\Delta E_{loss} = \frac{1}{2}\cdot (0.5\,kg\cdot m^{2})\cdot \left(7\,\frac{rad}{s} \right)^{2} - \frac{1}{2}\cdot (0.5\,kg\cdot m^{2} + 0.375\,kg\cdot m^{2})\cdot \left(4\,\frac{rad}{s} \right)^{2}[/tex]
[tex]\Delta E_{loss} = 5.25\,J[/tex]
The amount of mechanical energy lost due to friction is 5.25 joules.
The two arms of a U-tube are not identical, one having 2.5 times the diameter of the other. A cork in the narrow arm requires a force of 16 N to remove it. The tube is filled with water and the wide arm is fitted with a piston. The minimum force that must be applied to the piston to push the cork out is:
Answer:
F₁ = 100 N
Explanation:
The pressure must be equally transmitted from the piston to the narrow arm. Therefore,
P₁ = P₂
F₁/A₁ = F₂/A₂
F₁/F₂ = A₁/A₂
where,
F₁ = Force Required to be applied to piston = ?
F₂ = Force to push cork at narrow arm = 16 N
A₁ = Area of wider arm = πd₁²/4
A₂ = Area of narrow arm = πd₂²/4
Therefore,
F₁/16 N = (πd₁²/4)/(πd₂²/4)
F₁ = (16 N)(d₁²/d₂²)
but, it is given that the diameter of wider arm is 2.5 times the diameter of the narrow arm.
d₁ = 2.5 d₂
Therefore,
F₁ = (16 N)[(2.5 d₂)²/d₂²]
F₁ = (16 N)(6.25)
F₁ = 100 N
At a playground, two young children are on identical swings. One child appears to be about twice as heavy as the other. Part A If you pull them back together the same distance and release them to start them swinging, what will you notice about the oscillations of the two children
Answer:
The motion of the lighter child would look faster than that of the heavier child, but both have the same period of oscillation.
Explanation:
Oscillation is a type of simple harmonic motion which involves the to and fro movement of an object. The oscillation takes place at a required time called the period of oscillation.
Since the swings are similar, the period of oscillation of the two children are the same and they would complete one oscillation in the same time. Though the oscillation of the lighter child seems faster than that of the heavy child, their masses does not affect the period of oscillation.
When a heavy object oscillates, its mass increases the drag or damping force, but not the period of oscillation. Thus, it oscillate slowly.
The New England Merchants Bank Building in Boston is 152 m high. On windy days it sways with a frequency of 0.17 Hz, and the acceleration of the top of the building can reach 2.0% of the free-fall acceleration, enough to cause discomfort for occupants. What is the total distance, side to side, that the top of the building moves during such an oscillation?
Answer:
d = 8.4 cm
Explanation:
In order to calculate the amplitude of oscillation of the top of the building, you use the following formula for the max acceleration of as simple harmonic motion:
[tex]a_{max}=A\omega^2[/tex] (1)
A: amplitude of the oscillation
w: angular speed of the oscillation = 2[tex]\pi[/tex]f
f: frequency = 0.17Hz
The maximum acceleration of the top of the building is a 2.0% of the free-fall acceleration. Then, you have:
[tex]a_{max}=0.02(9.8m/s^2)=0.196\frac{m}{s^2}[/tex]
Then, you solve for A in the equation (1) and replace the values of the parameters:
[tex]A=\frac{a_{max}}{\omega^2}=\frac{a_{max}}{4\p^2i f^2}\\\\A=\frac{0.196m/s^2}{16\pi^2(0.17Hz)^2}\\\\A=0.042m=4.2cm[/tex]
The total distance, side to side, of the oscilation of the top of the building is twice the amplitude A. Then you obtain:
d = 2A = 2(4.2cm) = 8.4cm
The total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.
Given data:
The height of building is, h = 152 m.
The frequency on windy days is, f = 0.17 Hz.
The acceleration on the top of building is, a = 2/100g (Here g is gravitational acceleration).
