Final answer:
In an inelastic collision, two lumps of matter collide and stick together. The final speed of the combined lump is 0, meaning it comes to a complete stop after the collision.
Explanation:
In an inelastic collision, two lumps of matter collide and stick together. The final speed of the combined lump can be determined by applying the principles of conservation of momentum and the relativistic addition of velocities.
Let's calculate the final speed of the combined lump using the given information:
Mass of each lump (m) = 1.500 kg
Speed of each lump (v) = 0.930c
Using the relativistic addition of velocities formula:
vf = (v1 + v2) / (1 + (v1 * v2 / c2))
Plugging in the given values:
vf = (0.930c + (-0.930c)) / (1 + (0.930c * (-0.930c) / c2))
vf = 0
Therefore, the final speed (vf) of the combined lump is 0, which means it comes to a complete stop after the inelastic collision.
An LC circuit consists of a 3.400 capacitor and a coil with self-inductance 0.080 H and no appreciable resistance. At t = 0 the capacitor is fully charged so the potential between the plates is 1.588 V and the current in the inductor is zero. What is the charge on the plates? How long after t = 0 will the current in the circuit be maximum? What will be the maximum current? What is the total energy in the system?
Answer:
Explanation:
charge on the capacitor = capacitance x potential
= 1.588 x 3.4
= 5.4 C
Energy of capacitor = 1 / 2 C V ² , C is capacitance , V is potential
= .5 x 3.4 x 1.588²
= 4.29 J
If I be maximum current
energy of inductor = 1/2 L I² , L is inductance of inductor .
energy of inductance = Energy of capacitor
1/2 L I² = 4.29
I² = 107.25
I = 10.35 A
Time period of oscillation
T = 2π √ LC
=2π √ .08 X 3.4
= 3.275 s
current in the inductor will be maximum in T / 4 time
= 3.275 / 4
= .819 s.
Total energy of the system
= initial energy of the capacitor
= 4.29 J
Which of these is a measure of force?
100 newtons
O 100 meters per second
100 watts
O 100 joules
Answer:
100 newtons
Explanation:
Recall force is the product of mass and acceleration from Newton's 2nd law given base units kgms^-2 which is equivalent to a newton( N)
A radio transmitting station operating at a frequency of 115 MHzMHz has two identical antennas that radiate in phase. Antenna BB is 9.05 mm to the right of antenna AA. Consider point PP between the antennas and along the line connecting them, a horizontal distance xx to the right of antenna A For what values of x will constructive interference occur at point P?
Final answer:
Constructive interference at point P occurs when the path length difference between the two antennas is an integer multiple of the wavelength. The values of x where this occurs can be found using the equation path length difference = n * wavelength. Solving for n, we find that constructive interference occurs at x = 0.00906 m.
Explanation:
In order to have constructive interference at point P, the path length difference between the two antennas must be an integer multiple of the wavelength. Since the antennas are identical and radiating in phase, the path length difference is simply the distance between the two antennas, which is 9.05 mm.
The wavelength can be calculated using the formula: wavelength = speed of light / frequency. For a frequency of 115 MHz, the wavelength is approximately 2.6 m.
To find the values of x where constructive interference occurs, we can set up an equation: path length difference = n * wavelength, where n is an integer. So, 9.05 mm = n * 2.6 m.
Solving for n, we get:
n = (9.05 mm) / (2.6 m) = 0.00348
Therefore, constructive interference will occur at point P when x = n * wavelength. Substituting n = 0.00348 and wavelength = 2.6 m, we get:
x = (0.00348)(2.6 m) = 0.00906 m
Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and the velocity of the medium (the air or the string itself). Consider a transverse wave traveling in a string. The mathematical form of the wave is y(x,t)=Asin(kx−ωt). a. Find the speed of propagation vp of this wave.b. Find the y velocity vy(x,t) of a point on the string as a function of x and t.
Answer with Explanation:
We are given that the mathematical form of wave
[tex]y(x,t)=Asin(kx-\omega t)[/tex]
a.We have to find the speed of propagation of this wave
In given mathematical form
k=Wave number
[tex]\omega[/tex]=Angular frequency
A=Amplitude
We know that
Speed of propagation of the wave=[tex]v_p=\frac{\omega}{k}[/tex]
b.Differentiate w.r.r t
[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]
The velocity [tex]v_y(x,t)[/tex] of a point on the string as a function of x and t is given by
[tex]v_y(x,t)=-A\omega cos(kx-\omega t)[/tex]
(a) The speed of propagation of the wave is [tex]\frac{\omega}{k}[/tex]
(b) (b) The velocity of a point on the string as a function of x and t is [tex]v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]
The given parameters;
[tex]y(x, t) = Asin(kx - \omega t)[/tex]
where;
t is the time of motiony is the displacement k is the wave number ω is the angular frequency A is the amplitude of the wave(a) The speed of propagation of the wave is calculated as follows;
[tex]v_p = \frac{\omega }{k}[/tex]
(b) The velocity of a point on the string as a function of x and t is calculated as follows;
[tex]v = \frac{dy}{dt} \\\\v_y (x, t) = -A\omega cos(kx - \omega t)[/tex]
Learn more here:https://brainly.com/question/18493636
A laser beam has diameter 1.30 mm . The beam exerts a force of 3.6×10−9 N on a totally reflecting surface.What is the amplitude of the electric field of the electromagnetic radiation in this beam?
