Answer:
Explanation:
Given
Mechanical Energy of Spring-Block system is 1 J
Maximum Amplitude is [tex]A=10 cm[/tex]
maximum speed [tex]v_{max}=1.2 m/s[/tex]
Suppose [tex]x=A\sin \omega t [/tex]be general equation of motion of spring-mass system
where A=max amplitude
[tex]\omega [/tex]=Natural frequency of oscillation
t=time
[tex]v_{max}=A\omega =1.2[/tex]
[tex]0.1\cdot \omega =1.2[/tex]
[tex]\omega =12 rad/s[/tex]
maximum kinetic Energy must be equal to total Mechanical Energy when spring is un deformed i.e. at starting Position
[tex]\frac{1}{2}mv_{max}^2=1[/tex]
[tex]m=\frac{2}{1.2^2}=\frac{2}{1.44}=1.38 kg[/tex]
Also [tex]\omega [/tex]is also given by
[tex]\omega =\sqrt{\frac{k}{m}}[/tex]
[tex]k=\omega ^2\cdot m[/tex]
where k= spring constant
[tex]k=12^2\cdot 1.38 [/tex]
[tex]k=200 N/m[/tex]
One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is 148 kg and its initial temperature is 22.3°C, how much energy must the water transfer to its surroundings in order to freeze completely?
The energy [tex]6.311 \times 10^{7} \mathrm{J}[/tex] is transferred to the surroundings from water in order to freeze completely.
Explanation:
Water will transfer to surrounding will come from cooling energy from 22.3°C to 0°C and then freezing energy is
[tex]\mathrm{E}_{\mathrm{t}}=\mathrm{E}_{\text {cooling }}+\mathrm{E}_{\text {freezing }}[/tex]
[tex]\mathrm{E}_{\mathrm{t}}=\mathrm{M}\left(\mathrm{C}_{\mathrm{w}} \Delta \mathrm{t}+\mathrm{L}_{\mathrm{f}}\right)[/tex]
We know that,
\mathrm{C}_{\mathrm{W}}=4190 \mathrm{J} / \mathrm{kgk}
[tex]\mathrm{L}_{\mathrm{f}}=333 \times 10^{3} \mathrm{J} / \mathrm{kg}[/tex]
As per given question,
M = 148 kg
[tex]\Delta \mathrm{t}=22.3^{\circ} \mathrm{C}[/tex]
Substitute the values in the above formula,
[tex]\mathrm{E}_{\mathrm{t}}=148\left(4190 \times 22.3+333 \times 10^{3}\right)[/tex]
[tex]E_{t}=148\left(93437+333 \times 10^{3}\right)[/tex]
[tex]E_{t}=148 \times 426437[/tex]
[tex]\mathrm{E}_{\mathrm{t}}=6.311 \times 10^{7} \mathrm{J}[/tex]
The energy [tex]6.311 \times 10^{7} \mathrm{J}[/tex] is transferred to the surroundings from water in order to freeze completely.
To calculate the energy required to freeze a tub of water completely, we need to account for the energy needed to both cool the water to its freezing point and then to convert it to ice. This involves the specific heat capacity of water and the latent heat of fusion.
We use the formula Q = mc extDelta T to calculate the energy to change the temperature of water to 0°C, where:
Q is the heat transfer,m is the mass of the water,c is the specific heat capacity of water (4.18 kJ/kg°C), andextDelta T is the temperature change (22.3°C to 0°C).Then, we calculate the energy needed for the phase change (freezing) using the formula Q = mLf, where:
Lf is the latent heat of fusion of water (333.55 kJ/kg).The total energy is the sum of the energy to cool the water and the energy to freeze it.
Calculations:
To cool to 0°C: Q = (148 kg) * (4.18 kJ/kg°C) * (22.3°C) = 13854 kJTo freeze: Q = (148 kg) * (333.55 kJ/kg) = 49365 kJTotal energy: Q_total = 13854 kJ + 49365 kJ = 63219 kJAt what speed, as a fraction of c, must a rocket travel on a journey to and from a distant star so that the astronauts age 15 years while the Mission Control workers on earth age 130 years?
Answer:
The velocity as a fraction of c is 0.986 c m/s
Solution:
As per the question:
Time measured by the astronaut, t = 15 yrs
Time measured in the frame of mission control, t' = 130 yrs
Now,
Using the formula of time dilation:
[tex]t' = \frac{t}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]
Substituting appropriate values in the above eqn:
[tex]130 = \frac{15}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]
[tex]\sqrt{1 - \frac{v^{2}}{c^{2}}}= \frac{15}{130}[/tex]
Squaring both the sides we get:
[tex]1 - \frac{v^{2}}{c^{2}}= (\frac{15}{130})^{2}[/tex]
[tex]\frac{v^{2}}{c^{2}} = 1 - (\frac{15}{130})^{2}[/tex]
v = 0.986 c m/s
The rocket must travel at about 99.45% of the speed of light for the astronauts aboard to age 15 years while those on Earth age 130 years, according to the theory of special relativity and the phenomenon called time dilation.
Explanation:The question asks about the speed a rocket needs to travel as a fraction of the speed of light (c) for the astronaut aboard to age 15 years while those back on Earth age 130 years - an application of time dilation in special relativity. The solution to this is in the theory of relativity proposed by Albert Einstein, which states that time passes at different speeds for people depending on their relative motion.
