Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
A damped oscillator is released from rest with an initial displacement of 10.00 cm. At the end of the first complete oscillation, the displacement reaches 9.05 cm. When 4 more oscillations are completed, what is the displacement reached
Answer:
The displacement is [tex]A_r = 6.071 \ cm[/tex]
Explanation:
From the question we are told that
The initial displacement is [tex]A_o = 10 \ cm[/tex]
The displacement at the end of first oscillation is [tex]A_d = 9.05 \ cm[/tex]
Generally the damping constant of this damped oscillator is mathematically represented as
[tex]\eta = \frac{A_d}{A_o}[/tex]
substituting values
[tex]\eta = \frac{9.05}{10}[/tex]
[tex]\eta = 0.905[/tex]
The displacement after 4 more oscillation is mathematically represented as
[tex]A_r = \eta^4 * A_d[/tex]
substituting values
[tex]A_r = (0.905)^4 * (9.05)[/tex]
[tex]A_r = 6.071 \ cm[/tex]
Answer:
Displacement reached is 6.0708 cm
Explanation:
Formula for damping Constant "C"
[tex]C^n=\frac{A_2}{A_1}[/tex] where n=1,2,3,........n
Where:
[tex]A_2[/tex] is the displacement after first oscillation
[tex]A_1\\[/tex] is the initial Displacement
[tex]A_1=10\ cm\\A_2=9.05\ cm\\[/tex]
In our case, n=1.
[tex]C=\frac{9.05}{10}\\C=0.905[/tex]
After 4 more oscillation, n=4:
[tex]C^4=\frac{A_6}{A_2}[/tex]
Where:
[tex]A_6[/tex] is the final Displacement after 4 more oscillations.
[tex]A_6=(0.905)^4*(9.05)\\A_6=6.0708\ cm[/tex]
Displacement reached is 6.0708 cm
Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength is [tex]\lambda = 1805 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]
The distance of the screen is D = 3.0 m
The fringe width is [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]
Generally the fringe width for a bright fringe is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
=> [tex]d = \frac{ \lambda * D }{ y }[/tex]
=> [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]
=> [tex]d = 0.000373 \ m[/tex]
Generally the fringe width for a dark fringe is mathematically represented as
[tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
Here m = 0 for first order dark fringe
So
[tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
looking at which we see that [tex]y_d = y[/tex]
[tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]
=> [tex]\lambda = 1805 *10^{-9} \ m[/tex]
=> [tex]\lambda = 1805 nm[/tex]
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
A projectile is shot from the edge of a cliff 80 m above ground level with an initial speed of 60 m/sec at an angle of 30° with the horizontal. Determine the time taken by the projectile to hit the ground below.
Answer:
8 seconds
Explanation:
Answer:
Explanation:
Going up
Time taken to reach maximum height= usin∅/g
=3 secs
Maximum height= H+[(usin∅)²/2g]
=80+[(60sin30)²/20]
=125 meters
Coming Down
Maximum height= ½gt²
125= ½(10)(t²)
t=5 secs
What is the displacement of the object after 3 seconds?
Answer:
3 meters
Explanation:
An object is placed in a fluid and then released. Assume that the object either floats to the surface (settling so that the object is partly above and partly below the fluid surface) or sinks to the bottom. Answer the following question:
The magnitude of the buoyant force is equal to the weight of fluid displaced by the object. Under what circumstances is this statement true?
a. for every object submerged partially or completely in a fluid
b. only for an object that is floating
c. only for an object that is fully submerged and is sinking.
d. for no object submerged in a fluid
Answer:
Answer:
A. for every object submerged partially or completely in a fluid
Explanation:
This is following Archimedes principle which states that the upward buoyant force that is exerted on a body immersed in a fluid, whether FULLYor PARTIALLY submerged, is equal to the weight of the fluid that the body displaces.
Explanation:
In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit
Answer:
Impedance increases for frequencies below resonance and decreases for the frequencies above resonance
Explanation:
See attached file
Explanation:
The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?
