Answer:
Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.
Explanation:
Amount of energy reaching 1 square meter of area = 2278 kJ/hour
Total area of the United States = [tex]9,158,960 km^2=9,158,960\times 10^6 m^2[/tex]
[tex](1 km^2=10^6 m^2)[/tex]
Amount of energy reaching [tex]9,158,960\times 10^6 m^2[/tex] of area:
[tex]2278 kJ/hour\times 9,158,960\times 10^6 kJ/hour[/tex]
[tex]=2.0864\times 10^{16} kJ/Hour[/tex]
Total solar energy reaches the entire United States per hour is[tex]2.0864\times 10^{16} [/tex] kilo Joules.
Final answer:
The total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.
Explanation:
To calculate the total solar energy reaching the entire United States per hour, given that 2278 kJ of energy reaches a square meter in one hour, we would first convert the entire area of the United States into square meters. Since the area is given as 9,158,960 km², we convert this to square meters by multiplying by (1000 m/km)²:
9,158,960 km² * (1000 m/km)² = 9,158,960,000,000 m²
Next, we multiply the area in square meters by the energy received per square meter:
9,158,960,000,000 m² * 2278 kJ/m² = 20,866,319,680,000 kJ
To generate an accurate answer, we can express this in megajoules (MJ) by dividing by 1,000 (since 1 MJ = 1,000 kJ):
20,866,319,680,000 kJ / 1,000 = 20,866,319,680 MJ
Therefore, the total solar energy that reaches the entire United States per hour is 20,866,319,680 MJ.
A uniform solid sphere of mass M and radius R rotates with an angular speed ω about an axis through its center. A uniform solid cylinder of mass M, radius R, and length 2R rotates through an axis running through the central axis of the cylinder. What must be the angular speed of the cylinder so it will have the same rotational kinetic energy as the sphere?
Answer:
[tex]\omega' = 0.89\omega[/tex]
Explanation:
Rotational inertia of uniform solid sphere is given as
[tex]I = \frac{2}{5}MR^2[/tex]
now we have its angular speed given as
angular speed = [tex]\omega[/tex]
now we have its final rotational kinetic energy as
[tex]KE = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2[/tex]
now the rotational inertia of solid cylinder about its axis is given by
[tex]I = \frac{1}{2}MR^2[/tex]
now let say its angular speed is given as
angular speed = [tex]\omega'[/tex]
now its rotational kinetic energy is given by
[tex]KE = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]
now if rotational kinetic energy of solid sphere is same as rotational kinetic energy of solid sphere then
[tex]\frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]
[tex]\frac{2}{5}\omega^2 = \frac{1}{2}\omega'^2[/tex]
[tex]\omega' = 0.89\omega[/tex]
Answer:
w_cyl = ±√(4/5) ω
Explanation:
Kinetic energy
E = (1/2)Iw²
where I is the moment of inertia and w the angular frequency of rotation.
The moment of inertia of a solid sphere of mass M and radius R is:
I = (2/5)MR²,
Solid cylinder is of mass M and radius R
I = (1/2)MR²
Equate the energies through
(1/2)×(2/5) M R²× (w_sphere)² = (1/2)× (1/2) MR² × (w_cyl)²
(w_cyl)² = (4/5)(w_sphere)²
w_cyl = ±√(4/5) ω
The energy of rotation is independent of the direction of rotation
An automobile tire has a volume of 1.63 x 10-2 m3 and contains air at a gauge pressure (pressure above atmospheric pressure) of 165 kPa when the temperature is 0.00°C. What is the gauge pressure of the air in the tires when its temperature rises to 27.3°C and its volume increases to 1.70 x 10-2 m3
Answer: The gauge pressure of the air in the tires is 179.5 kPa.
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = Atmospheric pressure + gauge pressure = 101 kPa + 165 kPa = 266 kPa
[tex]P_2[/tex] = final pressure of gas = ?
[tex]V_1[/tex] = initial volume of gas = [tex]1.63\times 10^{-2}m^3[/tex]
[tex]V_2[/tex] = final volume of gas = [tex]1.70\times 10^{-2}m^3[/tex]
[tex]T_1[/tex] = initial temperature of gas = [tex]0^oC=273+0=273K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]27.3^oC=273+27.3=300.3K[/tex]
Now put all the given values in the above equation, we get the final pressure of gas.
[tex]\frac{266\times 1.63\times 10^{-2}}{273K}=\frac{P_2\times 1.70\times 10^{-2}}{300.3K}[/tex]
[tex]P_2=280.5kPa[/tex]
Gauge pressure = Absolute pressure - atmospheric pressure = (280.5 - 101) kPa= 179.5 kPa
Therefore, the gauge pressure of the air in the tires is 179.5 kPa.