The time dilation formula is given by Δt = γΔτ, where Δτ is the 'proper time' experienced by the moving astronaut, Δt is the time experienced by the stationary observers on Earth, and γ is the Lorentz factor, defined as γ = 1 / sqrt(1 - (v^2/c^2)).
Here, Δt is 130 years, Δτ is 15 years, and we are asked to solve for v/c. Plugging in the given values and solving for v/c, we find that v/c = sqrt(1 - (15/130)^2), approximately equals to 0.9945 or 99.45% of the speed of light. Therefore, the rocket must travel at about 99.45% of the speed of light for the astronaut to age 15 years while those on Earth age 130 years.
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In a Compton scattering experiment, a beam of x-rays of wavelength 0.07 nm scatters of nearly free electrons. The scattered radiation is investigated as a function of the angle theta between it and the primary beam. The scattered radiation contains__________.
a) a component shifted in wavelength by an amount that decreases as the scattering angle increases.
b) a component shifted in wavelength by an amount that is independent of the scattering angle.
c) a component shifted in wavelength by an amount that is independent of the wavelength of the primary beam.
d) two components with wavelength shifted up and down by an equal amount.
e) no wavelength other than that of the primary beam.
Answer:
Option (e).
Explanation:
The Compton's equation is:
[tex] \lambda^{'} = \frac{h}{m_{e} c} \cdot (1 - Cos(\theta)) + \lambda = \lambda_{c} (1 - Cos(\theta)) + \lambda [/tex] (1)
where λ': is the wavelength scattered, λ: is the initial wavelength, h: is Planck's constant, [tex] m_{e} [/tex]: is the electron rest mass, c: speed of light, Θ: scattering angle (between λ and λ'), and [tex] \lambda_{c} [/tex]: is a constant known as the Compton wavelength of the electron = 0.00243 nm.
From equation (1), the scattered radiation is directly proportional to the scattering angle, having the maximum value when Θ=90°:
[tex] \theta = 90 ^{ \circ} \rightarrow \lambda^{'} = \lambda_{c} (1-Cos(90)) + \lambda = \lambda_{c} + \lambda [/tex]
And the minimum value when Θ=0°:
[tex] \theta = 0 ^{ \circ} \rightarrow \lambda^{'} = \lambda_{c} (1-Cos(0)) + \lambda = \lambda [/tex]
Hence, the options (a) and (b) are incorrect.
Similarly, we can see from equation (1) that the scattered radiation depends also on the wavelength of the primary beam and no wavelength other than that (since [tex] \lambda_{c} [/tex] is a constant), so the correct option is (e).
Have a nice day!
A 94.0 N grocery cart is pushed 17.6 m along an aisle by a shopper who exerts a constant horizontal force of 42.6 N. The acceleration of gravity is 9.81 m/s 2 . If all frictional forces are neglected and the cart starts from rest, what is the grocery cart’s final speed
Answer:
vf = 12.51 m/s
Explanation:
Newton's second law to the grocery cart:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
We define the x-axis in the direction parallel to the movement of the grocery cart and the y-axis in the direction perpendicular to it.
Forces acting on the grocery cart
W: Weight of the block : In vertical direction downward
N : Normal force : In vertical direction upward
F : horizontal force
Calculated of the mas of the grocery cart (m)
W = m*g
m = W/g
W = 94.0 N , g = 9.81 m/s²
m = 94/9.81
m = 9.58 Kg
Calculated of the acceleration of the grocery cart (a)
∑F = m*a
F = m*a
42.6 = (9.58)*a
a = (42.6) / (9.58)
a = 4.45 m/s²
Kinematics Equation of the grocery cart
Because the grocery cart moves with uniformly accelerated movement we apply the following formula to calculate its final speed :
vf²=v₀²+2*a*d Formula (2)
Where:
d:displacement (m)
v₀: initial speed (m/s)
vf: final speed (m/s)
a: acceleration (m/s²)
Data:
v₀ = 0
a = 4.45 m/s²
d = 17.6 m
We replace data in the formula (2) :
vf²=v₀²+2*a*d
vf² = 0+2*(4.45)*(17.6)
vf² = 156.64
[tex]v_{f} = \sqrt{156.64}[/tex]
vf = 12.51 m/s
(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h?
(b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction?
(c) What is the average force of friction if the hill has a slope of 2.5º above the horizontal?
Answer:
Explanation:
a ) Let the height achieved be h .
We shall apply law of conservation of mechanical energy.
1 /2 mv² = mgh
h = v² / 2g
v = 110 km/h
= 30.55 m /s
h = [tex]\frac{30.55\times30.55}{2\times9.8}[/tex]
h = 47.61 m
b )
Kinetic energy of car in the beginning
= 1/2 x 750 x (30.55)²
= 349988.43 J
Potential energy at 22 m height
= 750 x 9.8 x 22
= 161700 J
Energy lost due to frictional force
= 349988.43 - 161700
= 188288.43 J
c )
Distance covered along the slope
d = 22 / sin2.5
= 22 / 0.043619
d = 504.36 m
If F be average frictional force
work done by friction
F x d
= F x 504.36
so
F x 504.36 = 188288.43
F = 188288.43 / 504.36
= 373.32 N
Final answer:
Using the conservation of energy principle, the car's initial kinetic energy is converted into potential energy as it ascends the hill. The potential energy at the given height can be used to determine the thermal energy generated by friction. Calculating the average force of friction requires knowing this thermal energy and the distance traveled, considering the slope angle.