Explanation:
F = ma
85 N = m (15 m/s²)
m ≈ 5.7 kg
What is 3/4 of 12 and 24
Answer:
3/4 of 12 = 16
3/4 of 24 = 32
Those are the answers based on how your question sounded
Explanation:
Answer:
27
Explanation:
3/4 of (12 and 24)
of means ×
and means +
therefore,3/4× (12+24)
3/4×(36)
3×9
27
A 25 cm diameter circular saw blade spins at 3500 rpm. How fast would you have to push a straight hand saw to have the teeth move through the wood at the same rate as the circular saw teeth
Answer:
The answer is "45.79 m/s"
Explanation:
Given values:
diameter= 25 cm
w= 3500 rpm
Formula:
[tex]\boxed{v=w \times r} \ \ \ \ \ \ _{where} \ \ \ w = \frac{rad}{s} \ \ \ and \ \ \ r = meters[/tex]
Calculating r:
[tex]r= \frac{diameter}{2}[/tex]
[tex]=\frac{25}{2}\\\\=12.5 \ cm[/tex]
converting value into meters: [tex]12.5 \times 10^{-2} \ \ meter[/tex]
calculating w:
[tex]w= diameter \times \frac{2\pi}{60}\\[/tex]
[tex]= 3500 \times \frac{2\times 3.14}{60}\\\\= 3500 \times \frac{2\times 314}{6000}\\\\= 35 \times \frac{314}{30}\\\\= 35 \times \frac{314}{30}\\\\=\frac{10990}{30}\\\\=\frac{1099}{3}\\\\=366.33[/tex]
w= 366.33 [tex]\ \ \frac{rad}{s}[/tex]
Calculating v:
[tex]v= w\times r\\[/tex]
[tex]= 366.33 \times 12.5 \times 10^{-2}\\\\= 366.33 \times 12.5 \times 10^{-2}\\\\= 4579.125 \times 10^{-2}\\\\\boxed{=45.79 \ \ \frac{m}{s}}[/tex]
Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
Answer:
T = 3.14 hours
Explanation:
We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
We know that the radius of Mars is 3,389.5 km.
So, r = 1,787 + 3,389.5 = 5176.5 km
Using Kepler's law,
[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]
M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]
So,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]
or
T = 3.14 hours
So, the orbital period is 3.14 hours
At TTT = 14 ∘C∘C, how long must an open organ pipe be to have a fundamental frequency of 262 HzHz ? The speed of sound in air is v≈(331+0.60T)m/sv≈(331+0.60T)m/s, where TT is the temperature in ∘C∘C.
Answer:
Length of pipe organ(L) = 0647 m (Approx)
Explanation:
Given:
Temperature (T) = 14°C
Fundamental frequency (F) = 262 Hz
Speed of sound (v) = 331 + 0.60(T) m/s
Find:
Length of pipe organ(L)
Computation:
Speed of sound (v) = 331 + 0.60(14) m/s
Speed of sound (v) = 339.4
Length of pipe organ(L) = Speed of sound (v) / 2(Fundamental frequency)
Length of pipe organ(L) = 339.4 / 2 (262)
Length of pipe organ(L) = 0647 m (Approx)
Heat and thermodynamics Numerical
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
A radar pulse returns 3.0 x 10-4 seconds after it is sent out, having been reflected by an object. What is the distance between the radar antenna and the object
Answer:
The distance is [tex]D = 45000 \ m[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 3.0 *10^{-4 } \ s[/tex]
Generally the speed of the radar is equal to the speed of light and this has a value
[tex]c = 3.0*10^{8} \ m /s[/tex]
Now the distance covered by the to and fro movement of the radar is mathematically evaluated as
[tex]d = c * t[/tex]
=> [tex]d = 3.0*10^{8} * 3.0*10^{-4}[/tex]
=> [tex]d = 90000 \ m[/tex]
Therefore the distance between the radar antenna and the object is
[tex]D = \frac{d}{2}[/tex]
[tex]D = \frac{ 90000}{2}[/tex]
[tex]D = 45000 \ m[/tex]
The distance between the radar antenna and the object will be 45000 m.
What is a radar antenna?A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.