This problem can be resolved using the concept of amplitude and angular frequency. The expression for the magnitude of acceleration at the top pf building is given as,
[tex]a = A \times \omega^{2}\\\\a = A \times (2 \pi f)^{2}\\\\\dfrac{2}{100} \times g=A \times (2 \pi \times 0.17)^{2}\\\\\dfrac{2}{100} \times 9.8=A \times (2 \pi \times 0.17)^{2}\\\\A =0.042 \;\rm m[/tex]
And, the total distance, side to side, of the oscillation of the top of the building is twice the amplitude A. Then,
]s = 2A
s = 2 (0.042)
s = 0.084 m
s = 8.4 cm
Thus, we can conclude that the total side to side distance at the top of the building, covered during such an oscillation is 8.4 cm.
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suppose the ball has the smallest possible frequency that allows it to go all the way around the circle. what tension in the string when the ball is at the highest point
The complete question is missing, so i have attached the complete question.
Answer:
A) FBD is attached.
B) The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
Explanation:
A) I've attached the image of the free body diagram.
B) The formula for the net force is given as;
F_net = mv²/r
We know that angular velocity;ω = v/r
Thus;
F_net = mω²r
Now, the minimum downward force is the weight and so;
mg = m(ω_min)²r
m will cancel out to give;
g = (ω_min)²r
(ω_min)² = g/r
ω_min = √(g/r)
The condition that must be satisfied is for ω_min = √(g/r)
C) The tension in the string would be zero. This is because at the smallest frequency, the only radially inward force at that point is the weight(force of gravity).
A uniform electric field stack \rightarrow Ei subscript I with rightwards arrow on top is present in the region between infinite parallel plane plates A and B and a uniform electric field stack\rightarrow Eii subscript I I end subscript with rightwards arrow on top is present in the region between infinite parallel plane plates B and C. When the plates are vertical, stack \rightarrow Ei subscript I with rightwards arrow on top is directed to the right and stack \rightarrow Eii subscript I I end subscript with rightwards arrow on top to the left. The signs of the charges on plates A, B and C may be
a. ?, ?, ?.
b. +, ?, ?.
c. +, ?, +.
d. +, +, +.
e. any one of the above
Answer:
e. any one of the above
Explanation:
Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what is the value of the spring constant if the spring is compressed from its rest length to a maximum depth of 2.4 m by the train
Answer:
187,500N/mExplanation:
From the question, the kinectic energy of the train will be equal to the energy stored in the spring.
Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².
Equating both we will have;
1/2 mv² = 1/2ke²
mv² = ke²
m is the mass of the train
v is the velocity of then train
k is the spring constant
e is the extension caused by the spring.
Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m
Substituting this values into the formula will give;
30000*4² = k*1.6²
[tex]k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}[/tex]
The value of the spring constant is 187,500N/m
The coefficient of static friction is usually
Answer:
Higher than the coefficient of kinetic friction.
Explanation:
Hope it helps u.. :)
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high does it go
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
The ball reaches the maximum height of 54 feet
The question is about projectile motion,
the ball is shot at an angle α = 45°, and
the initial velocity u = 46 ft/s.
Under the projectile motion, the maximum height H is given by:
[tex]H=\frac{u^2sin^2\alpha }{2g} [/tex]
where, g = 9.8 m/s²
substituting the given values we get:
[tex]H=\frac{46^2sin^{2}(45)}{2*9.8}\\ \\ H=\frac{46*46*(1/2)}{2*9.8}\\ \\ H=54 feet[/tex]
Hence, the maximum height is 54 feet.
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A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine shaft by a cable connected to a winch. The shaft is inclined at 28.0° above the horizontal. The car accelerates uniformly to a speed of 2.35 m/s in 14.0 s and then continues at constant speed.(A) What power must the winch motor provide when the car is moving at constant speed? kW(B) What maximum power must the motor provide? kW(C) What total energy transfers out of the motor by work by the time the car moves off the end of the track, which is of length 1,250 m?