Explanation:
To calculate the height a car can coast up with negligible work done by friction, we can use the principle of conservation of energy. Specifically, the car's initial kinetic energy (due to its initial speed) will be converted into potential energy (due to gaining height) as it coasts uphill until it comes to a stop.
(a) The Height a Car Can Coast Up
Initial kinetic energy (KE) is given by the equation KE = \(\frac{1}{2}mv^2\), where m is the mass of the car and v is its speed. If we convert 110 km/h to meters per second (30.56 m/s), we can calculate the available kinetic energy. The potential energy (PE) at the height h is given by PE = mgh, where g is the acceleration due to gravity (9.81 m/s^2) and h is the height. Setting KE equal to PE, we can solve for h.
(b) Thermal Energy Generated by Friction
If the car actually reaches a height of 22.0 m, we can calculate the difference in the theoretical and actual potential energy to find the thermal energy due to friction. Subtracting the actual potential energy from the total initial kinetic energy gives us the energy lost to friction.
(c) Average Force of Friction
To calculate the average force of friction on a slope of 2.5°, we will use the energy lost to friction divided by the distance traveled along the slope and correct for the slope angle. This gives the component of the friction force that acts parallel to the hill's surface.
A canister is released from a helicopter 500 m above the ground. The canister is designed to withstand an impact speed of up to 100 m/s.
(a) Ignoring air resistance, find an equation of the height at any time t.
(b) Find the impact speed of the canister.
(c) Your answer to (b) should be less than 100 m/s.
Rather than just releasing it, let’s see if we could break it by throwing it down with an initial velocity. Re-do parts (a) and (b) with an initial velocity v0, then find the value of v0 required to break the canister.
Answer:
(a) h = 500 - 4.9t²
(b) 98.99 m/s
(c) h = 500 - Vot - 4.9t²
14.14 m/s
Explanation:
(a) You can use the equation of linear motion s = vi.t + 0.5gt²
(b) [tex]v_{f} ^{2} = v_{i} ^{2} +2ax\\v = \sqrt{0^{2} + 2*9.8*500 } = 98.99 m/s[/tex]
(c) [tex]v_{f} ^{2} = v_{i} ^{2} + 2ax\\v_{i} = \sqrt{100^{2} -2*9.8*500}=14.14 m/s[/tex]
The initial velocity mucg be greater than 14.14 m/s to break the cannister
A 96.1 kg man sits on the stern of a 6 m long boat. The prow of the boat touches the pier, but the boat isn’t tied. The man notices his mistake, stands up and walks to the boat’s prow, but by the time he reaches the prow, it’s moved 3.66 m away from the pier. Assuming no water resistance to the boat’s motion, calculate the boat’s mass (not counting the man). Answer in units of kg.
Info:
weight: 96.1 kg
Length: 6 m
moved: 3.66 m
Answer:
m L = m d + M d
m L − m d = M d
m (L − x) = M x
M = m (L − x) / x
M = 96.1 kg (6 m − 3.66 m) / 3.66 m
M = 61.44098361 kg
The boats mass.
A 96.1 kg of man sits on a stem of a 6-meter longboat. The boat ouches the pier and but is not tied. The man notices a mistake and walks to the prow of the boat he moves a distance of 3.66 meters away from the pier. Thus assuming the given information the bot offers no resistance.
The mass of the boat will be equal to 61.4 kg.
As all the given information states that the boat is not tied and is at a distance f 3.66 m. The mass of boat will be m L = m d + M dm L − m d = M d m (L − x) = M xM = m (L − x) / x then the M = 96.1 kg (6 m − 3.66 m) / 3.66 mHence the M = 61.44098361 kg.Learn more about the man sits on the stern.
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You need to move a 105 kg sofa to a different location in the room. It takes a force of 102 N to start it moving. What is the coefficient of static friction between the sofa and the carpet? (use g = to about -10 m/s/s)
Answer:
0.0971
Explanation:
we know that
[tex]f max = kn[/tex]
that fmax is maximum of static friction , k is coefficient of friction and n is surface reaction force
so we know that from newtons second law
mg=n
so
kmg = 102
k = 102/mg = 102/(10*105) = 0.0971
Final answer:
The coefficient of static friction (μs) between the sofa and carpet is calculated by dividing the force required to start the sofa moving by the normal force, yielding μs = 102 N / 1050 N.
Explanation:
The coefficient of static friction μs is the ratio of the force of static friction Fs to the normal force N. It can be calculated using the equation μs = Fs / N, where Fs is the force required to start moving the object and N is the weight of the object acting perpendicular to the surface. Here, the normal force is equivalent to the weight of the sofa, which is the mass of the sofa multiplied by the acceleration due to gravity (g). With a mass of 105 kg and g approximated as -10 m/s2, the normal force is N = 105 kg × 10 m/s2 = 1050 N. The provided force to start the sofa moving is 102 N, so the coefficient of static friction is μs = 102 N / 1050 N.
The neurons that select a particular motor program are the __________.
Answer:
upper motor neurons in the premotor cortex
Explanation:
Motor neurons are a type of nervous system cells that are located in the brain and in the spinal cord. They have the function of producing the stimuli that cause the contraction of the different muscle groups of the organism. They are therefore essential for daily activities that require muscle contraction: walking, talking, moving hands and in general all body movements.