An express subway train passes through an underground station. It enters at t = 0 with an initial velocity of 23.0 m/s and decelerates at a rate of 0.150 m/s^2 as it goes through. The station in 205 m long (a) How long is the nose of the train in the station? (b) How fast is it going when the nose leaves the station? (c) If the train is 130 m long, at what time t does the end of the train leave the station? (d) What is the velocity of the end of the train as it leaves?
Answer:
a) Train's nose will be present 9.19 seconds in the station.
b) The nose leaves the station at 21.62 m/s.
c) Train's end will leave after 15.33 seconds from station.
d) The end leaves the station at 20.70 m/s.
Explanation:
a) We have equation of motion s = ut + 0.5at²
Here u = 23 m/s, a = -0.15 m/s², s = 205 m
Substituting
205 = 23t + 0.5 x (-0.15) x t²
0.075t² -23 t +205 = 0
We will get t = 9.19 or t = 297.47
We have to consider the minimum time
So train's nose will be present 9.19 seconds in the station.
b) We have equation of motion v= u + at
Here u = 23 m/s, a = -0.15 m/s², t = 9.19
Substituting
v= 23 - 0.15 x 9.19 = 21.62 m/s
The nose leaves the station at 21.62 m/s.
c) We have equation of motion s = ut + 0.5at²
Here u = 23 m/s, a = -0.15 m/s², s = 205 + 130 = 335 m
Substituting
335 = 23t + 0.5 x (-0.15) x t²
0.075t² -23 t +335 = 0
We will get t = 15.33 or t = 291.33
We have to consider the minimum time
So train's end will leave after 15.33 seconds from station.
d) We have equation of motion v= u + at
Here u = 23 m/s, a = -0.15 m/s², t = 15.33
Substituting
v= 23 - 0.15 x 15.33 = 20.70 m/s
The end leaves the station at 20.70 m/s.
The radius of a right circular cone is increasing at a rate of 1.5 in/s while its height is decreasing at a rate of 2.4 in/s. At what rate is the volume of the cone changing when the radius is 150 in. and the height is 144 in.?
Answer:
[tex]11304 \frac{in^{3}}{s}[/tex]
Explanation:
r = radius of right circular cone = 150 in
h = height of right circular cone = 144 in
[tex]\frac{dr}{dt}[/tex] = rate at which radius increase = 1.5 in/s
[tex]\frac{dh}{dt}[/tex] = rate at which height decrease = - 2.4 in/s
Volume of the right circular cone is given as
[tex]V = \frac{\pi r^{2}h}{3}[/tex]
Taking derivative both side relative to "t"
[tex]\frac{dV}{dt} = \frac{\pi }{3}\left ( r^{2}\frac{dh}{dt} \right + 2rh\frac{dr}{dt})[/tex]
[tex]\frac{dV}{dt} = \frac{3.14 }{3}\left ( (150)^{2}(- 2.4) + 2(150)(144)(1.5))[/tex]
[tex]\frac{dV}{dt} = 11304 \frac{in^{3}}{s}[/tex]
An astronaut aboard the International Space Station, which is orbiting at an altitude of 4.00 x 105 m above the Earth's surface, has a gravitational potential energy of 2.94 x 106 J. What is the weight of the astronaut when he returns to the Earth's surface
Answer:
The weight of the astronaut is 0.4802 N.
Explanation:
Gravitational potential energy, [tex]U=2.94\times 10^6\ J[/tex]
Distance above earth, [tex]d=4\times 10^5\ m[/tex]
The gravitational potential energy is given by :
[tex]U=\dfrac{GMm}{R}[/tex]
G is universal gravitational constant
M is the mass of Earth, [tex]M=5.97\times 10^{24}\ kg[/tex]
m is mass of astronaut
R is the radius of earth, R = R + d
[tex]R=6.37\times 10^6\ m+4\times 10^5\ m=6770000\ m[/tex]
[tex]m=\dfrac{U(R+d)^2}{GM}[/tex]
[tex]m=\dfrac{2.94\times 10^6\ J\times (6770000\ m)}{6.67\times 10^{-11}\times 5.97\times 10^{24}\ kg}[/tex]
m = 0.049 kg
The weight of the astronaut is given by :
W = mg
[tex]W=0.049\ kg\times 9.8\ m/s^2[/tex]
W = 0.4802 N
So, the weight of the astronaut when he returns to the earth surface is 0.4802 N. Hence, this is the required solution.
Final answer:
The weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.