Answer:
a) P = 10.27 kW
b) Pmax = 10.65 kW
c) E = 5.47 MJ
Explanation:
Mass of the loaded car, m = 950 kg
Angle of inclination of the shaft, θ = 28°
Acceleration due to gravity, g = 9.8 m/s²
The speed of the car, v = 2.35 m/s
Change in time, t = 14.0 s
a) The power that must be provided by the winch motor when the car is moving at constant speed.
P = Fv
The force exerted by the motor, F = mg sinθ
P = mgv sinθ
P = 950 * 9.8 *2.35* sin28°
P = 10,271.3 W
P = 10.27 kW
b) Maximum power that the motor must provide:
[tex]P = mv\frac{dv}{dt} + mgvsin \theta\\dv/dt = \frac{2.35 - 0}{14} \\dv/dt = 0.168 m/s^2\\P = (950*2.35*0.168) + (950*9.8*2.35* sin28)\\P = 374.74 + 10271.3\\P = 10646.04 W\\10.65 kW[/tex]
c) Total energy transferred:
Length of the track, d = 1250 m
[tex]E = 0.5 mv^2 + mgd sin \theta\\E = (0.5 * 950 * 2.35^2) + (950 * 9.8 * 1250 * sin 28)\\E = 2623.19 + 5463475.31\\E = 5466098.50 J\\E = 5.47 MJ[/tex]
A person is riding on a Ferris wheel of radius R. He starts at the lowest point of the wheel. When the wheel makes one complete revolution, is the net work done by the gravitational force positive, zero or negative? Do you need to know how the speed of the person changed before you can answer the question?
Answering the two questions in reverse order:
-- No. I don't need to know how the speed of the person changed before I can answer the question. I can answer it now.
-- The NET work done by the gravitational force is zero.
-- As the person and his girl-friend go up the first half of the wheel, the motor does positive work and gravity does negative work.
-- After they pass the peak at the top and come down the second half of the wheel, the motor does negative work and gravity does positive work, even though the couple may be interested in other things during that time.
-- The total work done by gravity in one complete revolution is zero.
-- The total work done by the motor in one complete revolution is only what it takes to pay back the energy robbed by friction and air resistance.
The work done by the gravitational force is zero.
Work Done:Work done by a conservative force is path independent. Which means it only depends on the initial and final position of the body. The gravitational force is a conservational force and the gravitational potential energy depends only upon the height of the body.
Let the lowest point of the body is at some height h, then the initial gravitational potential energy of the person is:
PE(initial) = mgh
The final position of the person is also at a height h, thus, the final gravitational potential energy :
PE(final) = mgh
According to the work-energy theorem:
work done = - change in potential energy
work done = -(mgh - mgh) = 0
Thus, the work done is zero in the given case.
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two 200 pound lead balls are separated by a distance 1m. both balls have the same positive charge q. what charge will produce an electrostatic force.between the balls that is of the same order of magnitude as the weight of one ball?
Answer:
The charge is [tex]q = 3.14 *10^{-4} \ C[/tex]
Explanation:
From the question we are told that
The mass of each ball is [tex]m = 200 \ lb = \frac{200}{2.205} = 90.70 \ kg[/tex]
The distance of separation is [tex]d = 1 \ m[/tex]
Generally the weight of the each ball is mathematically represented as
[tex]W = m * g[/tex]
where g is the acceleration due to gravity with a value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]W = 90.70 * 9.8[/tex]
[tex]W = 889 \ N[/tex]
Generally the electrostatic force between this balls is mathematically represented as
[tex]F_e = \frac{k * q_1* q_2 }{d^2}[/tex]
given that the the charges are equal we have
[tex]q_1= q_2 = q[/tex]
So
[tex]F_e = \frac{k * q^2 }{d^2}[/tex]
Now from the question we are told to find the charge when the weight of one ball is equal to the electrostatic force
So we have
[tex]889 = \frac{9*10^9 * q^2}{1^2}[/tex]
=> [tex]q = 3.14 *10^{-4} \ C[/tex]
The magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
Given data:
The masses of two lead balls are, m = 200 lb = 200/2.205 = 90.70 kg.