The given data in the problem is;
t is the time= 3.0 x 10⁻⁴
d is the distance between the radar antenna and the object=?
c is the peed of light=3×10⁸ m/sec
The radar's speed is usually equal to the speed of light, and this has a value. The distance covered by the radars to and fro movement is now calculated mathematically as
[tex]\rm d= c \times t \\\\ \rm d= 3.0 \times 10^8 \times 3.0 \times 10^{-4} \\\\ d=90000 \ m[/tex]
As a result, the radar antenna's distance from the target is
[tex]\rm D=\frac{d}{2} \\\\ \rm D=\frac{90000}{2} \\\\ \rm D=\ 45000 \ m[/tex]
Hence the distance between the radar antenna and the object will be 45000 m.
To learn more about the radar antena refer to the link;
https://brainly.com/question/24067190
A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contract- ing when it is cooled from 120.0°C to 10.0°C?
Answer:
42000N
Explanation:
First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.
1) linear thermal expansion coef brass 19e-6 /K
∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity.
for brass, young's modulus is 15e6 psi or 100 GPa
cross-sectional area of rod is π(0.008)² = 0.0002 m²
F = EA∆L/L
F = (100e9)(0.0002)(0.00387) / (1.85)
F = 42000 or 42 kN
In order to waken a sleeping child, the volume on an alarm clock is doubled. Under this new scenario, how much more energy will be striking the child's ear drums each second?
Answer:4 times more energy will be striking the childbearing
Explanation:
Because Volume is directly proportional to amplitude of sound. Energy is proportional to amplitude squared. If you triple the amplitude, you multiply the energy by 4
A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 540N . What are the frequencies of the fundamental and first two overtones? Express your answers using two significant figures. Enter your answers in ascending order separated by commas.
Answer:
[tex]v_1 = 301 Hz[/tex]
[tex]v_2 = 601 \ \ Hz[/tex]
[tex]v_3 = 901 \ Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 90 \ cm = 0.9 \ m[/tex]
The mass of the string is [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]
The distance from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]
The tension is [tex]T = 540 \ N[/tex]
Generally the frequency is mathematically represented as
[tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]
Where n is and integer that defines that overtones
i.e n = 1 is for fundamental frequency
n = 2 first overtone
n =3 second overtone
Also [tex]\mu[/tex] is the linear density of the string which is mathematically represented as
[tex]\mu = \frac{m_s}{l}[/tex]
=> [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]
=> [tex]\mu = 0.003889 \ kg/m[/tex]
So for n = 1
[tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_1 = 301 \ Hz[/tex]
So for n = 2
[tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_2 = 601 \ Hz[/tex]
So for n = 3
[tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v =901 \ Hz[/tex]
Suppose a certain laser can provide 82 TW of power in 1.1 ns pulses at a wavelength of 0.24 μm. How much energy is contained in a single pulse?
Answer:
The energy contained in a single pulse is 90,200 J.
Explanation:
Given;
power of the laser, P = 82 TW = 82 x 10¹² W
time taken by the laser to provide the power, t = 1.1 ns = 1.1 x 10⁻⁹ s
the wavelength of the laser, λ = 0.24 μm = 0.24 x 10⁻⁶ m
The energy contained in a single pulse is calculated as;
E = Pt
where;
P is the power of each laser
t is the time to generate the power
E = (82 x 10¹²)(1.1 x 10⁻⁹)
E = 90,200 J
Therefore, the energy contained in a single pulse is 90,200 J
A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium. a) What is the total mechanical energy of the motion
Answer:
Explanation:i think this would help u
Currents in DC transmission lines can be 100 A or higher. Some people are concerned that the electromagnetic fields from such lines near their homes could pose health dangers.
A. For a line that has current 150 A and a height of 8.0 m above the ground, what magnetic field does the line produce at ground level? Express your answer in teslas.
B. What magnetic field does the line produce at ground level as a percent of earth's magnetic field which is 0.50 G?
C. Is this value of magnetic field cause for worry? Choose your answer below.
i. Yes. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
iii. Yes. Since this field is much greater than the earth's magnetic field, it would be expected to have more effect than the earth's field.
iv. No. Since this field does not differ a lot from the earth's magnetic field, it would be expected to have almost the same effect as the earth's field.