Explanation:
To calculate the weight of the astronaut when he returns to the Earth's surface, we can use the formula for gravitational potential energy:
PE = mgh
where PE is the gravitational potential energy, m is the mass of the astronaut, g is the acceleration due to gravity, and h is the altitude of the astronaut.
Given that the gravitational potential energy is 2.94 x 10^6 J and the altitude is 4.00 x 10^5 m, we can rearrange the formula to solve for m:
m = PE / (gh)
Substituting the values, we get:
m = (2.94 x 10^6 J) / ((9.80 m/s^2) * (4.00 x 10^5 m))
Calculating this, we find that the mass of the astronaut is approximately 7.49 kg.
Now, to find the weight of the astronaut when he returns to the Earth's surface, we can use the formula:
Weight = mg
Substituting the mass we just calculated, we get:
Weight = (7.49 kg) * (9.80 m/s^2)
Calculating this, we find that the weight of the astronaut when he returns to the Earth's surface is approximately 73.32 N.
An undamped 1.23 kg horizontal spring oscillator has a spring constant of 37.4 N/m. While oscillating, it is found to have a speed of 2.48 m/s as it passes through its equilibrium position. What is its amplitude of oscillation?
Answer:
0.45 m
Explanation:
m = 1.23 kg, k = 37.4 N/m, vmax = 2.48 m/s
velocity is maximum when it passes through the mean position.
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]
[tex]T = 2\pi \sqrt{\frac{1.23}{37.4}}[/tex]
T = 1.139 sec
w = 2 π / T
w = 2 x 3.14 / 1.139
w = 5.51 rad / s
Vmax = w A
Where, A be the amplitude
2.48 = 5.51 A
A = 2.48 / 5.51 = 0.45 m
Final answer:
The amplitude of oscillation for the given spring oscillator is 0.58 m.
Explanation:
The amplitude of oscillation can be calculated using the equation:
A = vmax/ω
where A is the amplitude, vmax is the maximum speed, and ω is the angular frequency.
The angular frequency can be calculated using the equation:
ω = sqrt(k/m)
where k is the spring constant and m is the mass of the oscillator.
Substituting the given values into the equations:
ω = sqrt(37.4 N/m / 1.23 kg) = 4.29 rad/s
A = 2.48 m/s / 4.29 rad/s = 0.58 m
Therefore, the amplitude of oscillation is 0.58 m.
A 2.3 kg block is dropped from rest from a height of 4.6 m above the top of the spring. When the block is momentarily at rest, the spring is compressed by 25.0 cm. What is the speed of the block when the compression of the spring is 15.0 cm?
Answer:
7.6 m/s
Explanation:
m = 2.3 kg, h = 4.6 m, x = 25 cm = 0.25 m
Use the conservation of energy
Potential energy of the block = Elastic potential energy of the spring
m g h = 1/2 k x^2
Where, k be the spring constant
2.3 x 9.8 x 4.6 = 0.5 x k x (0.25)^2
k = 3317.88 N/m
Now, let v be the velocity of the block, when the compression is 15 cm.
Again use the conservation of energy
Potential energy of the block = Kinetic energy of block + Elastic potential
energy of the spring
m g h = 1/2 m v^2 + 1/2 k x^2
2.3 x 9.8 x 4.6 = 0.5 x 2.3 x v^2 + 0.5 x 3317.88 x (0.15)^2
103.684 = 1.15 v^2 + 37.33
v = 7.6 m/s
Emperor penguins are known to dive down to a depth of about 530 meter to hunt. If atmospheric pressure is 1.013 x 10^5 Pa, and the density of seawater is about 1025 kg/m^3, what is the absolute pressure a penguin experiences at that depth?
Answer:
5.4 x 10⁶ Pa
Explanation:
h = depth to which penguins dive under seawater = 530 m
P₀ = Atmospheric pressure = 1.013 x 10⁵ pa
ρ = density of seawater = 1025 kg/m³
P = absolute pressure experienced by penguin at that depth
Absolute pressure is given as
P = P₀ + ρgh
Inserting the values
P = 1.013 x 10⁵ + (1025) (9.8) (530)
P = 5.4 x 10⁶ Pa
A satellite is in a circular orbit around the Earth at an altitude of 3.18x10 m. Find the period and th orbital speed of the satellite? A. T= 2.94 h, B. T= 3.23 h, v 5610 m/s C. T= 1.75 h, v = 5920 m/s D. T 1.12 h, v 4980 m/s E. T 2.58 h, v 6460 m/s
Answer:
112.17 m/s
56.427 years
Explanation:
h = 3.18 x 10^10 m
R = 6.4 x 10^6 m
r = R + h = 3.18064 x 10^10 m
M = 6 x 10^24 kg
The formula for the orbital velocity is given by
[tex]v = \sqrt{\frac{G M }{r}}[/tex]
[tex]v = \sqrt{\frac{6.67 \times 10^{-11}\times 6\times 10^{24} }{3.18064\times 10^{10}}}[/tex]
v = 112.17 m/s
Orbital period, T = 2 x 3.14 x 3.18064 x 10^10 / 112.17
T = 0.178 x 10^10 s
T = 56.427 years
Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. How does the molecular mass of A compare to that of B? (a) It is twice that of B. (b) It is one half that of B (c) It is four times that of B (d) It is one fourth that of B. (e) It is 1.4 times that of B.
