Answer:
the value of the boltzman constant 8.61 73303 into 10 ratio by - 5 EV upon knev
Aluminium has a work function of 4. 08 ev. (a) find the cutoff wavelength and cutoff frequency for the photoelectric effect?
Aluminium has a work function of 4. 08 eV. The cutoff wavelength and cutoff frequency for the photoelectric effect is 303.9nm and 911.7× 10¹⁷ s⁻¹ respectively.
Work Function is the minimum energy required to eject an electron from a photoelectric material.
Cutoff Wavelength is the maximum wavelength below which electron will be ejected
Cutoff Frequency is the minimum frequency required to eject electron.
Let the work function of Aluminium be Φ
Given, work function Φ = 4.08eV
We know that hc/λ = Φ
where, h is Planck constant
c is speed of light
λ is wavelength of light used
Hence, on subsitution
1240 / λ = 4.08 eV (hc = 1240)
⇒ λ = 303.9 nm
Hence, cutoff wavelength used is 303.9 nm
We know that ν = c/λ
ν = 3 × 10⁸ / 303.9
⇒ ν = 911.7 × 10¹⁷ s⁻¹
Hence, cutoff frequency used is 911.7 × 10¹⁷ s⁻¹
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If the signal light changes from green to yellow as you enter the intersection, you should:_______
When the light turns from green to yellow while you are already in the intersection, proceed through the intersection at a safe speed.
To find the answer, we have to study more about the importance of traffic signals.
What are the importance of yellow signal light in traffic?Recognize that you have the right-of-way when a yellow light is flashing. Reduce your speed as you get close to a flashing yellow light and keep an eye out for other cars that might not be paying attention to or respecting your right-of-way. You should safely stop your car at any intersection where there is a constant yellow light. When the light turns from green to yellow while you are already in the intersection, proceed through the intersection at a safe speed.Thus, we can conclude that, When the light turns from green to yellow while you are already in the intersection, proceed through the intersection at a safe speed.
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I need help with these 2 physics questions ASAP thanks (Note, ignore problem 6 as I have solved it)
The final velocity of the ball is 20.8 m/s. Option C
What is the final speed?Now we have the equation;
v^2 = u^2 + 2gh
v = final velocity
u = initial velocity
a = acceleration
s = distance
v^2 = (11.4)^2 + (2 * 10 * 15.5)
v^2 = 129.96 + 310
v = √ 129.96 + 310
v = 20.8 m/s
b)
Given that;
R = u^2sin2θ/g
Where;
g = acceleration due to gravity
θ = acute angle
u = velocity
R = (60)^2 sin2(30)/10
R = 310 m
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As a gaseous element condenses, the atoms become ________ and they have ________ attraction for one another.
Answer:
blank 1: close together
blank 2: more
An electron moving perpendicular to a magnetic field of 3. 2 × 10-2 t moves in a circle of radius 0. 40 cm. how fast is this electron moving?
An electron moving perpendicular to a magnetic field of 3. 2 × 10-2 t moves in a circle of radius 0. 40 cm. this electron is moving with 2.25 x 10⁷ m/s
The magnetic influence on moving electric charges, electric currents, and magnetic materials is described by a magnetic field, which is a vector field. A force perpendicular to the charge's own velocity and the magnetic field acts on it when the charge is traveling through a magnetic field.
A magnetic field is defined as a location in space close to a magnet or an electric current where a physical field is formed by a moving electric charge acting as a force on another moving electric charge. The magnetic field of the Earth is an illustration of a magnetic field. The region around a magnet where the effects of magnetism are felt is known as the magnetic field.
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Which method of measurement would be accurate but lack precision?
A. Measuring a volume of water by estimating the size of the
container
B. Measuring a volume of water using a graduated cylinder that can
be read to nearest mL
C. Measuring air temperature with a thermometer that has just been
taken out of hot water
D. Measuring air temperature with a thermometer that can be read to
the nearest degree and is calibrated to the correct temperature
Answer:
B
Explanation:
reading the volume of water in a graduated cylinder which can be read to the nearest mL is accurate, it lacks precision due to the bottom meniscus formed.
the bottom meniscus may cause a wrong reading due to refraction of light
Approximating the eye as a single thin lens 2. 70 cm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cm?
The focal Length of the eye is 0.37cm.
The retina, which is always 2.70 cm away from the lens, serves as the image's primary imaging medium. Image distance is 2.70 cm as a result.
The object is located 265 cm away from the eye's lens.
Based on lens formula:
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
where the object distance is u, the image distance is v, and the focal length is f.
Consequently, u is 265.00 cm and v is 2.70 cm.