The distance of separation of two balls is, d = 1 m.
First of all we need to obtain the weight of ball. The weight of the ball is expressed as,
W = mg
Here,
g is the gravitational acceleration.
Solving as,
W = 90.70 × 9.8
W = 888.86 N
The expression for the electrostatic force between this balls is mathematically represented as,
[tex]F = \dfrac{k \times q_{1} \times q_{2}}{d^{2}}[/tex]
Since, the charges are equal then,
[tex]q_{1} =q_{2}=q[/tex]
Also, the magnitude of force between the balls is same as the weight of one ball. Then,
F = W
Solving as,
[tex]F =W= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\889= \dfrac{(9 \times 10^{9}) \times q^{2}}{1^{2}}\\\\q = 3.14 \times 10^{-4} \;\rm C[/tex]
Thus, we can conclude that the magnitude of charge on the balls is [tex]3.14 \times 10^{-4} \;\rm C[/tex].
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What are the two types of long-term memory?
A. sensory and short-term
B. iconic and echoic
C. explicit and implicit
D. recency and primacy
Answer:
C. explicit and implicit
Explanation:
E20
If you were to experimentally determine the length of the pendulum, why would you not get the same length in Iowa?
Answer:
The length of the pendulum depends on acceleration due to gravity (g) which varies in different Earth's location beacuse Earth is not perfectly spherical.
Explanation:
The period of oscillation is calculated as;
[tex]T = 2\pi\sqrt{\frac{l}{g} }[/tex]
where;
L is the length of the pendulum bob
g is acceleration due to gravity
If we make L the subject of the formula in the equation above, we will have;
[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\\sqrt{\frac{l}{g} } = \frac{T}{2\pi} \\\\\frac{l}{g} = (\frac{T}{2\pi} \)^2\\\\\frac{l}{g} =\frac{T^2}{4\pi^2}\\\\L = \frac{gT^2}{4\pi^2}[/tex]
The length of the pendulum depends on acceleration due to gravity (g).
Acceleration due to gravity is often assumed to be the same everywhere on Earth, but it varies because Earth is not perfectly spherical. The variation of acceleration due to gravity (g) as a result of Earth's geometry, will also cause the length of the pendulum to vary.
Find the period of revolution for the planet Mercury, whose average distance from the Sun is 5.79 x 1010 m.
Answer:
T = 7.61*10^6 s
Explanation:
In order to calculate the Mercury's period. in its orbit around the sun, you take into account on the Kepler's law. You use the following formula:
[tex]T=\sqrt{\frac{4\pi^2r^3}{GM_s}}[/tex] (1)
T: period of Mercury
r: distance between Mercury and Sun
Ms: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.674*10^-11 m^3 kg^-1 s^-2
You replace the values of all parameters in the equation (1):
[tex]T=\sqrt{\frac{4\pi^2(5.79*10^{10}m)^3}{(6.674*10^{-11}m^3kg^{-1}s^{-2})(1.98*10^{30}kg)}}\\\\T=7.61*10^6s*\frac{1h}{3600s}*\frac{1d}{24h}=88.13\ days[/tex]
The period of Mercury is 7.61*10^6 s, which is approximately 88.13 Earth's days
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lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens
The nearpoint of an eye is 151 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it. What should be the focal length of this lens?
Answer:
29.96cm
Explanation:
Using the corrective lens, the image should be formed at the front of the eye and be upright and virtual.