A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor at 25◦C. To achieve the desired amount, we cool the tank while we fill it slowly, keeping the tank and its content at 30◦C. Find the final pressure to reach before closing the valve and the heat transfer.
Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.
Final answer:
The question is about the angles of maxima and minima in a double-slit interference pattern for two identical loudspeakers placed 1.00 m apart.
Explanation:
The distance between two identical loudspeakers placed 1.00 m apart is given as 'd'.
In the given situation, the listener stands 4.00 m from the wall directly in front of one of the speakers. This distance can be considered as 'L'.
The formula to calculate the distance between adjacent maxima and minima in a double-slit interference pattern is:
d*sin(theta) = m*lambda
- Where 'd' is the distance between the speakers
- 'theta' is the angle at which the maxima or minima occur
- 'm' is the order of the maxima or minima
- 'lambda' is the wavelength of the sound
Applying this formula, we can calculate the angles at which the maxima and minima occur for the given situation.
The phase difference between the waves reaching the observer is approximately 0.66 radians. The frequency closest to 300 Hz for minimal sound due to destructive interference is approximately 286 Hz.
To solve the problem, we need to find the phase difference and the frequency closest to 300 Hz for minimal sound.
a.) The speed of sound, v, is typically about 343 m/s in air. Using the formula for wavelength, λ = v/f, we get:
λ = 343 m/s / 300 Hz λ = 1.143 mFor the listener standing 4.00 m from one speaker and 4.12 m from the second speaker, the difference in the path length, ΔL, is:
ΔL = sqrt((4.00 m)² + (1.00 m)²) - 4.00 m ΔL = 4.12 m - 4.00 m = 0.12 mThe phase difference, φ, is given by:
φ = (2π/λ) * ΔL φ = (2π/1.143 m) * 0.12 m φ ≈ 0.66 radiansb.) For destructive interference, the path difference should be half-integer multiples of the wavelength:
(2n + 1)λ/2 = ΔLSimplifying, we find the frequency:
0.12 m = (2n + 1) * (343 m/s) / (2f)Solving for f when n = 0 (closest to 300 Hz):
f = 343 m/s / (2 * 0.12 m) f ≈ 286 HzThus, the frequency closest to 300 Hz for minimal sound is approximately 286 Hz.
complete question:
Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz.
(a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.)
rad
(b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?in Hz
Draw a free-body diagram of the rod ab. Assume the contact surface at b is smooth.
Answer:
See attachment
Explanation:
A person is using a rope to lower a 5.0-n bucket into a well with a constant speed of 2.0 m/s. What is the magnitude of the force exerted by the rope on the bucket?
Answer:
5 N
Explanation:
The bucket is moving at a constant speed of 2m/s Therefore F=ma is 0 N for this to be correct the magnitude of the force exerted by the rope must be equal to the weight of the bucket which is 5 N
Answer:
Magnitude of force on the rope=5N
Explanation:
The bucket is lowered by a constant speed,therefore the tension on the rope must be equal to the weight of the bucket.
Sound waves are created by _____.
a. the compression of air molecules
b. the rarefaction of air molecules.
c. undulating displacement of air molecules caused by pressure changes.
d. None of the answers is correct.
Answer:
C. undulating displacement of air molecules caused by pressure changes.
Explanation:
Sound is a mechanical wave arising from the movement back and forth of the media objects through which the sound wave travels.If a sound wave moves through air from left to right, then particles of air will move both to the right and to the left as the sound wave's energy passes through it.
The sound wave is formed by air molecules that are displaced due to pressure changes.
Therefore the answer is C.
Sound waves are caused by the undulating displacement of air molecules due to pressure changes. Hence, option(c) is correct.
1. When an object like a speaker cone vibrates, it alternately compresses and expands the air around it.
2. The regions where air molecules are pushed together are called compressions, and the regions where they are spread apart are called rarefactions.
3. Sound waves travel through the air as longitudinal waves, consisting of alternating high-pressure (compression) and low-pressure (rarefaction) regions.
4. These pressure changes propagate at a speed of approximately 340 m/s in air, creating the disturbance known as a sound wave.
Compression: Area of high pressure where molecules are close together.Rarefaction: Area of low pressure where molecules are spread apart.A(n) _____ generator collects charges on a charge-carrying belt, transfers the charges onto a large metal sphere, and discharges the charges, thereby releasing up to 5,000,000 volts of electricity or more.
Answer:
Explanation:
This generator is called Vande Graff generator.
It collects the charge from the belts and accumulate on a large sphere.
It is used to accelerate the charge particles and based on the principle of corona discharge and charge resides on the outer surface.
The sides of a composite-cone volcano may be interrupted with which feature?
Answer:
fumaroles or smokers
Explanation:
Fumaroles: they are gaseous emissions from lavas in craters at more or less elevated temperatures. Its composition varies according to the temperature of the lavas, in such a way that it changes since the fumaroles appear until their extinction.
Water flows in the form of steam continuously through the cracks of a volcano or volcanic surface because its temperature exceeds 100 °C.
Groups of fumaroles
- Dry fumaroles: They are those that come off the melting lava, in the vicinity of the crater. Its temperature is higher than 500oC. They are mainly composed of sodium, potassium and sulfur dioxide and carbon dioxide.
- Acid fumaroles: with temperatures between 300oC and 400oC, they are constituted by a large amount of water vapor, with hydrochloric acid and sulphurous anhydride.