Answer:
Explanation:
magnetic field due to an infinite current carrying conductor
B = k x 2I / r where k = 10⁻⁷ , I is current in conductor and r is distance from wire
putting the given data
B = 10⁻⁷ x 2 x 100 / 8
= 25 x 10⁻⁷ T .
B )
earth's magnetic field = .5 gauss
= .5 x 10⁻⁴ T
= 5 x 10⁻⁵ T
percent required = (25 x 10⁻⁷ / 5 x 10⁻⁵) x 100
= 5 %
C )
ii. No. Since this field is much lesser than the earth's magnetic field, it would be expected to have less effect than the earth's field.
an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate
Answer:
The electric field can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.
Explanation:
Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.
g A solenoid 63.5 cm long has 960 turns and a radius of 2.77 cm. If it carries a current of 2.28 A, find the magnetic field along the axis at its center.Find the magnetic field on the solenoidal axis at the end of the solenoid.
Answer:
The value is [tex]B = 0.0043 \ T[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 63.5 = 0.635 \ m[/tex]
The number of turns is [tex]N = 960 \ turns[/tex]
The current is [tex]I = 2.28 \ A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \mu _o * n * I[/tex]
Where n is the number of turn per unit length which is mathematically evaluated as
[tex]n = \frac{N}{l}[/tex]
[tex]n = \frac{960}{0.635}[/tex]
[tex]n = 1512 \ turns /m[/tex]
and [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 4\pi * 10^{-7} * 1512 * 2.28[/tex]
[tex]B = 0.0043 \ T[/tex]
How could a country benefit from making it into space?
Answer:
space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.
A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?
Answer:
The work done on the system is -616 kJ
Explanation:
Given;
Quantity of heat absorbed by the system, Q = 767 kJ
change in the internal energy of the system, ΔU = +151 kJ
Apply the first law of thermodynamics;
ΔU = W + Q
Where;
ΔU is the change in internal energy
W is the work done
Q is the heat gained
W = ΔU - Q
W = 151 - 767
W = -616 kJ (The negative sign indicates that the work is done on the system)
Therefore, the work done on the system is -616 kJ
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room
Answer:
28.3°C
Explanation:
Using
T(t) = (T(0) - 20)*(e^(-k*t)) + 20
for some positive number k, and some initial temperature T(0).
Boundary conditions:
T(4) = T(0) - 5 _______ (i)
T(8) = T(0) - 7 _______ (ii)
==> solving for T(0) and k :
(i):
(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>
(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5
(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5
5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))
5 = (T(0) - 20) * ( 1 - e^(-k*4) )
(ii):
(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7
(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7
7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))
7 = (T(0) - 20) * (1 - e^(-k*8))
In both results, subsitute x = e^(-4k) and C = (T(0) - 20)
(i): 5 = C * (1 - x)
(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)
Substitute C*(1-x) from (i) into (ii):
(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5
back into (i):
(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3
C = T(0) - 20 ==>
T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3
= 28.3°C
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387 N. Between the box of mass m2 and the box of mass m3, the force meter reads F23=2304 N. Assume that the ropes and force meters are massless.
The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:
[tex]F_{pull}=m_{T}.a[/tex]
[tex]m_{T}=\frac{F_{pull}}{a}[/tex]
[tex]m_{T}=\frac{3615}{1.516}[/tex]
[tex]m_{T}=2384.5[/tex]
Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
For [tex]m_{1}[/tex]:
The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.
[tex]m_{1} = \frac{F_{12}}{a}[/tex]
[tex]m_{1} = \frac{1387}{1.516}[/tex]
[tex]m_{1}[/tex] = 915kg
For [tex]m_{2}[/tex]:
There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:
[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]
[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]
[tex]m_{2}[/tex] = 605kg
For [tex]m_{3}[/tex]:
[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]
[tex]m_{3} = 2384.5-1520.0[/tex]
[tex]m_{3}[/tex] = 864.5kg
IMPORTANT ANSWER ALL 3 PLEASE!
Answer:
4. Liters
5. Celsius
6. Grams