Answer:
[tex]1.75\cdot 10^4 V/m[/tex]
Explanation:
For a totally reflecting surface, the radiation pressure is related to the intensity of radiation by
[tex]p=2\frac{I}{c}[/tex] (1)
where
I is the intensity of radiation
c is the speed of light
The radiation pressure is given by
[tex]p=\frac{F}{A}[/tex]
where in this case:
[tex]F=3.6\cdot 10^{-9}N[/tex] is the force exerted by the beam
A is the area on which the force is exerted
The beam has a diameter of d = 1.30 mm, so its area is
[tex]A=\pi (\frac{d}{2})^2=\pi (\frac{0.0013 m}{2})^2=1.33\cdot 10^{-6} m^2[/tex]
Now we can write eq(1) as
[tex]\frac{F}{A}=\frac{2I}{c}[/tex]
From which we find the intensity of radiation, I:
[tex]I=\frac{Fc}{2A}=\frac{(3.6\cdot 10^{-9})(3.0\cdot 10^8)}{2(1.33\cdot 10^{-6})}=4.06\cdot 10^5 W/m^2[/tex]
Now we can find the amplitude of the electric field using the equation
[tex]I=\frac{1}{2}c\epsilon_0 E^2[/tex]
where
[tex]\epsilon_0=8.85\cdot 10^{-12}F/m[/tex] is the vacuum permittivity
E is the amplitude of the electric field
And solving for E,
[tex]E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(4.06\cdot 10^5)}{(3\cdot 10^8)(8.85\cdot 10^{-12})}}=1.75\cdot 10^4 V/m[/tex]
In a solar system far far away, the sun's intensity is 100 W/m2 for a planet located a distance R away. What is the sun's intensity for a planet located at a distance 5 R from the Sun
Answer:
The sun's intensity for a planet located at a distance 5 R from the Sun is 4 W/m²
Explanation:
The intensity of Sun located at a distance of R away is [tex]100\ W/m^2[/tex].
We need to find the sun's intensity for a planet located at a distance 5 R from the Sun. Intensity is given by :
[tex]I\propto \dfrac{1}{r^2}[/tex]
So,
[tex]\dfrac{I_1}{I_2}=(\dfrac{r_2}{r_1})^2[/tex]
Here, I₁ = 100 W/m² and r₁ = R and r₂ = 5R
[tex]I_2=\dfrac{I_1r_1^2}{r_2^2}[/tex]
[tex]I_2=\dfrac{100(R^2)}{(5R)^2}\\\\I_2=\dfrac{100}{25}\\\\I_2=4\ W/m^2[/tex]
So, the sun's intensity for a planet located at a distance 5 R from the Sun is 4 W/m².
The current in a long solenoid of radius 5 cm and 17 turns/cm is varied with time at a rate of 5 A/s. A circular loop of wire of radius 7 cm and resistance 5 Ω surrounds the solenoid. Find the electrical current induced in the loop (in µA).
Answer:
The current is [tex]I =0.2mA[/tex]
Explanation:
From the question we are told that
The first radius is [tex]R_1 = 5cm = \frac{5}{100} = 0.05cm[/tex]
The number of turns is [tex]N = 17 \ turn/cm[/tex]
The current rate is [tex]\frac{dI}{dt} = 5 A/s[/tex]
The second radius is [tex]R_2 = 7cm = \frac{7}{100} = 0.07m[/tex]
The resistance is [tex]r = 5 \Omega[/tex]
Generally the magnetic flux induced in the solenoid is mathematically represented as
[tex]\O = B A[/tex]
Where is the magnetic field mathematically represented as
[tex]B = N \mu_o I[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with a value of [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
and A is the area mathematically represented as
[tex]A = \pi (R_2 - R_1)^2[/tex]
So
[tex]\O = N \mu I * \pi R^2[/tex]
Substituting values
[tex]\O = 17 * 4\pi *10^{-7} * \pi (7-5)^2I[/tex]
[tex]\O = 2.68*10^{-4}I[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- |\frac{d\O}{dt}|[/tex]
[tex]\epsilon = 2.68*10^{-4 } \frac{dI}{dt}[/tex]
substituting values
[tex]\epsilon =2.68 *10^{-4} * 5[/tex]
[tex]=1.3 *10^{-3} V[/tex]
From Ohm law
[tex]I = \frac{\epsilon }{r}[/tex]
Substituting values
[tex]I = \frac{1.3*0^{-3}}{5}[/tex]
[tex]I =0.2mA[/tex]
Enter the measured values of the angles of incidence and refraction below. Angle of incidence θi = 76.5 Correct: Your answer is correct. Your value is acceptable.° Angle of refraction θr = 76.5 Incorrect: Your answer is incorrect. Your value is too high.° Calculate the index of refraction using Snell's Law and the measured values of the angles of incidence and refraction. nacrylic = The accepted value of the index of refraction for acrylic is 1.49. What is the percent error between the accepted value and the experimental value of n? Hint Percent error = %
Answer:
Explanation:
Given that the inputted angle of incidence is accepted
Angle of incidence θi = 76.5°
But angle of refraction is not acceptable
Angle of refraction θr = 76.5°
We are told that the value is too high
Then, θr < 76.5°
We want to calculate index of refraction n?