- Alkaline fumaroles: Temperature close to 100oC, contain water vapor with hydrogen sulfide and ammonium chloride.
7. An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where the Earth’s horizontal magnetic field is 5.5×10−5T. What is the maximum possible emf induced in the antenna due to this motion?
The maximum possible emf induced in the antenna due to the motion of the automobile can be calculated using Faraday's law of electromagnetic induction. Plugging in the given values, the maximum possible emf is calculated as 2.4 x 10^-8 V.
Explanation:The maximum possible emf induced in the antenna due to the motion of the automobile can be calculated using Faraday's law of electromagnetic induction. The equation for calculating the emf is given by μoθlvB, where μo is the permeability of free space, l is the length of the antenna, v is the velocity of the automobile, and B is the magnetic field strength. Plugging in the given values, we have:
μo = 4π x 10^-7 Tm/A
l = 1.0 m
v = 100.0 km/h = 27.78 m/s
B = 5.5 x 10^-5 T
Using these values, we can calculate the maximum possible emf induced in the antenna as:
emf = (μoθlvB) = (4π x 10^-7 Tm/A)(1.0 m)(27.78 m/s)(5.5 x 10^-5 T) = 2.4 x 10^-8 V
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The maximum possible emf induced in the 1.0 m long antenna moving at 100.0 km/h in a 5.5×10⁻⁵ T magnetic field is approximately 1.53 mV. The calculation involves using the motional emf formula: emf = vBL. Converting the velocity to meters per second and plugging in the values gives the result.
Motional EMF Calculation
The question is about finding the electromotive force (emf) induced in a radio antenna moving through the Earth's magnetic field. This situation involves motional emf, which can be determined using the formula:
emf = vBL
where:
v = velocity of the antenna = 100.0 km/h = 27.78 m/s (converted to meters per second)B = magnetic field strength = 5.5 × 10⁻⁵ TL = length of the antenna = 1.0 mNow, substituting these values into the formula:
emf = 27.78 m/s × 5.5 × 10⁻⁵ T × 1.0 m
emf ≈ 1.53 × 10⁻³ V
Therefore, the maximum possible induced emf in the antenna due to its motion is approximately 1.53 mV.
The red light from a helium-neon laser has a wavelength of 727.7 nm in air. Find the speed, wavelength, and frequency of helium-neon laser light in air, water, and glass. (The glass has an index of refraction equal to 1.50.)
Answer:
speed 2.997 10⁸, 2,254 10⁸, 1,999 10⁸ m / s
[tex]\lambda_{n}[/tex] 727.7, 547, 485 nm
Explanation:
The index of refraction is defined as the relationship between the speed of light in a vacuum and in a material medium
n = c / v
Let's calculate the speed of light in the media
Air
The refractive index is very close to that of the vacuum n = 1,00029
In most experiments they are considered equal
v = c/ n
v = 2,998 10⁸ / 1,00029
v = 2,997 10⁸ m / s
Water
n = 1.33
v = 2.998 10⁸ / 1.33
v = 2,254 10⁸ m / s
Glass
n = 1.50
v = 2,998 10⁸ / 1,50
v = 1,999 10⁸ m / s
Frequency and wavelength are related by the equation
c = λ f
When a beam with a given frequency hits excites the electrons of the material and induces forced oscillations, which has the same frequency of the incident, so the frequency of the beam does not change when passing from one medium to the other.
As speed changes the only way that equality is maintained is that the wavelength changes
[tex]\lambda_{n}[/tex] = λ₀ / n
Air
[tex]\lambda_{n}[/tex] =727.7 nm
Water
[tex]\lambda_{n}[/tex] = 727.7 / 1.33
[tex]\lambda_{n}[/tex] = 547.14 nm
Glass
[tex]\lambda_{n}[/tex] = 727.7 / 1.50
[tex]\lambda_{n}[/tex] = 485.13 nm
Which of the following is true of Giovanni Pierluigi da Palestrina’s Pope Marcellus Mass?
a. It demonstrates the Protestant practice of simple congregational singing
b. The text is in Palestrina’s native language of Italian
c. All the parts were originally sung by men and boys
d. It was banned by the Council of Trent 1
Answer: c is true
Explanation: the boy choirstars sang the soprano and alto. Although Palestrina choir is made up of six voice part of soprano, alto, tenor, baritone and bass and the choir sings without instruments.
In the case of Giovanni Pierluigi da Palestrina’s Pope Marcellus Mass, the correct statement is that all the parts were originally sung by men and boys. It was not written in Italian, does not represent Protestant music tactics and was not banned by the Council of Trent.
Explanation:Out of the provided options, the statement that is true of Giovanni Pierluigi da Palestrina’s Pope Marcellus Mass is that all the parts were originally sung by men and boys (option c). The work is a significant one in the history of polyphonic choral music and it marked a departure from the complex style of the early Renaissance period. Instead of Italian, the mass is actually written in Latin, which was the language conventionally used in the liturgy of the Catholic Church (thus discrediting option b). Unlike Protestant music of the time (which negates option a), it does not emphasize congregational singing but instead focuses on the choir. It was also never banned by the Council of Trent. (disproving option d)
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A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is parallel to the ground. The mass of the stuntman is 119 kg, and the coefficient of kinetic friction between the road and him is 0.697. Find the tension in the cable.