Answer:
option (d)
Explanation:
The relation between the rms velocity and the molecular mass is given by
v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant
So for two gases
[tex]\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]
[tex]\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}[/tex]
[tex]{\frac{M_{B}}{M_{A}}} = 4[/tex]
[tex]{\frac{M_{B}}{4}} = M_{A}[/tex]
If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.
How does the molecular mass of A compare to that of B?We know that the RMS velocity of the molecule is given as,
[tex]V = \dfrac{1}{\sqrt{M}}[/tex]
Given to us
RMS speed of the molecules of gas A is twice that of gas B, therefore, [tex]V_A = 2 V_B[/tex]
Substitute the value of RMS in the equation [tex]V_A = 2 V_B[/tex],
[tex]\dfrac{1}{\sqrt{M_A}} = \dfrac{2}{\sqrt{M_B}}\\\\\\\sqrt{\dfrac{M_B}{M_A}} = 2\\\\\\\dfrac{M_B}{M_A} = 4[/tex]
Hence, If the RMS speed of the molecules of gas A is twice that of gas B then the molecular mass of A compared to that of B is one-fourth that of B.
Learn more about RMS speed:
https://brainly.com/question/15514391
A 4.80 Kg watermelon is dropped from rest from the roof of an 18.0 m building. Calculate the work done by gravity on the watermelon from the roof to the ground.
Answer:
Work, W = 846.72 Joules
Explanation:
Given that,
Mass of the watermelon, m = 4.8 kg
It is dropped from rest from the roof of 18 m building. We need to find the work done by the gravity on the watermelon from the roof to the ground. It is same as gravitational potential energy i.e.
W = mgh
[tex]W=4.8\ kg\times 9.8\ m/s^2\times 18\ m[/tex]
W = 846.72 Joules
So, the work done by the gravity on the watermelon is 846.72 Joules. Hence, this is the required solution.
A helium nucleus contains two protons and two neutrons. The mass of the helium nucleus is greater than the combined masses of two protons and two neutrons because binding energy has been added. True False
Answer:
False
Explanation:
Actually, the converse is true. The mass number would be lower than the sum of the mass of the individual nucleons combined. According to Einstein’s equation of E=MC², this will be due to a phenomenon called mass defect. This ‘anomaly’ is due to the loss of some energy (now the nuclear binding energy) when the nucleons were brought in together to form the nucleus.
At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzle velocity of 600 m/s. Determine the time after A is fired, as to when bullet B passes bullet A. At what altitude does this occur?
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
[tex]s=ut+0.5at^2 \\ [/tex]
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
[tex]h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9 [/tex]
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
[tex]h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\ [/tex]
Solving both equations
[tex]600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\ [/tex]
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
[tex]h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\ [/tex]
Passing of B occurs at 4108.31 height.
The mass of mars is 6.38x10^23 kg and its radius is 3.38 x10^6m. Mars rotates on its axis with a period of 1.026 days.(G=6.67x10^-11 Nm^2/kg^2). calculate the orbital speed for a satellite at an altitude of 1.62x10^6 m.
Answer:
v = 2917.35 m/s
Explanation:
let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.
at any point in the orbit the forces acting on the satelite are balanced such that:
Fc = Fg
mv^2/r = GmM/r^2
v^2 = GM/r
v = \sqrt{GM/r}
= \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}
= 2917.35 m/s
Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 m/s.
The Eiffel tower in Paris is 300 meters tall on a cold day (T = -24 degrees Celsius), what is its height on a hot day when the temperature is 35 degrees? (It is made of iron which has a coefficient of linear expansion of 0.000012 per degree Celsius.)