The acceptable value of refraction index of acrylic is 1.49
So the true value is 1.49
So, let calculate the measure value
Refractive index is given as
n = Sin(i) / Sin(r)
Then,
n = Sin(76.5) / Sin(76.5)
Then, n = 1
Now, percentage error of the refractive index,
Percentage error is
%error= |true value—measure value| / true value × 100
The true value is 1.49
The measure value is 1
Then,
%error = ( |1.49—1| / 1.49 ) × 100
% error = ( 0.49 / 1.49 ) × 100
%error = 0.3289 × 100
%error = 32.89%
Answer:
Explanation:
32.89%
Below is a schematic of a vapor power plant in which water steadily circulates through the four components. The water flows through the boiler and condenser at constant pressure and through the turbine and pump adiabatically. Kinetic and potential energy effects can be ignored. Process data follow: Process 4-1: constant pressure at 8 MPa from saturated liquid to saturated vapor Process 2-3: constant pressure at 8 kPa from x2
Complete Question:
the first attached file shows the schematic
Answer:
a. irreversible cycle
b. 0.182
c. 0.2646
Explanation:
From the saturated vapour pressure table
When the pressure is 1 MPa
Temperature, T₄₁ = 179.9 °C = 452.9 °k
When Pressure is 20 kPa
Temperature, T₃₂ = 60.06 °C = 333.06 °K
attached images 2, 3, 4 and 5 shows a comprehensive solution to the questions
A vapor power plant uses water as the working substance, which goes through a cycle of becoming vaporized and condensed in order to drive a turbine. The schematic provided includes four components: the boiler, condenser, turbine, and pump. Two specific processes mentioned include 4-1 and 2-3, both of which occur at constant pressures.
Explanation:A vapor power plant operates by using water as the working substance, which goes through a cycle of becoming vaporized, driving a turbine, and then condensing back into a liquid state. In this specific schematic, the water flows through four components: the boiler, condenser, turbine, and pump. The process data provided includes two specific processes: 4-1 which occurs at a constant pressure of 8 MPa, and 2-3 which also occurs at a constant pressure of 8 kPa.
Learn more about Vapor power plant here:https://brainly.com/question/32027269
#SPJ3
A motorcycle starts from rest at and travels along a straight road with a constant acceleration of 2 6 / ft s until it reaches a speed of 50 / ft s . Afterwards it maintains this speed. Also, when t s 0 , a car located 6000 ft down the road is traveling toward the motorcycle at a constant speed of 30 / ft s . Determine the time and the distance traveled by the motorcycle when they pass each other.
Answer:
Time = 77.63 s
Distance = 3673.3 ft
Explanation:
Using equation of motion
v = u + at'
50 = 6 * t'
t' = 50 / 6
t' = 8.33 s
v² = u² + 2a(s - s•)
50² = 0² + 2 * 6 * (s - 0)
2500 = 12 * s
s' = 2500 / 12
s' = 208.3 ft
At t' = 8.33 the distance traveled by the car is
s'' = v• * t'
s'' = 30 * 8.33
s'' = 250 ft
Now, the distance between the motorcycle and car is
6000 - 208.33 - 250 = 5541.67
For motorcycle, when passing occurs,
s = v• * t, x = 50 * t''
For the car, when passing occurs,
s = v• * t, 5541.67 - x = 30 * t''
Solving both simultaneously, we have
5541.67 - [50 * t''] = 30 * t''
5541.67 = 30t''+ 50t''
5541.67 = 80t''
t''= 5541.67 / 80
t''= 69.3 s, substituting t'' = 69.3 back, we have
x = 50 * 69.3
x = 3465 ft
Therefore for motorcycle,
t = 69.3 + 8.33
t = 77.63 s
S = 3465 + 208.3
s = 3673.3 ft
Final Answer:
The total time [tex]\( t \)[/tex] for the motorcycle to pass the car is calculated to be [tex]\( 77.63 \)[/tex] seconds, and the distance traveled by the motorcycle when passing the car is approximately [tex]\( 3673.3 \)[/tex]feet.
Explanation:
The calculation begins by using the equation of motion ( v = u + at ), where (v ) represents final velocity, ( u ) is initial velocity, ( a ) is acceleration, and (t) is time. Substituting the given values, the time ( t' ) for the motorcycle to reach a speed of ( 50 ) ft/s with a constant acceleration of [tex]( 6\) ft/s\(^2\) )[/tex]is found to be approximately ( 8.33 ) seconds.
Next, the equation of motion [tex]\( v^2 = u^2 + 2a(s - s_0) \)[/tex] is used, where [tex]\( s \)[/tex] represents distance, [tex]\( s_0 \)[/tex] is initial distance, and other variables have the same meanings as before. Solving for [tex]\( s' \),[/tex] the distance traveled by the motorcycle during this time is approximately [tex]\( 208.3 \)[/tex] feet.
At [tex]\( t' = 8.33 \)[/tex] seconds, the distance traveled by the car is calculated to be [tex]\( 250 \)[/tex]feet.
The distance between the motorcycle and car when they pass each other is found to be approximately [tex]\( 5541.67 \)[/tex] feet.
To determine the time [tex]\( t'' \)[/tex] when the passing occurs, simultaneous equations for the motorcycle and car are set up using their respective distances. Solving these equations yields [tex]\( t'' = 69.3 \)[/tex] seconds. Substituting this value back, the distance traveled by the motorcycle when passing the car is approximately [tex]\( 3465 \)[/tex]feet.