Now using the lens equation as follows;
[tex]\frac{1}{f} = \frac{1}{v} + \frac{1}{u}[/tex] -------------(i)
Where;
f = focal length of the lens
v = image distance as seen by the lens
u = object distance from the lens
From the question;
v = -151cm [-ve since the image formed is virtual]
u = 25cm
Rewrite equation (i) to have;
[tex]f = \frac{uv}{u+v}[/tex]
Substitute the values of v and u into the equation;
[tex]f = \frac{25*(-151)}{25-151}[/tex]
[tex]f = \frac{-3775}{-126}[/tex]
f = 29.96cm
The focal length should be 29.96cm
In an RC circuit, how many time constants must elapse if an initially uncharged capacitor is to reach 80% of its final potential difference
Answer:
1.6 time constants must elapse
Explanation:
voltage on a cap, charging is given as
v = v₀[1–e^(–t/τ)]
Where R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
v = v₀[1–e^(–t/τ)]
1–e^(–t/τ) = 0.8
e^(–t/τ) = 0.2
–t/τ = –1.609
t = 1.609τ
Imagine two free electrons that collide elastically in an acidic solution where one electron was moving and the other electron was stationary. When the electrons separate the moving electron now has a velocity of 400 m/s and the stationary electron now has a velocity of 200 m/s. What was the initial kinetic energy of the moving electron
Answer: 9.1 × 10^-26 Joule
Explanation:
Since the collision is elastic. The kinetic energy will be conserved. That is, the sum of kinetic energy before collision will be the same as the sum of the
energy after collision.
Mass of an electron = 9.1 × 10^-31 kg
Given that the velocity of the moving electron = 400 m/s and the stationary electron now has a velocity = 200 m/s.
K.E = 1/2mv^2
Add the two kinetic energies
1/2mV1^2 + 1/2mV2^2
1/2m( V1^2 + V2^2 )
Since they both have common mass
Substitute m and the two velocities
1/2 × 9.1×10^-31( 400^2 + 200^2)
4.55×10^-31 ( 160000 + 40000 )
4.55×10^-31 × 200000
K.E = 9.1 × 10^-26 Joule
Therefore, the initial kinetic energy of the moving electron is 9.1×10^-26 J
How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller
Answer:
The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
Explanation:
Let us first consider the initial characteristics of the angular motion of the disk
moment of inertia = [tex]I[/tex]
angular speed = ω
For the second case, we consider the characteristics to now be
moment of inertia = [tex]5I[/tex] (five times larger)
angular speed = ω/5 (five times smaller)
Recall that the kinetic energy of a spinning body is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
therefore,
for the first case, the K.E. is given as
[tex]KE = \frac{1}{2}Iw^{2}[/tex]
and for the second case, the K.E. is given as
[tex]KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2} = \frac{5}{50}Iw^{2}[/tex]
[tex]KE = \frac{1}{10}Iw^{2}[/tex]
this is one-tenth the kinetic energy before its spinning characteristics were changed.
This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.
A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.
Relation between Kinetic energy and Moment of Inertia:
Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.Now, let's consider moment of inertia = I and angular speed = ω
It is asked that what would be change in Kinetic energy if
moment of inertia = (five times larger)
angular speed = ω/5 (five times smaller)
The kinetic energy of a spinning body is given as:
[tex]K.E.=\frac{1}{2} I. w^2[/tex]
On substituting the values, we will get:
[tex]K.E.= \frac{1}{2} (5I) (\frac{w}{5} )^2 \\\\K.E. =\frac{1}{10} I. w^2[/tex]
Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.
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If a diver below the water's surface shines a light up at the bottom of the oil film, at what wavelength (as measured in water) would there be constructive interference in the light that reflects back downward
Answer:
see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference
Explanation:
This is an interference problem in thin films, the refractive index of water is 1.33 and the refractive index of oil is 1.5
Let's analyze the light beam path emitted by the diver.
* when the beam passes from the water to the oil with the highest refractive index, it has a phase change of 180º
* also the wavelength of light in a material medium changes
λ_n = λ / n
where λ_n is the wavelength in the material and λ the wavelength in the vacuum air and n the refractive index.