Answer:
Length of Eiffel tower, when the temperature is 35 degrees = 300.21 m
Explanation:
Thermal expansion is given by the expression
[tex]\Delta L=L\alpha \Delta T \\ [/tex]
Here length of Eiffel tower, L = 300 m
Coefficient of thermal expansion, α = 0.000012 per degree Celsius
Change in temperature, = 35 - (-24) = 59degrees Celsius
Substituting
[tex]\Delta L=L\alpha \Delta T= 300\times 0.000012\times 59=0.2124m \\ [/tex]
Length of Eiffel tower, when the temperature is 35 degrees = 300 + 0.2124 = 300.21 m
A circular loop 40 cm in diameter is made froma flexible conductor and lies at right angles to a uniform 12-T magnetic field. At time t = 0 the loop starts to expand, its radius increasing at the rate of 5.0 mm/s Find the induced emf in the loop: a) at t 1.0 s and b) at t 10 s.
Answer:
Explanation:
As we know that magnetic flux is given by
[tex]\phi = B.A[/tex]
[tex]\phi = B.\pi r^2[/tex]
now from Faraday's law
[tex]EMF = \frac{d\phi}{dt}[/tex]
[tex]EMF = \frac{d(B. \pi r^2)}{dt}[/tex]
[tex]EMF = 2\pi r B \frac{dr}{dt}[/tex]
now we have
[tex]r = 40/2 = 20 cm[/tex]
B = 12 T
[tex]\frac{dr}{dt} = 5 \times 10^{-3} m/s[/tex]
Part a)
now at t = 1 s
r = 20 + 0.5 = 20.5 cm
[tex]EMF = (2\pi (0.205))(12)(5 \times 10^{-3})[/tex]
[tex]EMF = 0.077 Volts[/tex]
Part b)
now at t = 10 s
r = 20 + 0.5(10) = 25 cm
[tex]EMF = (2\pi (0.25))(12)(5 \times 10^{-3})[/tex]
[tex]EMF = 0.094 Volts[/tex]
A piano tuner stretches a steel piano wire with a tension of 1070 N . The wire is 0.400 m long and has a mass of 4.00 g . A. What is the frequency of its fundamental mode of vibration?
B. What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to 1.00×10^4 Hz ?
A. 409 Hz
The fundamental frequency of a string is given by:
[tex]f_1=\frac{1}{2L}\sqrt{\frac{T}{m/L}}[/tex]
where
L is the length of the wire
T is the tension in the wire
m is the mass of the wire
For the piano wire in this problem,
L = 0.400 m
T = 1070 N
m = 4.00 g = 0.004 kg
So the fundamental frequency is
[tex]f_1=\frac{1}{2(0.400)}\sqrt{\frac{1070}{(0.004)/(0.400)}}=409 Hz[/tex]
B. 24
For this part, we need to analyze the different harmonics of the piano wire. The nth-harmonic of a string is given by
[tex]f_n = nf_1[/tex]
where [tex]f_1[/tex] is the fundamental frequency.
Here in this case
[tex]f_1 = 409 Hz[/tex]
A person is capable to hear frequencies up to
[tex]f = 1.00 \cdot 10^4 Hz[/tex]
So the highest harmonics that can be heard by a human can be found as follows:
[tex]f=nf_1\\n= \frac{f}{f_1}=\frac{1.00\cdot 10^4}{409}=24.5 \sim 24[/tex]
The fundamental frequency of the piano wire is approximately 409 Hz, based on the given mass, length, and tension. The highest harmonic that could be heard by a person capable of hearing frequencies up to 10,000 Hz would be the 24th harmonic.
Explanation:To find the fundamental frequency (f1) of the piano wire, we can use the formula for the fundamental frequency of a stretched string, which is f1 = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the mass per unit length (linear mass density).
First, we need to calculate the linear mass density of the piano wire:
Mass (m) = 4.00 g = 0.004 kgLength (L) = 0.400 mLinear mass density (μ) = mass / length = 0.004 kg / 0.400 m = 0.010 kg/mNow, we plug in the values into the formula:
Tension (T) = 1070 NLength (L) = 0.400 mμ = 0.010 kg/mf1 = (1 / (2 * 0.400 m)) * sqrt(1070 N / 0.010 kg/m) = (1 / 0.8 m) * sqrt(107000 N/m) = (1 / 0.8) * 327.16 Hz ≈ 409 Hz.
For part B, we need to determine the highest harmonic that can be heard if the upper limit of human hearing is 10,000 Hz. Since the fundamental frequency is 409 Hz, the harmonics will be integer multiples of this value. The highest harmonic number (n) will be the largest integer such that:
n * f1 ≤ 10,000 Hz
Doing the division gives us n ≤ 10,000 / 409, and we round down to the nearest whole number because we cannot have a fraction of a harmonic. So, the highest harmonic heard would be:
n = floor(10,000/409) = 24 (since 24 * 409 ≈ 9816 Hz which is below the 10,000 Hz threshold).
Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit. The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm. What is the width of the slit?
Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, [tex]\lambda=608\ nm=608\times 10^{-9}\ m[/tex]
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
[tex]y=\dfrac{mD\lambda}{a}[/tex]
where
a = width of the slit
[tex]a=\dfrac{mD\lambda}{y}[/tex]
[tex]a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}[/tex]
a = 0.000167 m
[tex]a=1.67\times 10^{-4}\ m[/tex]
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
A circular rod with a gage length of 4 m and a diameter of 2.3 cm is subjected to an axial load of 70 kN . If the modulus of elasticity is 200 GPa , what is the change in length?
Answer:
The change in length is 3.4 mm.
Explanation:
Given that,
Length = 4 m
Diameter = 2.3 cm
Load = 70 kN
Modulus of elasticity = 200 GPa
We need to calculate the change in length
Using formula of modulus of elasticity
[tex]E=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]
[tex]\Delta l=\dfrac{Fl}{AE}[/tex]
Where, F = force
A = area
L = length
E = modulus elasticity
Put the value into the formula
[tex]\Delta l=\dfrac{70\times10^{3}\times4}{\pi\times(1.15\times10^{-2})^2\times200\times10^{9}}[/tex]
[tex]\Delta l=0.00336\ m[/tex]
[tex]\Delta l=3.4\ mm[/tex]
Hence, The change in length is 3.4 mm.
An electric dryer with an equivalent series resistance and inductance of values 13 ΩΩ and 46 mHmH is plugged into a standard 240V(RMS) 60Hz wall socket. You may assume the phase of the source voltage is zero. a) Calculate the load current.
Given:
R = 13 Ω
L = 46 mH
V = 240 V(rms)
f = 60 Hz
Formula used:
[tex]I_{L} = \frac{V}{R + jX_{L} }[/tex]
[tex]X_{L} = 2\pi fL[/tex]
Solution:
Now, using the above formula for [tex]X_{L}[/tex]:
[tex]X_{L} = 2\pi\times 60\times 46\times 10^{-3} [/tex] = 17.34 Ω
From the above formula for [tex]I_{L}[/tex]:
[tex]I_{L} = \frac{240\angle0}{13 + j17.34 }[/tex]
[tex]I_{L}[/tex] = (6.64 - j8.86) A = [tex]11.07\angle-53.14^{\circ}[/tex] A
[tex]i_{L}(t)[/tex] = [tex]\sqrt{2}\times 11.07cos(2\pi \times 60t - 53.14)[/tex] A
[tex]i_{L}(t)[/tex] = 15.65cos(376.99t - 53.14)A
Load current in the given AC circuit can be obtained by using the concept of Impedance and Ohm's law in its AC variant where current, I = V/Z. Impedance, Z is calculated taking into account both Resistance and Reactance.
Explanation:In the case of the mentioned electric dryer, we are dealing with an AC circuit that contains both resistance and inductance. In such cases, we should leverage the concept of Impedance, which is the effective resistance in an AC circuit resulting from combined effect of resistors, inductors and capacitors.
Impedance is defined by the relation Z = √ (R^2 + (XL)^2), where R is the resistance (13 Ω) and XL = 2πfL is the Reactance, f is the frequency (60Hz), and L the inductance (46mmH). However, when calculating the load current, we use Ohm's law in its AC version, I = V/Z, where I denotes the current, V the voltage supplied (240V), and Z the impedance defined earlier.
By substitizing all the values and calculating we can get the desired load current.
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How much energy is stored by the electric field between two square plates, 8.5 cm on a side, separated by a 2.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 14 nC .
The energy stored by the electric field between two plates, known as a parallel-plate capacitor, is calculated using the respective formulas for voltage, capacitance and energy in a capacitor, factoring in the charge of the plates, surface area, and separation distance.
Explanation:The energy stored by the electric field between two parallel plates (known as a parallel-plate capacitor) can be calculated using the formula for the energy stored in a capacitor, which is U = 0.5 * C * V^2. Here, V is the voltage across the plates and C is the capacitance of the capacitor.
To calculate V, we use the relation V = E * d, where E is the electric field, which can be found using the formula E = Q/A (Charge per Area), and d is the distance separating the plates. In this case, the charge Q is 14 nC, the area A is (8.5 cm)^2 and the distance d is 2.5 mm.
Finally, to find C, we use the formula C = permittivity * (A/d), where the permittivity of free space is approximately 8.85 × 10^-12 F/m. With these variables, C, V and U can be calculated accordingly.