Antonio wants to replace the lightbulbs in his house. How much money would he save in one year by replacing 10 incandescent bulbs with 10 CFL bulbs?
incandescent$1.47 per bulb CFL $3.99
annual operating cost incandescent $32.86 CFL $7.67
Answer:
226.70
Explanation:
Answer: $226.70 on edg
Explanation:
A uniform ladder of length l rests against a smooth, vertical wall. If the mass of the ladder is m and the coefficient of static friction between the ladder and the ground is 0.40, find the minimum angle at which the ladder does not slip.
What If? What if a person begins to climb the ladder when the angle is 51°? Will the presence of a person on the ladder make it more or less likely to slip?
Answer:
51.3 Degrees
The presence of person above the center of mass of ladder will make the ladder more likely to slip.
Explanation:
Explanation in the attachment.
51.3 Degrees, The presence of person above the centre of mass of ladder will make the ladder more likely to slip.
What is uniform ladder?The middle of a uniform ladder is where the center of mass is located. A smooth wall signifies that there is no friction felt on the wall, whereas rough ground indicates that there is friction there.
How to use static equilibrium to determine the coefficient of friction between the bottom of the ladder and the ground is shown in the Ladder Problem from Static Equilibrium. The torque, the x-direction forces, and the y-direction forces are added together and set to zero.
It follows that the force exerted on the ladder as a result of the resultant moment of the forces must also equal zero due to a random point.
Thus, it is 51.3 Degrees.
For more information about uniform ladder, click here:
https://brainly.com/question/29768495
#SPJ2
An incompressible fluid flows steadily through two pipes of diameter 0.15 m and 0.2 m which combine to discharge in a pipe of 0.3 diameter. If the average velocities in the 0.15 m and 0.2 m diameter pipes are 2 m/s and 3 m/s respectively, then find the average velocity in the 0.3 m diameter pipe.
Answer:
Average velocity = 1.835 m/s
Explanation:
Detailed explanation and calculation is shown in the image below
The average velocity in the 0.3 m diameter pipe. The result is 1.833 m/s.
To find the average velocity in the 0.3 m diameter pipe, we'll use the principle of continuity for incompressible fluid flow. The flow rate (Q) must be the same at all points in the system.
First, we calculate the flow rates in the two initial pipes:
For the 0.15 m diameter pipe: Q1 = A₁ v₁ = π (0.075)² 2 m/s = π × 0.005625 m² × 2 m/s = 0.01125π m³/s
For the 0.2 m diameter pipe: Q2 = A2 v2 = π (0.1)² 3 m/s = π × 0.01 m² × 3 m/s = 0.03π m³/s
The combined flow rate entering the 0.3 m diameter pipe is:
[tex]Q_t = Q_1 + Q_2[/tex] = 0.01125π m³/s + 0.03π m³/s = 0.04125π m³/s
Next, we find the area of the 0.3 m diameter pipe:
A₃ = π (0.15)² = π 0.0225 m²
The average velocity in the 0.3 m diameter pipe is then:
v₃ = Qtotal / A₃ = (0.04125π m³/s) / (π 0.0225 m²) ≈ 1.833 m/s
Thus, the average velocity in the 0.3 m diameter pipe is 1.833 m/s.
What pushes against gravity in: a main sequence star, a white dwarf, a neutron star, and a black hole? electron degeneracy, neutron degeneracy, nothing, and heat pressure nothing in all cases. Gravity always wins. nothing, heat pressure, electron degeneracy, and neutron degeneracy heat pressure, neutron degeneracy, electron degeneracy, and nothing heat pressure, electron degeneracy, neutron degeneracy, and nothing
A main sequence star relies on heat pressure to counteract gravity, a white dwarf uses electron degeneracy pressure, a neutron star uses neutron degeneracy pressure, and in a black hole, gravity ultimately triumphs.
Explanation:In a main sequence star, it is the heat pressure, due to nuclear fusion, which pushes against gravity and maintains the star's stability. A white dwarf is stabilized against gravity by what is known as electron degeneracy pressure, a quantum mechanical effect that arises because no two electrons can be in the same place doing the same thing at the same time (the Pauli exclusion principle).
The neutron star is supported against gravity by neutron degeneracy pressure, the same quantum effect but explained by neutrons instead of electrons. However, if the mass of a star's core is high enough (more than about three times that of the Sun), no known force can stop it from collapsing into a black hole; gravity ultimately overcomes all other forces.
Learn more about Counteracting Forces against Gravity in Stars here:https://brainly.com/question/30267591
#SPJ3
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headlights that are 0.691 m apart. At what distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source? Take the wavelength of the light to be 539 nm and your pupil diameter as 5.11 mm. ______________km
To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,
[tex]\theta = \frac{1.22 \lambda }{D}[/tex]
Here,
D is diameter of the eye
[tex]D = \frac{1.22 (539nm)}{5.11 mm}[/tex]
[tex]D= 1.287*10^{-4}m[/tex]
The angle that relates the distance between the lights and the distance to the lamp is given by,
[tex]Sin\theta = \frac{d}{L}[/tex]
For small angle, [tex]sin\theta = \theta[/tex]
[tex]sin \theta = \frac{d}{L}[/tex]
Here,
d = Distance between lights
L = Distance from eye to lamp
For small angle [tex]sin \theta = \theta[/tex]
Therefore,
[tex]L = \frac{d}{sin\theta}[/tex]
[tex]L = \frac{0.691m}{1.287*10^{-4}}[/tex]
[tex]L = 5367m[/tex]
Therefore the distance is 5.367km.