If we include these aspects, the constructive interference equation is
2t = (m + ½) λ_n
2nt = (m + ½) λ
let's apply this equation to our case
λ = 2nt / (m + ½)
The incidence of replacement of the oil with respect to water is
n = n_oil / n_water = 1.5 / 1.33
n = 1,128
let's calculate
λ = 2 1,128 t / (m + ½)
λ = 2,256 t / (m + ½)
In your statement you do not include the value of the oil layer that is the thin film, suppose a value to finish the calculation
t = 0.001 mm = 1 10⁻⁶ m
the formula remains
λ = 2,256 10⁻⁶ / (m + ½)
Let's find what values of m we have to cut light in the visible range (400 to 700) 10⁻⁹ m
m + ½ = 2,256 10⁻⁶ / λ
m = 2,256 10⁻⁶ / λ - ½
light purple lan = 400 10⁻⁹m
m = 2,256 10-6 / 400 10⁻⁹ - ½
m = 5.64 - 0.5
m = 5.14
m = 5
red light λ = 700 10⁻⁹m
m = 2,256 1-6 / 700 10⁻⁹ - ½
m = 3.22 - 0.5
m = 2.72
m = 3
we see that the entire spectrum of the visible is between the integers from 3 to 5 so only three wavelengths are reflected with constructive interference
The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the ground. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held
Answer:
The tube should be held vertically, perpendicular to the ground.
Explanation:
As the power lines of ground are equal, so its electrical field is perpendicular to the ground and the equipotential surface is cylindrical. Therefore, if we put the position fluorescent tube parallel to the ground so the both ends of the tube lie on the same equipotential surface and the difference is zero when its potential.
And the ends of the tube must be on separate equipotential surfaces to optimize potential. The surface near the power line has a greater potential value and the surface farther from the line has a lower potential value, so the tube must be placed perpendicular to the floor to maximize the potential difference.
Consider a loop of wire placed in a uniform magnetic field. Which factors affect the magnetic flux Φm through the loop?
Answer:
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
Explanation:
Magnetic flux is the scalar product of the magnetic field over the area
Ф = ∫ B. dA
where B is the magnetic field and A is the area
Let's look at stationary, for which factors affect flow
* The value of the magnetic field changes either in time or space
* The waxed area changes, the bow is fitting in size
* The angle between the field and the area changes
A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below. a. Find the work done on the sack of flour (in J). b. Find the distance you must push with the 150 N force-on the left side (in m). c. Briefly explain the benefit of using a lever to lift a heavy object.
You are trying to overhear a juicy conversation, but from your distance of 20.0 m, it sounds like only an average whisper of 20.0 dB. So you decide to move closer to give the conversation a sound level of 70.0 dB instead. How close should you come?
Given that,
Distance = 20.0 m
Average whisper = 20.0 dB
Sound level = 70.0 dB
We know that,
The minimum intensity is
[tex]I_{o}=10^{-12}\ W/m^2[/tex]
We need to calculate the sound intensity in the distance of 20 m
Using formula of sound intensity
[tex]dB=10\log(\dfrac{I_{a}}{I_{o}})[/tex]
Put the value into the formula
[tex]20=10\log(\dfrac{I_{a}}{10^{-12}})[/tex]
[tex]10^{2}=\dfrac{I_{a}}{10^{-12}}[/tex]
[tex]I_{a}=10^{-10}\ W/m^2[/tex]
If the conversation a sound level of 70.0 dB instead
We need to calculate the sound intensity
Using formula of sound intensity
[tex]dB=10\log(\dfrac{I_{b}}{I_{o}})[/tex]
Put the value into the formula
[tex]70=10\log(\dfrac{I_{a}}{10^{-12}})[/tex]
[tex]10^{7}=\dfrac{I_{b}}{10^{-12}}[/tex]
[tex]I_{b}=10^{-5}\ W/m^2[/tex]
We know that,
The intensity is inversely proportional with the square of the distance.
We need to calculate the distance
Using formula of intensity
[tex]\dfrac{I_{a}}{I_{b}}=\dfrac{R_{b}^2}{R_{a}^2}[/tex]
Put the value into the formula
[tex]\dfrac{10^{-10}}{10^{-5}}=\dfrac{R_{b}^2}{20^2}[/tex]
[tex]R_{b}^2=20^2\times\dfrac{10^{-10}}{10^{-5}}[/tex]
[tex]R_{b}=\sqrt{20^2\times\dfrac{10^{-10}}{10^{-5}}}[/tex]
[tex]R_{b}=0.063\ m[/tex]
Hence, The distance from the conversation should be 0.063 m.