Answer:
T = 812.8414 N
Explanation:
Using the law of newton we found the vertical(y) and horizontal(x) forces as:
∑[tex]F_x[/tex] = T - [tex]F_k[/tex] = ma
Where T is the tension, [tex]F_k[/tex] is the friction force, m is the mass of the stuntman and a is the aceleration of the stuntman.
but a is equal to 0 because he is moving at a constant velocity, so:
T - [tex]F_k[/tex] = 0
T = [tex]F_k[/tex]
Also,
[tex]F_k[/tex] = [tex]U_kN[/tex]
where [tex]U_k[/tex] is the coefficient of kinetic friction and N is the normal force.
For find N we use:
∑[tex]F_y =[/tex] N - mg = 0
N = mg
N = (119)(9.8)
N = 1166.2
Finally we solve for T as:
T = [tex]U_kN[/tex]
T = (0.697)(1166.2)
T = 812.8414 N
The height of a cylinder is decreasing at a constant rate of 8 inches per minute, and the volume is decreasing at a rate of 161 cubic inches per minute. At the instant when the height of the cylinder is 66 inches and the volume is 919 cubic inches, what is the rate of change of the radius? The volume of a cylinder can be found with the equation V=\pi r^2 h.V=πr 2 h. Round your answer to three decimal places.
Answer:
0.056 inches per minute
Explanation:
dh/dt = 8 inches per minute
dV/dt = 161 cubic inch per minute
h = 66 inches
V = 919 cubic inch
dr/dt = ? rate of change of radius
The volume of cylinder is given by
V = πr²h
where, r be the radius of cylinder
Differentiate both sides with respect to t
dV/dt = πr² x dh/dt + πh x 2r dr/dt .... (1)
When h = 66 inches, V = 919 cubic inches
So, 919 = 3.14 x r² x 66
r = 2.11 inch
Substitute the values in equation (1)
161 = 3.14 x 2.11 x 2.11 x 8 + 3.14 x 66 x 2 x 2.11 x dr/dt
dr/dt = 0.056 inches per minute
The rate of change of the radius of the cylinder, given the provided rates of change of volume and height and current volume and height, is approximately -0.123 inches/minute.
Explanation:
This problem is a classic example of related rates in calculus, specifically focusing on the cylinder. Given the volume of a cylinder is V = πr^2h, we know that the rate at which the volume is changing (dV/dt) is related to the rate at which the radius is changing (dr/dt) and the rate at which the height is changing (dh/dt). In this problem, we are given dV/dt (-161 cubic inches/minute) and dh/dt (-8 inches/minute). Using these given rates, the current volume and height, we can differentiate the volume formula with respect to time to find dr/dt.
Applying the chain rule, we get dV/dt = πr^2 dh/dt + 2πrh dr/dt. Substituting all given values and solving for dr/dt, we find that the rate of change of the radius when the height of the cylinder is 66 inches comes out to be approximately -0.123 inches/minute (rounded to three decimal places).
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You are asked to construct a mobile with four equal m = 141 kg masses, and three light rods of negligible mass and equal lengths. The rods are of length 55 cm. (a) At what location on the level 1 rod should the free end of rod 2 be attached?
Answer:
The free end must be attached at a distance of 27.5 cm
Solution:
Mass, m = 141 kg
Length of the rods, L= 55 cm
Now,
As clear from fig. 1:
The free end of the rod 2 must be attached at:
F = 2 W
[tex]WL = W(55 - L)[/tex]
2L = 55
L = 27.5 cm
What are the major steps of solar system formation in the correct order?
Answer:(1) Pre-solar nebula
(2). Planet formation
Explanation:
Nebular hypothesis says that the Solar System formed from the gravitational collapse of a fragment of a giant molecular cloud.
One of the collapsing fragments called the pre-solar nebula formed what became the Solar System.
The planet can also be formed by accretion. Accretion is a process in which planets began as dust grains in orbit around the central protostar.
When the terrestrial planets were forming, they remained immersed in a disk of gas and dust.
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance of 2.7×1011 m from the center of the sun, what is its speed when at a distance of 4.8×1010 m?
Answer:
v₂ = 7.6 x 10⁴ m/s
Explanation:
given,
speed of comet(v₁) = 1.6 x 10⁴ m/s
distance (d₁)= 2.7 x 10¹¹ m
to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m
v₂ = ?
speed of planet can be determine using conservation of energy
K.E₁ + P.E₁ = K.E₂ + P.E₂
[tex]\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}[/tex]
[tex]\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}[/tex]
[tex]v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}[/tex]
[tex]v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}[/tex]
[tex]v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}[/tex]
v₂ = 7.6 x 10⁴ m/s
The speed when at a distance of [tex]4.8*10^{10} m[/tex] is mathematically given as
vf = 6.92 * 10^(4) m/s
What is the speed when at a distance of [tex]4.8*10^{10} m[/tex]?Generally, the equation for the conservation of energy is mathematically given as
E = (1/2)mv^2 - GmM/r
Where
E_i = E_f
Hence
(1/2)mv_i^2 - [tex]\frac{GmM}{(r_i)}[/tex] = (1/2)mv_f^2 - [tex]\frac{GmM}{(r_f)}[/tex]
[tex]v_f = \sqrt{[(v_i)^2 + 2MG((1/r_f) - (1/r_i))]}[/tex]
Therefore
[tex]v_f = \sqrt{(2.1*10^4)^2 + 2(1.9891 * 10^{31})*(6.67 * 10^{-11})(\frac{1}{4.9} * 10^{10}) - (\frac{1}{2.5} * 10^{11}))} *20.408 *10^{12}[/tex]
[tex]v_f = \sqrt{(441000000) + 435.38 * 10^{7}}[/tex]
vf = 6.92 * 10^(4) m/s
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In a double-slit interference experiment, a special lamp emitting yellow light from heated sodium atoms is used to produce an interference pattern on a screen located 1.49 m from a pair of slits separated by 0.12 mm. If the distance between adjacent bright regions in the resulting pattern is 7.32 mm, what is the wavelength (in nm) of the sodium light?