(a) How long in seconds does it take a radio signal to travel 160 km from a transmitter to a receiving antenna? (b) We see a full Moon by reflected sunlight. How much earlier did the light that enters our eye leave the Sun? The Earth – Moon and Earth – Sun distances are 3.8x105 km and 1.5 × 108 km, respectively. (c) What is the round-trip travel time in seconds for light between Earth and a spaceship at a 3.6 × 108 km distance from Earth? (d) Suppose astronomers observe a supernova about 9300 light-years (ly) distant. How long ago in years did the explosion actually occur?
Explanation:
Given:
Distance between the transmitter and receiver = 160 km
Distance between Earth and Sun (D) = 150000000 km
Distance between Earth and Moon (D') = 380000 km
Distance between Earth and space ship (D") = 360000000 km
Distance between Earth and Supernova (d) = 9300 LY
We know that,
Speed of Light = 300000 km/sec
Lets us solve each problem now:
a) We know that the radio waves travle with speed of light. So the time(T) taken by the signal from transmitter to receiving antenna will be,
[tex]Time = \frac{Distance}{Speed}[/tex]
[tex]T = \frac{160}{300000}[/tex]
T = 0.53 millisecond
b) Time taken by sunlight to reach Earth is,
[tex]T_{1} = \frac{150000000}{300000} seconds[/tex]
T₁ = 500 seconds = 08 min 20 seconds
Time taken by light from Moon to reach Earth will be,
[tex]T_{2} = \frac{380000}{300000} seconds[/tex]
T₂ = 1.26 seconds
Let us assume that in the given situation Earth and Moon are at the same distance from the Sun. This means that Sun light will take the same time to reach Moon as it will take to reach Earth. So Sun light reaches the Moon in 500 seconds. After that it will be reflected towards Earth and will take 1.26 seconds more to reach Earth.
Thus the light left the Sun 501.26 seconds earlier.
c) Time taken by the light from space ship to reach Earth will be,
[tex]T = \frac{360000000}{300000} seconds[/tex]
T = 1200 seconds.
Thus round trip time will be = 2T = 2400 seconds
d) Lightyear is a unit of distance and is equal to the distance travelled by light in a year. If a supernova is at a distance of 9300 LY it means that light has taken 9300 years to reach Earth from that supernova. It means that the light we see today was actually generated from the supernova 9300 years ago.
It is evident that the explosion actually occurred 9300 years back.
We have calculated the times light and radio signals take to travel given distances at the speed of light; these times depend directly on the distances. For the observed supernova, we use the reverse calculation, determining the time light has traveled given the distance in light-years.
Explanation:(a) The radio signal is essentially light, so it travels at the speed of light, which is about 300,000 kilometers per second (3 × 10⁸ m/s). To find the time it takes to travel 160 km, we simply divide the distance by the speed: 160 km / 300000 km/s = 0.00053 seconds, or about 0.53 milliseconds.
(b) The light we see from the Moon has indeed traveled from the Sun, bounced off the Moon, and then hit our eyes. Therefore, it has traveled a distance of the Earth–Sun distance plus the Earth–Moon distance. Adding these distances, the light has traveled a total of 1.5 × 10⁸ km + 3.8 × 10⁵ km = 1.50038 × 10⁸ km. Dividing this distance by the speed of light gives the time this light has traveled: 1.50038 × 10⁸ km / 300000 km/s = 500.13 seconds, or about 8 minutes and 20 seconds.
(c) The round-trip travel time of light between Earth and a spaceship would be twice the time it would take light to make a one-way trip. The one-way trip time is calculated as the distance divided by the speed of light, which is 3.6 × 10⁸ km / 300000 km/s = 1200 seconds. Therefore, the round-trip travel time would be 2400 seconds or 40 minutes.
(d) A light-year is the distance that light travels in one year, so if a supernova is observed to be 9300 light-years away, this means the light from the supernova has been traveling for 9300 years before reaching Earth. Therefore, the supernova actually happened 9300 years ago.
Learn more about Light Travel Time here:https://brainly.com/question/8031947
#SPJ11
A 2 kg sample of air contained in a 2 m3 rigid container undergoes a process in which its pressure reduces from 100.45 kPa to 86.10 kPa while the volume remains the same. Please calculate (a) how much work was done, (b) by how much did the internal energy of the air change, (c) how much heat was transferred (please specify the direction of heat transfer).
Answer:
Explanation:
The pictures attached below shows the solution to the problem and i hope its explanatory enough. Thank you
Two strain gauges are mounted so that they sense axial strain on a steel member in uniaxial tension. The 120 Ω gauges form two legs of a Wheatstone bridge, and are mounted on opposite arms (e.g., arms 1 and 4). The gauge factor for each of the strain gauges is 2 and Em for this steel is 29 × 106 psi. For a bridge excitation voltage of 4 V and a bridge output voltage of 120 μV under load (i.e., Ei = 4 V and 0 E =120 μV ):
Answer:
a. The maximum strain is 60 * 10^-6
b. Resistance change = 0.014395 ohms
Explanation:
The complete question is as follows;
Two strain gauges are mounted so that they sense axial strain on a steel member in uniaxial tension. The 120 V gauges form two legs of a Wheatstone bridge, and are mounted on opposite arms (e.g., arms 1 and 4). The gauge factor for each of the strain gauges is 2 and E m for this steel is 29 × 106 psi. For a bridge excitation voltage of 4 V and a bridge output voltage of 120 μV under load (i.e., Ei = 4 V and 0 E =120 μV ):
(a) Estimate the maximum strain.