The distance you should move to achieve a loudness of 70 dB is; 0.0633 m
We are given;
Average distance 1; R_1 = 20 m
Average whisper 1; dB1 = 20 dB
Average whisper 2; dB2 = 70 dB
Now, the formula for loudness in dB is;
dB = 10log(I/I_o)
Where;
I is the intensity of the sound
I_o is the minimum intensity that a human ear can detect = 10^(-12) W/m²
Thus, for dB1 = 20 dB;20 = 10log (I/10^(-12))
20/10 = log (I/10^(-12))
log (I/10^(-12)) = 2
10² = (I/10^(-12))
I = 10² × 10^(-12)
I_1 = 10^(-10) W/m²
Similarly, for dB2 = 70 dB;70 = 10log (I/10^(-12))
70/10 = log (I/10^(-12))
log (I/10^(-12)) = 7
10^(7) = (I/10^(-12))
I = 10^(7) × 10^(-12)
I_2 = 10^(-5) W/m²
The relationship between their intensities and distance is;
I_1/I_2 = (R_2/R_1)²
Where R_2 is the distance you should move to achieve a loudness of 70 dB.
Thus;
(10^(-10))/(10^(-5)) = R_2/20
(R_2)/20 = √(10^(-5))
(R_2)/20 = 0.00316227766
R_2 = 20 × 0.00316227766
R_2 = 0.0632 m
Read more at; https://brainly.com/question/21195959
The vector indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at , its velocity and acceleration vectors are and . Which statement is correct?
Answer:
The only force acts on a projectile is gravitational force {Fg}, therefore its acceleration a=Fg/m will always directed towards the direction of force i.e. vertically downwards. Therefore it will always be perpendicular to the x direction or here we can say that a is always perpendicular to Vx}.
Explanation:
The vector r indicates the instantaneous displacement of a projectile from the origin. At the instant when the projectile is at r , its velocity and acceleration vectors are v and a . Which statement is correct?
A 100 cm length of nichrome wire has a radius of 0.50 mm, a resistivity LaTeX: \rho_0ρ 0= 1.0 × 10-6 Ω ∙ m , and a temperature coefficient LaTeX: \alphaα = 0.4 × 10-3 (oC)-1. At T0 = 20 oC the wire carries current of 0.50 A. How much power does the wire dissipate at a temperature T = 350 oC? Assume the potential difference across the ends of the wire remains constant. Group of answer choices
Answer:
P₃₅₀ = 0.28 watt
Explanation:
First we find the resistance of the wire at 20°C:
R₀ = ρL/A
where,
ρ = resistivity = 1 x 10⁻⁶ Ωm
L = Length of wire = 100 cm = 1 m
A = cross-sectional area of wire = πr² = π(0.5 x 10⁻³ m)² = 0.785 x 10⁻⁶ m²
Therefore,
R₀ = (1 x 10⁻⁶ Ωm)(1 m)/(0.785 x 10⁻⁶ m²)
R₀ = 1.27 Ω
Now, from Ohm's Law:
V = I₀R₀
where,
V = Potential Difference = ?
I₀ = Current Passing at 20°C = 0.5 A
Therefore,
V = (0.5 A)(1.27 Ω)
V = 0.64 volts
Now, we need to find the resistance at 350°C:
R₃₅₀ = R₀(1 + αΔT)
where,
R₃₅₀ = Resistance at 350°C = ?
α = temperature coefficient of resistance = 0.4 x 10⁻³ °C⁻¹
ΔT = Difference in Temperature = 350°C - 20°C = 330°C
Therefore,
R₃₅₀ = (1.27 Ω)[1 + (0.4 x 10⁻³ °C⁻¹)(330°C)]
R₃₅₀ = 1.44 Ω
Now, for power at 350°C:
P₃₅₀ = VI₃₅₀
where,
P₃₅₀ = Power dissipation at 350°C = ?