Answer:
λ=589nm
Explanation:
The wavelength of the sodium light can be calculated using the next equation:
[tex] \lambda = \frac{\Lambda D}{L} [/tex]
where Λ: is the distance between the adjacent bright regions, λ: is the wavelength of the sodium light, L: is the distance between slits and screen, and D: is the distance between slits.
[tex] \lambda = \frac{7.32mm \cdot 0.12mm}{1490mm} = 5.89\cdot 10^{-4}mm = 589nm [/tex]
Therefore, the λ of the sodium light is 589 nm.
I hope it helps you!
The bicycle wheel is spinning counterclockwise when viewed from above (so as you view the picture, the right hand side of the wheel is moving away from you). The man and stool are initially at rest. If the man now rotates the wheel’s axis 180˚ so that his right hand is above the wheel, the wheel/man/stool system will:__________
Answer:
a) rotate counterclockwise as viewed from above.
Explanation:
The text tells us it is spinning counterclockwise.
A nonuniform, 80.0-g, meterstick balances when the support is placed at the 51.0-cm mark. At what location on the meterstick should a 5.00-g tack be placed so that the stick will balance at the 50.0 cm mark?
Answer:34 cm
Explanation:
Given
mass of meter stick m=80 gm
stick is balanced when support is placed at 51 cm mark
Let us take 5 gm tack is placed at x cm on meter stick so that balancing occurs at x=50 cm mark
balancing torque
[tex]80\times 10^{-3}(51-50)=5\times 10^{-3}(50-x)[/tex]
[tex]80=5(50-x)[/tex]
[tex]80=250-5x[/tex]
[tex]5x=170[/tex]
[tex]x=\frac{170}{5}[/tex]
[tex]x=34 cm[/tex]
To balance the nonuniform meterstick, the 5.00-g tack should be placed at approximately 50.4 cm from the pivot point.
Explanation:To find the location on the meterstick where the 5.00-g tack should be placed so that the stick will balance at the 50.0 cm mark, we can use the principle of torque balance.
The torque balance equation is given by:
Torque(counter-clockwise) = Torque(clockwise).
In this case, the torque is equal to the force multiplied by the distance from the pivot point. The force exerted by the 5.00-g tack can be calculated using the formula:
Force = mass * acceleration due to gravity.
By substituting the known values into the equation and solving for the distance, we can determine where the tack should be placed on the meterstick.
In this case, the tack should be placed at approximately 50.4 cm from the pivot point.
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A typical car has 16 L of liquid coolant circulating at a temperature of 95 ∘C through the engine's cooling system. Assume that, in this normal condition, the coolant completely fills the 2.0 −L volume of the aluminum radiator and the 14.0 −L internal cavities within the steel engine. When a car overheats, the radiator, engine, and coolant expand and a small reservoir connected to the radiator catches any resultant coolant overflow. Estimate how much coolant overflows to the reservoir if the system is heated from 95 ∘C to 105 ∘C. Model the radiator and engine as hollow shells of aluminum and steel, respectively. The coefficient of volume expansion for coolant is β=410⋅10−6/C∘
Answer:
[tex]\Delta V=0.0592\ L[/tex]
Explanation:
Given:
Initial temperature of the coolant, [tex]T_f=95^{\circ}C[/tex]final temperature of the coolant, [tex]T_f=105^{\circ}C[/tex]total volume of the coolant, [tex]V_c=16\ L[/tex]coefficient of volume expansion for coolant, [tex]\beta_c=410\times 10^{-6}\ ^{\circ}C^{-1}[/tex]volume of Al radiator, [tex]V_a=2\ L[/tex]volume of steel radiator, [tex]V_s=14\ L[/tex]We have:
coefficient of volume expansion for Aluminium, [tex]\beta_a=75\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
coefficient of volume expansion for steel, [tex]\beta_a=35\times 10^{-6}\ ^{\circ}C^{-1}[/tex]
Now, change in volume of the coolant after temperature rises:
[tex]\Delta V_c=V_c.\beta_c.\Delta T[/tex]
[tex]\Delta V_c=16\times 410\times 10^{-6}\times (105-95)[/tex]
[tex]\Delta V_c=0.0656\ L[/tex]
Now, volumetric expansion in Aluminium radiant:
[tex]\Delta V_a=V_a.\beta_a.\Delta T[/tex]
[tex]\Delta V_a=2\times 75\times 10^{-6}\times (105-95)[/tex]
[tex]\Delta V_a=0.0015\ L[/tex]
Now, volumetric expansion in steel radiant:
[tex]\Delta V_s=V_s.\beta_s.\Delta T[/tex]
[tex]\Delta V_s=14\times 35\times 10^{-6}\times (105-95)[/tex]
[tex]\Delta V_s=0.0049\ L[/tex]
∴Total extra accommodation volume created after the expansion:
[tex]V_X=\Delta V_s+\Delta V_a[/tex]
[tex]V_X=0.0049+0.0015[/tex]
[tex]V_X=0.0064\ L[/tex]
Hence, the volume that will overflow into the small reservoir will be the volume of coolant that will be extra after the expanded accommodation in the radiator.