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The energy stored in the electric field of a parallel-plate capacitor can be calculated using the dimensions of the plates, their separation, and charges. For this particular scenario, the energy stored would be approximately 0.44 μJ.
Explanation:In this scenario, we have a parallel-plate capacitor where each plate is a square with side 8.5 cm and has a charge of 14 nC. The plates are separated by a 2.5mm gap. Using this information, we can calculate the energy stored in the capacitor.
Firstly, the surface area A of the plates can be calculated using the formula A = s², where s is the side of the square. Converting 8.5 cm to meters, we get s = 0.085 m. So, A = (0.085 m)² = 0.007225 m².
Next, we calculate the electric field E between the plates using the formula E = Q/ε₀A, where Q is the charge on one plate and ε₀ is the permittivity of free space (8.85 x 10^-12 F/m). Converting 14 nC to coulombs gives Q = 14 x 10^-9 C. Substituting these values into the formula, we find E ≈ 224.717 kV/m.
The energy U stored in an electric field can be calculated using the formula U = 0.5ε₀EA². Substituting the earlier calculated values, we find U ≈ 0.44 μJ.
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An engine flywheel initially rotates counterclockwise at 5.85 rotations/s. Then, during 21.3 s, its rotation rate changes to 3.31 rotations/s clockwise. Find the flywheel's average angular acceleration (including its sign) in radians per second squared. Define counterclockwise rotation as positive.
Answer:
- 2.7 rad/s^2
Explanation:
f0 = + 5.85 rotations per second (counter clockwise)
t = 21.3 s
f = - 3.31 rotations per second (clockwise)
w0 = 2 x 3.14 x 5.85 = + 36.738 rad/s
w = - 2 x 3.14 x 3.31 = - 20.79 rad/s
Let α be teh angular acceleration.
α = (w - w0) / t
α = (-20.79 - 36.738) / 21.3
α = - 2.7 rad/s^2
The flywheel's average angular acceleration is -0.43 rad/s².
Average angular accelerationThe average angular acceleration of the flywheel is determined by applying the following kinematic equation as shown below;
α = (ωf - ωi)/t
where;
ωf is the final angular speed = -3.31 rad/sωi is the initial angulra speed = 5.85 rad/st is the time of motion, = 21.3 sα = (-3.31 - 5.85)/21.3
α = -0.43 rad/s²
Thus, the flywheel's average angular acceleration is -0.43 rad/s².
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A proton in a certain particle accelerator has a kinetic energy that is equal to its rest energy. What is the TOTAL energy of the proton as measured by a physicist working with the accelerator? (c = 3.00 × 108 m/s, mproton = 1.67 × 10-27 kg)
Answer:
Total energy, [tex]TE=3.006\times 10^{-10}\ J[/tex]
Explanation:
It is given that, a proton in a certain particle accelerator has a kinetic energy that is equal to its rest energy. Let KE is the kinetic energy of the proton and E₀ is its rest energy. So,
[tex]KE=E_o[/tex]
The total energy of the proton is equal to the sun of kinetic energy and the rest mass energy.
[tex]TE=KE+E_o[/tex]
[tex]TE=2E_o[/tex]
[tex]TE=2m_{proton}c^2[/tex]
[tex]TE=2\times 1.67\times 10^{-27}\ kg\times (3\times 10^8\ m/s)^2[/tex]
[tex]TE=3.006\times 10^{-10}\ J[/tex]
So, the total energy of the proton as measured by a physicist working with the accelerator is [tex]3.006\times 10^{-10}\ J[/tex]
A flashlight bulb operating at a voltage of 14.4 V has a resistance of 11.0 Ω . How many electrons pass through the bulb filament per second (e = 1.6 ´ 10-19 C)? (Give your answer to two significant figures)
For a DC circuit, the following equation relates the voltage, resistance, and current:
V = IR
V is the total voltage supplied, I is the total current, and R is the total resistance.
Given values:
V = 14.4V
R = 11.0Ω
Plug in the values and solve for I:
14.4 = I×11.0
I = 1.309A
Since one electron carries 1.6×10⁻¹⁹C of charge, divide the current by this number.
1.309/(1.6×10⁻¹⁹) = 8.182×10¹⁸
Round this value to 2 significant figures:
8.2×10¹⁸ electrons per second.
Define a scalar. Give two examples of scalars that you have used recently
In Mathematics and Physics, scalar is a quantity or a single number that shows the measurement of a medium in magnitude only (It does not include direction as vectors do); examples of scalars are voltage, mass, temperature, speed, volume, power, energy, and time.