(b) What is the resistance change experienced by each gauge?
solution;
Please check attachment for complete solution and step by step explanation
are 1.8 mm apart is illuminated by a monochromatic coherent light source. A fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 bright fringes/cm on the screen, what is the wavelength of the monochromatic light
The question is incomplete! Complete question along with answer and step by step explanation is provided below.
Question:
A pair of narrow slits that are 1.8 mm apart is illuminated by a monochromatic coherent light source. A fringe pattern is observed on a screen 4.8 m from the slits. If there are 5.0 bright fringes/cm on the screen, what is the wavelength of the monochromatic light?
Given Information:
Distance from the double slits to the screen = D = 4.8 m
Double slit separation distance = d = 1.8 mm = 0.0018 m
Number of fringes = m = 5
Distance between fringes = y = 1 cm = 0.01 m
Required Information:
Wavelength of the monochromatic light = λ = ?
Answer:
Wavelength of the monochromatic light = λ = 7.5x10⁻⁷ m
Explanation:
The wavelength of the monochromatic light can be found using young's double slits formula,
λ = yd/mD
where d is the double slit separation distance, D is the distance from the double slits to the screen, y is the distance between bright fringes and m is number of fringes.
λ = 0.01*0.0018/5*4.8
λ = 0.00000075 m
λ = 7.5x10⁻⁷ m
Therefore, the wavelength of the monochromatic light is 7.5x10⁻⁷ m
in hooke's law is force directly proportional to extension?
Answer:
Explanation:
Hooke's Law is a principle of physics that states that the that the force needed to extend or compress a spring by some distance is proportional to that distance. ... In addition to governing the behavior of springs, Hooke's Law also applies in many other situations where an elastic body is deformed
Answer:
Yeah ,The extension of the spring is directly proportional to the force applied.
The constant pressure molar heat capacity, C_{p,m}C p,m , of nitrogen gas, N_2N 2 , is 29.125\text{ J K}^{-1}\text{ mol}^{-1}29.125 J K −1 mol −1 at 298\text{ K}298 K. Calculate the change in the internal energy when 20\text{ mol}20 mol of nitrogen gas at 298\text{ K}298 K is heated so that its temperature increases by 15.0^{\circ}\text{C}15.0 ∘ C. You may assume constant heat capacity over the temperature range.
Answer:
Explanation:
Constant pressure molar heat capacity Cp = 29.125 J /K.mol
If Cv be constant volume molar heat capacity
Cp - Cv = R
Cv = Cp - R
= 29.125 - 8.314 J
= 20.811 J
change in internal energy = n x Cv x Δ T
n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature
Putting the values
= 20 x 20.811 x 15
= 6243.3 J.
a correlation coefficient represents what two things
Answer:
A correlation coefficient represents the following:
1- The direction of the relationship
2- The strength of the relationship
A correlation coefficient represents the relationship between two variables. It quantifies the strength and direction of the relationship.
Explanation:A correlation coefficient represents the relationship between two variables. It quantifies the strength and direction of the relationship. The correlation coefficient ranges from -1 to 1, where a value close to -1 indicates a strong negative correlation, a value close to 1 indicates a strong positive correlation, and a value close to 0 indicates no or weak correlation.
Learn more about correlation coefficient here:https://brainly.com/question/33643115
#SPJ6
(NH4)2O compound name
The compound (NH4)2O is known as Ammonium Oxide, consisting of two ammonium ions and one oxide ion. It's produced via an acid-base reaction, with the ammonium ion acting as a weak acid and the oxide ion as a strong base. Such compounds have broad applications.
Explanation:The compound (NH4)2O is known as Ammonium Oxide. It consists of two ammonium ions (NH4+) and one oxide ion (O2-). In terms of acid-base reactions, the ammonium ion is considered a weak acid due to its ability to donate a proton to water, while the oxide ion as a strong base accepts protons from water.
The formation of Ammonium Oxide involves the transfer of H+ ions from water molecules to ammonia molecules, which then react with the oxide anions to create the final compound. Ammonium compounds, including Ammonium Oxide, have applications in areas like agriculture and commodity chemical synthesizing.
Learn more about Ammonium Oxide here:https://brainly.com/question/30639539
#SPJ2
Three liquids are at temperatures of 5 ◦C, 25◦C, and 34◦C, respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is 17◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 30.7 ◦C. Find the equilibrium temperature when equal masses of the first and third are mixed. Answer in units of ◦C.
Answer:
m * c1 * (9 - 6) = m * c2 * (19 - 9) → mass m cancels
c2 = c1 * 3 / 10 = 0.3*c1
Explanation:
However, for the second mix, there should be nothing from the first mix's answer except for 0.3c1. Essentially, the left side of the second mix equation should look like: m *0.3c1 * (31.1-19). Then, you can properly solve the equation.