[tex]\Delta V=\Delta V_c-\Delta V_X[/tex]
[tex]\Delta V=0.0656-0.0064[/tex]
[tex]\Delta V=0.0592\ L[/tex]
When the system is heated from 95 °C to 105 °C, approximately 0.066 liters of coolant would overflow to the reservoir, using the coefficient of volume expansion for the coolant.
Explanation:To estimate how much coolant overflows due to thermal expansion when the system is heated from 95 °C to 105 °C, we use the coefficient of volume expansion for coolant (given as β = 410×10-6/°C). Calculate the change in volume (ΔV) with the formula:
ΔV = βVΔ0ΔT
where:
β is the coefficient of volume expansionVΔ0 is the original volume of the coolant (16 L)ΔT is the change in temperature (105 °C - 95 °C = 10 °C)Plugging in the values, we get:
ΔV = (410×10-6/°C)(16 L)(10 °C)
ΔV = 0.066 L
Therefore, approximately 0.066 liters of coolant would overflow to the reservoir. Since metals like aluminum and steel also expand, this is an approximation as we are not factoring in the expansion of the radiator and engine housing.
In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 80 kg, and the collision on the floor lasts 0.081 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?
Answer:
Impulse, J = 250.4 kgm/s, Avg Force F=3091.4 N
Explanation:
Since we know that impulse is the change in momentum i.e. Δp and Δp[tex]=mv[/tex], therefore, to calculate the velocity we perform:
As the person has fallen by a 0.50m height, its potential energy changes into kinetic energy, therefore,
K.E.=P.E.
[tex]\frac{1}2}mv^{2}[/tex]=[tex]mgh[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2*9.8*0.50}[/tex]
[tex]v=3.13ms^{-1}[/tex]
(a) Impulse [tex]J[/tex] = Δp[tex]=mv[/tex]
[tex]J= 80*3.13[/tex]
[tex]J = 250.4 kgms^{-1}[/tex]
(b) Avg Force F = Δp/Δt
[tex]F=\frac{250.4}{0.081}[/tex]
[tex]F=3091.4[/tex] N
A paper filled capacitor is charged to a potential difference of 2.1 V and then disconnected from the charging circuit. The dielectric constant of the paper is 3.7. Keeping the plates insulated, the paper filling is withdrawn, allowing air to fill the capacitor instead. Find the resulting potential difference off the capacitor.
(volts)
While continuing to keep the capacitor's plates insulated, an unknown substance is inserted between them. The plates then attain a potential difference that is 0.59 times the original potential difference (when paper filled the capacitor). What is the substances dielectric constant?
Answer:
Part a)
[tex]V' = 7.77 Volts[/tex]
Part b)
[tex]k' = 6.27 [/tex]
Explanation:
As we know that capacitor plate is connected across 2.1 V and after charging it is disconnected from the battery
So here we can say that charge on the plates will remain conserved
So we will have
[tex]Q = kC(2.1)[/tex]
now dielectric is removed between the plates of capacitor
so new potential difference between the plates
[tex]V' = \frac{Q}{C'}[/tex]
[tex]V' = \frac{kC(2.1)}{C}[/tex]
[tex]V' = 3.7 \times 2.1 [/tex]
[tex]V' = 7.77 Volts[/tex]
Part b)
Now the capacitor plates are again isolated and unknown dielectric is inserted between the plates
So again charge is same so potential difference is given as
[tex]V" = \frac{Q}{k'C}[/tex]
[tex]0.59 V = \frac{kCV}{k'C}[/tex]
[tex]0.59 = \frac{3.7}{k'}[/tex]
[tex]k' = 6.27 [/tex]
Final answer:
The resulting potential difference of the capacitor after the paper is withdrawn is 7.77V. The dielectric constant of the unknown substance, where the potential decreased to 0.59 times the original, is approximately 6.27.
Explanation:
The potential difference across a capacitor is inversely proportional to the dielectric constant of the material between the plates. When a paper dielectric with a constant of 3.7 is removed from a charged capacitor, the potential difference increases because the dielectric constant of air is nearly 1.
Find the resulting potential difference of the capacitor
The initial potential difference (V1) with paper is 2.1V, and the dielectric constant for paper (k1) is 3.7. When the paper is removed, the dielectric constant becomes that of air, which is approximately 1 (k2). For an isolated capacitor, the charge (Q) remains constant, so we can use the relation that the product of the potential difference (V) and the dielectric constant (κ) is a constant (Q=CV and C=κC0). Therefore, V1k1 = V2k2. By substituting the known values: 2.1V * 3.7 = V2 * 1, we find that V2 = 2.1V * 3.7 = 7.77V.
Find the dielectric constant of the inserted unknown substance
If the potential difference with the new substance is 0.59 times the initial potential (when paper filled the capacitor), then V3 = 0.59 * 2.1V. The dielectric constant of the new substance (k3) can be found using the same principle where V1k1 = V3k3. With V3 = 0.59 * 2.1 = 1.239V, we get 2.1V * 3.7 = 1.239V * k3, leading to k3 = (2.1V * 3.7) / 1.239V ≈ 6.27.