V = constant potential difference = 0.64 volts
I₃₅₀ = Current at 350°C = V/R₃₅₀ (From Ohm's Law)
Therefore,
P₃₅₀ = V²/R₃₅₉
P₃₅₀ = (0.64 volts)²/(1.44 Ω)
P₃₅₀ = 0.28 watt
The mass of M1 = 12 Daltons and it has a speed of v1 = 200 m/s. The mass of M2 = 4 Daltons. What was the total momentum of the system consisting of both masses before the collision (in Dalton meters per second, assume positive to the right and negative to the left)?
Answer:
The total momentum is [tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
Explanation:
The diagram illustration this system is shown on the first uploaded image (From physics animation)
From the question we are told that
The mass of the first object is [tex]M_1 = 12 \ Dalton[/tex]
The speed of the first mass is [tex]v_1 = 200 \ m/s[/tex]
The mass of the second object is [tex]M_2 = 4 \ Dalton[/tex]
The speed of the second object is assumed to be [tex]- v_2[/tex]
The total momentum of the system is the combined momentum of both object which is mathematically represented as
[tex]p__{T }} = M_1 v_1 + M_2 v_2[/tex]
substituting values
[tex]p__{T }} = 12 * 200 + 4 * (-v_2)[/tex]
[tex]p__{T }} =(2400 -4 v_2) \ Dalton \cdot m/s[/tex]
The disks you will be using in the lab have the following parameters. All disks are solid with outer radius R1 = 0 0632 m and inner radius R2 = 0 0079 m The masses of the disks are: MT starless steel = 1357kg, MBottom starless stee,ã¼1.344kg. Calculate the moments of inertia of the disks.
Answer:
MT disc I = 2,752 10-3 kg m²
MB disc I = 2,726 10⁻³ kg m²
Explanation:
The moment of inertia given by the expression
I = ∫ r² dm
for bodies with high symmetry it is tabulated
for a hollow disk it is
I = ½ M (R₁² + R₂²)
let's apply this equation to our case
disc MT = 1,357 kg
I = ½ 1,357 (0.0079² + 0.0632²)
I = 2,752 10-3 kg m²
disk MB = 1,344 kg
I = ½ 1,344 (0.0079² + 0.0632²)
I = 2,726 10⁻³ kg m²
Determine the magnitude of the force between two 11 m-long parallel wires separated by 0.033 m, both carrying 5.2 A in the same direction.
Answer:
[tex]F=1.8\times 10^{-3}\ N[/tex]
Explanation:
We have,
Length of wires is 11 m
Separation between wires is 0.033 m
Current in both the wires is 5.2 A
It is required to find the magnitude of force between two wires. The force between wires is given by :
[tex]F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 5.2\times 5.2\times 11}{2\pi \times 0.033}\\\\F=1.8\times 10^{-3}\ N[/tex]
So, the magnitude of force between wires is [tex]1.8\times 10^{-3}\ N[/tex]
Although electromagnetic waves can always be represented as either photons or waves, in the radio part of the spectrum we typically do not discuss photons (like we do in the visible) because they are at such a low energy. Nevertheless. they exist. Consider such a photon in a radio wave from an AM station has a 1545 kHz broadcast frequency.
Required:
a. What is the energy, in joules, of the photon?
b. What is the energy, in electron volts. of the photon?
Answer:
a. E = 1.02*10^-27 J
b. E = 6.39*10^-9eV
Explanation:
a. In order to calculate the energy of the radio photon, you use the following formula:
[tex]E=hf[/tex] (1)
h: Planck's constant = 6.626*10^-34 Js
f: frequency of the photon = 1545kHz = 1.545*10^6 Hz
Then, by replacing you obtain the energy of the photon:
[tex]E=(6.626*10^{-34}Js)(1.545*10^6s^{-1})=1.02*10^{-27}J[/tex]
b. In electron volts, the energy of the photon is:
[tex]E=1.02*10^{-27}J*\frac{6.242*10^{18}eV}{1J}=6.39*10^{-9}eV[/tex]