Required information Problem 16.048 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. A uniform slender rod AB rests on a frictionless horizontal surface, and a force P of magnitude 0.25 lb is applied at A in a direction perpendicular to the rod. Assume that the rod weighs 2.2 lb. Problem 16.048.a - Acceleration A of slender rod on frictionless surface Determine the acceleration of Point A. (You must provide an answer before moving on to the next part.) The acceleration of Point A is 14.6 14.6 Correct ft/s2 →.
NOTE: The diagram is attached to this solution
Answer:
The acceleration of point A = 14.64 ft/s²
Explanation:
By proper analysis of the diagram, acceleration of point A will be: (Check the free body diagram attached)
[tex]a_{A} = \bar{a} + \frac{\alpha L}{2}[/tex]
Weight, W = mg
g = 32.2 ft/s²
m = W/g
[tex]p = m \bar{a}\\p = w \bar{a} /g[/tex]
[tex]\bar{a} = pg/w\\\bar{a} = 0.25g/2.2\\\bar{a} =3.66[/tex]
[tex]\frac{pL}{2} = I \alpha[/tex]
but [tex]I = \frac{wL^{2} }{12g}[/tex]
[tex]\frac{pL}{2} = \frac{\alpha wL^{2} }{12g}\\\alpha = \frac{6 g p}{wL}\\\alpha = \frac{6*g*0.25}{2.2L} \\\alpha = 21.96/L[/tex]
[tex]a_{A} = 3.66 + \frac{(21.96/L ) * L}{2}\\a_{A} = 3.66 + 10.98\\a_{A} = 14. 64 ft/s^{2}[/tex]
When light of wavelength 242 nm shines on a metal surface the maximum kinetic energy of the photoelectrons is 1.99 eV. What is the maximum wavelength (in nm) of light that will produce photoelectrons from this surface? (Use 1 eV = 1.602 ✕ 10^−19 J, e = 1.602 ✕ 10^−19 C, c = 2.998 ✕ 10^8 m/s, and h = 6.626 ✕ 10^−34 J · s = 4.136 ✕ 10^−15 eV · s as necessary.)
Answer:
Maximum wavelength will be [tex]3.96\times 10^{-7}m[/tex]
Explanation:
It is given wavelength [tex]\lambda =242nm=242\times 10^{-9}m[/tex]
Speed of light [tex]c=3\times 10^8m/sec[/tex]
Plank's constant [tex]h=6.6\times 10^{-4}Js[/tex]
So energy is equal to
[tex]E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{242\times 10^{-9}}=8.18\times 10^{-19}J[/tex]
Maximum kinetic energy is given
[tex]KE_{max}=1.99eV=1.99\times 1.6\times 10^{-19}=3.184\times 10^{-19}J[/tex]
Work function is equal to
[tex]w_0=E-KE_{max}=8.18\times 10^{-19}-3.184\times 10^{-19}=5\times 10^{-19}J[/tex]
[tex]\frac{hc}{\lambda _0}=5\times 10^{-19}[/tex]
[tex]\frac{6.6\times 10^{-34}\times 3\times 10^8}{\lambda _0}=5\times 10^{-19}[/tex]
[tex]\lambda _0=3.96\times 10^{-7}m[/tex]
Rotation of the lever OA is controlled by the motion of the contacting circular disk of radius r = 300 mm whose center is given a horizontal velocity v = 1.64 m/s. Determine the angular velocity ω (positive if counterclockwise, negative if clockwise) of the lever OA when x = 880 mm.
Answer:
The angular velocity is
5.64rad/s
Explanation:
This problem bothers on curvilinear motion
The angular velocity is defined as the rate of change of angular displacement it is expressed in rad/s
We know that the velocity v is given as
v= ωr
Where ω is the angular velocity
r is 300mm to meter = 0.3m
the radius of the circle
described by the level
v=1.64m/s
Making ω subject of the formula and solving we have
ω=v/r
ω=1.64/0.3
ω=5.46 rad/s
Consider the following descriptions of the vertical motion of an object subject only to the acceleration due to gravity. Begin with the acceleration equation a(t)equals=v primev′(t)equals=g, where gequals=minus−9.8 m divided by s squared9.8 m/s2. a. Find the velocity of the object for all relevant times. b. Find the position of the object for all relevant times. c. Find the time when the object reaches its highest point. What is the height? d. Find the time when the object strikes the ground. A softball is popped up vertically (from the ground) with a velocity of 30 m divided by s30 m/s.
Answer:
a) [tex]v = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex], b) [tex]s = s_{o} + v_{o}\cdot t - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex], c) [tex]t = \frac{v_{o}}{9.8\,\frac{m}{s^{2}} }[/tex], d) [tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)}+\frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]
Explanation:
a) The acceleration of the object is:
[tex]a = - 9.8\,\frac{m}{s^{2}}[/tex]
The velocity function is found by integration:
[tex]v = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex]
b) The position function is found by integrating the velocity function:
[tex]s = s_{o} + v_{o}\cdot t - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t^{2}[/tex]
c) The time when the object reaches its highest point ocurrs when speed is zero:
[tex]0\,m = v_{o} - \left(9.8\,\frac{m}{s^{2}} \right)\cdot t[/tex]
[tex]t = \frac{v_{o}}{9.8\,\frac{m}{s^{2}} }[/tex]
d) The time when the object hits the ground occurs when [tex]s = 0\,m[/tex]. The roots are found by solving the second-order polynomial:
[tex]t = \frac{-v_{o}\pm \sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot (-9.8\,\frac{m}{s^{2}} )}[/tex]
[tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)} \mp \frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]
Since time is a positive variable and [tex]v_{o} < \sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex], the only possible solution is:
[tex]t = \frac{v_{o}}{2\cdot \left(9.8\,\frac{m}{s^{2}}\right)}+\frac{\sqrt{v_{o}^{2}+4\cdot s_{o}\cdot \left(9.8\,\frac{m}{s^{2}} \right)} }{2\cdot \left(9.8\,\frac{m}{s^{2}} \right)}[/tex]
The physics of vertical motion under gravity deals with velocity, position, and time. Velocity over time can be calculated by integrating the acceleration function, while position over time is calculated by integrating the velocity function. The height and time at the peak of motion, as well as when the object hits the ground, are determined by setting these equations to zero at appropriate points.