Two examples of scalars I have used recently are Degrees Celsius to measure the temperature of my living room and Cubic Feet to measure the volume of my mug.
A scalar is a quantity represented only by a magnitude and devoid of direction. Examples of scalars include temperature and energy, which are significant in physics and everyday measurements. Unlike vectors, scalars do not change with coordinate system rotations.
Explanation:Definition of Scalar
A scalar is a physical quantity that is represented only by a magnitude (or numerical value) but does not involve any direction. In contrast to vectors, which have both magnitude and direction, scalars are not affected by coordinate system rotations or translations. Scalars come in handy when representing physical quantities where direction is non-applicable.
Two examples of scalars that I have used recently are:
Temperature: I noted the room temperature this morning, which was 20°C. The temperature is a scalar because it has no direction.Energy: I consumed a snack bar which had an energy content of 250 kilocalories, or 250 Calories. Energy is a scalar quantity as it does not possess a specific direction.When dealing with physics problems or everyday measurements, it's crucial to understand whether you're working with a scalar or a vector to correctly interpret the quantity in question.
An electromagnetic standing wave has a frequency of 120 MHz. (3396) Problem 2: What is the distance (in m) between adjacent antinodes of this standing wave?
Answer:
The distance between adjacent antinodes of the standing wave is 1.25 m.
Explanation:
Given that,
Frequency f= 120 MHz
We need to calculate the distance between adjacent antinodes of the standing wave
Using formula of distance
[tex]\Delta x=\dfrac{\lambda}{2}[/tex].....(I)
We know that,
[tex]\lambda=\dfrac{c}{f}[/tex]
Put the value of [tex]\lambda[/tex] in to the equation (I)
[tex]\Delta x=\dfrac{c}{2f}[/tex]
Where, c = speed of light
f = frequency
Put the all value into the formula
[tex]\Delta x=\dfrac{3\times10^{8}}{2\times120\times10^{6}}[/tex]
[tex]\Delta x=1.25\ m[/tex]
Hence, The distance between adjacent antinodes of the standing wave is 1.25 m.
A planetâs moon revolves around the planet with a period of 39 Earth days in an approximately circular orbit of radius of 4.8Ã10^8 m. How fast does the moon move?
Answer:
v = 895 m/s
Explanation:
Time period is given as 39 Earth Days
[tex]T = 39 days \times 24 hr \times 3600 s[/tex]
[tex]T = 3369600 s[/tex]
now the radius of the orbit is given as
[tex]r = 4.8 \times 10^8 m[/tex]
so the total path length is given as
[tex] L = 2 \pi r[/tex]
[tex]L = 2\pi (4.8 \times 10^8)[/tex]
[tex]L = 3.015 \times 10^9 [/tex]
now the speed will be given as
[tex]v = \frac{L}{T}[/tex]
[tex]v = \frac{3.015 \times 10^9}{3369600} [/tex]
[tex]v = 895 m/s[/tex]
What is the resistance of a 1000 m long copper wire with a 1 × 10-6 m² (1 mm²) cross-section assuming copper's resistivity is 1.68 × 10-8 Ω·m? Answer in the form: ##.#
Answer:
[tex]R=16.8[/tex]Ω
Explanation:
l=1000m
a=1mm2
p=1.68x10^-8
Electrical Resistivity Equation
[tex]R=p(\frac{L}{A} )\\[/tex]
where
p= proportional constant ρ (the Greek letter “rho”) is known as Resistivity.
L = is the length in metres (m)
A= is the area in square metres (m2),
R=1.68×10^-8×(1000/1×10^-6)
[tex]R=16.8[/tex]Ω
Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water line if it floats in sea water? b) What percentage of the ice floe is below the level of the liquid if it floats in mercury?
Answer:
(a) 92 %
(b) 6.76 %
Explanation:
length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3
density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3
(a) Let v be the volume of ice below water surface.
By the principle of flotation
Buoyant force = weight of ice block
Volume immersed x density of water x g = Total volume of ice block x density
of ice x g
v x 1000 x g = V x 920 x g
v / V = 0.92
% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %
(b) Let v be the volume of ice below the mercury.
By the principle of flotation
Buoyant force = weight of ice block
Volume immersed x density of mercury x g = Total volume of ice block x
density of ice x g
v x 13600 x g = V x 920 x g
v / V = 0.0676
% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %
One characteristic of mass wasting processes is that they a. only operate on steep slopes. b. move materials very slowly. c. operate only during nonfreezing months of the year. d. move materials relatively short distances compared to streams.
Answer:
One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.