Explanation:Considering the physics of vertical motion under gravity, the concepts of velocity, position, and time are important. The acceleration needs to be considered for handling such scenarios, which in your case, has been described by the equation a(t)=v′(t)=g. The value of g=−9.8 m/s², where 'g' represents the acceleration due to gravity.
a. Velocity over time: Using the equation of acceleration a=v′=g, and because g is a constant, the velocity at any point can be determined by integrating the acceleration function. This gives v(t)=gt+c, where the constant c is the initial velocity. For a softball with an initial vertical velocity of 30 m/s, this becomes v(t)=-9.8*t+30.
b. Position over time: The position equation can be found by integrating the velocity function, that is: s(t)=½gt²+vt+s₀. The s₀ term is the initial height, which is 0 because the softball is popped up from the ground. Therefore, s(t)=-4.9*t²+30*t.
c. Time and height at the highest point: The motion reaches its highest point when the velocity is zero. So, set v(t) equal to zero and solve for t. Substituting this time into the position equation gives the maximum height.
d. Time when the softball strikes the ground: The softball will hit the ground when its position is again zero. Set s(t) equal to zero and solve for t to find this time.
Learn more about Vertical Motion here:https://brainly.com/question/12640444
#SPJ3
The lens-makers’ equation applies to a lens immersed in a liquid if n in the equation is replaced by n2/n1. Here n2 refers to the index of refraction of the lens material and n1 is that of the medium surrounding the lens. (a) A certain lens has focal length 79.0 cm in air and index of refraction 1.55. Find its focal length in water. (b) A certain mirror has focal length 79.0 cm in air. Find its focal length in water.
Answer:
a
The focal length of the lens in water is [tex]f_{water} = 262.68 cm[/tex]
b
The focal length of the mirror in water is [tex]f =79.0cm[/tex]
Explanation:
From the question we are told that
The index of refraction of the lens material = [tex]n_2[/tex]
The index of refraction of the medium surrounding the lens = [tex]n_1[/tex]
The lens maker's formula is mathematically represented as
[tex]\frac{1}{f} = (n -1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
Where [tex]f[/tex] is the focal length
[tex]n[/tex] is the index of refraction
[tex]R_1 and R_2[/tex] are the radius of curvature of sphere 1 and 2 of the lens
From the question When the lens in air we have
[tex]\frac{1}{f_{air}} = (n-1) [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
When immersed in liquid the formula becomes
[tex]\frac{1}{f_{water}} = [\frac{n_2}{n_1} - 1 ] [\frac{1}{R_1} - \frac{1}{R_2} ][/tex]
The ratio of the focal length of the the two medium is mathematically evaluated as
[tex]\frac{f_water}{f_{air}} = \frac{n_2 -1}{[\frac{n_2}{n_1} - 1] }[/tex]
From the question
[tex]f_{air }[/tex]= 79.0 cm
[tex]n_2 = 1.55[/tex]
and the refractive index of water(material surrounding the lens) has a constant value of [tex]n_1 = 1.33[/tex]
[tex]\frac{f_{water}}{79} = \frac{1.55- 1}{\frac{1.55}{1.44} -1}[/tex]
[tex]f_{water} = 262.68 cm[/tex]
b
The focal length of a mirror is dependent on the concept of reflection which is not affected by medium around it.
A charge of +3.0 µC is distributed uniformly along the circumference of a circle with a radius of 20 cm. How much external energy is required to bring a charge of 25µC from infinity to the center of the circle?
Answer:
3.37x10^6J
Explanation:
The external energy is given as KQ1Q2/r
=(9*10^9)*(3*10^-3)*(25*10^-3)/(0.2)
=3.375 *10^6 J
The external energy required can be calculated using the electric potential energy formula. The external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.
Explanation:The external energy required to bring a charge of 25µC from infinity to the center of the circle can be calculated using the electric potential energy formula, which is given by:
U = k * (q1 * q2) / r
Where U is the electric potential energy, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, the charge at the center of the circle is 25µC, and the charge distributed uniformly along the circumference is 3.0µC. The radius of the circle is given as 20 cm. Plugging these values into the formula, we have:
U = (9 * 10^9 Nm^2/C^2) * (25 * 10^-6 C) * (3 * 10^-6 C) / (0.20 m) = 337.5 J
Therefore, the external energy required to bring the charge of 25µC from infinity to the center of the circle is 337.5 joules.
Learn more about Electric potential energy here:
https://brainly.com/question/33051432
#SPJ3