[tex]\frac{1}{f} = \frac{1}{265} + \frac{27}{10}[/tex]
[tex]\frac{1}{f} = \frac{7165}{2650}[/tex]
[tex]f = \frac{2650}{7165}[/tex]
f = 0.37
Thus, the focal length of the eye is 0.37cm.
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A small fan in a closed insulated room releases heat at the rate of 42 watts and pushes the air at the rate of 9 m/s with a mass flow rates of 1.2 kg/s. The room has a heater that produces heat at the rate of 0.30 kJ/s as well as a computer that produces 65 watts. Light bulbs in the room produce up to 125 watts. The room loses 0.32 kJ/s. Calculate the amount of heat maintained in the room.
The refrigerator is removing 300 watts of heat every hour.
Energy can only be changed in form; it cannot be created or destroyed, according to the basic law of thermodynamics. For any system, energy transfer examples include mass crossing the control boundary, external work, or heat transfer across the barrier. These have an impact on the energy reserves of the control volume.
The rate of heat removal from within the refrigerator may be calculated using the formula below thanks to the First Law of Thermodynamics and the definition of a refrigeration cycle.
Rate of heat transfer to the space, measured in watts.
Q=800 W W=500
The refrigerator removes heat at a rate of (800-500) 300 watts per hour.
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The aeroplane then continues at a CONSTANT VELOCITY for a further 800 m before leaving the runway. The length of the runway is 2 000 m. 8 m-s-1 1 .2 1.3 → Define the term vector. Convert 67 m-s¹ to km.h¹. 67 m-s¹ Calculate the: .3.1 3.2 1.3.3 3.4 30 s 800 m Acceleration of the aeroplane during the first 30 seconds Distance travelled by the aeroplane during the first 30 seconds Time taken by the aeroplane to travel the 800 m Length of the runway NOT USED when the aeroplane leaves the runway (2) (1 (4 (4 (3
800 = 0 + [tex]\frac{1}{2}[/tex]a(30)[tex]^{2}[/tex]
800 = 450a
a = 1.77m[tex]s^{-2}[/tex]
Distance = s = ut + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]s = 0+ [tex]\frac{1}{2}[/tex]1.77[tex]t^{2}[/tex]
s = [tex]\frac{1}{2}[/tex]1.77(30)[tex]^{2}[/tex]
s = 796.5m
Time taken = distance / velocityt = 800/8
t = 100 sec
Length of the runway not used = total length of runway-total length covered before leaving runway2000 -800
length = 1200m
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3. [-/2 Points]
A race car traveling northward on a straight, level track at a constant speed travels 0.760 km in 21.0 s. The return trip over the
same track is made in 26.0 s.
DETAILS SHIPPS13 2.E.010. MY NOTES
ASK YOUR TEACHER
(a) What is the average velocity of the car in m/s for the first leg of the run?
m/s
(b) What is the average velocity (in m/s) for the total trip?
m/s
PRACTICE ANOTHER
(a) The average velocity of the car in m/s for the first leg of the run is 36.2 m/s.
(b) The average velocity (in m/s) for the total trip is 0.
Average velocity
The average velocity of the car in m/s for the first leg of the run is calculated as follows;
Average velocity = total displacement/total time
Average velocity = (760) / (21) = 36.2 M/S
Average velocity for total tripThe average velocity (in m/s) for the total trip is calculated as follows;
v = total displacement/total time
v = 0/(26 + 21)
v = 0
Thus, the average velocity of the car in m/s for the first leg of the run is 36.2 m/s.
The average velocity (in m/s) for the total trip is 0.
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A satellite is 1200 k above the earth (radius = 6400 k); it completes one revolution in 90min. what is the linear speed?
A satellite is 1200 k above the earth (radius = 6400 k); it completes one revolution in 90min. what is the linear speed?
530.58 (1200+6400)*pi*2/90
What is linear speed?The measurement of a moving object's actual distance traveled is called linear speed. Linear speed is the rate of motion of an object along a straight line. In plain English, it is the distance traveled along a linear path in the allotted amount of time.
A pseudovector used in physics to express how quickly the angular location or orientation of an item changes over time is called an angular velocity or rotational velocity The pseudovector's direction is normal to the instantaneous plane of rotation or angular displacement, and its magnitude corresponds to the object's angular speed, or the rate at which it rotates or revolves. The right-hand rule is a common way to specify the direction of angular velocity.
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An electron is traveling at 0. 85c. what is its mass? (the rest mass is 9. 11 × 10-31 kg. )
The mass of the electron is [tex]1.72\times 10^{-30}[/tex] Kg.
How can we calculate the mass of the electron?To calculate the mass of the electron we are using the formula,
[tex]m=\frac{m_{0} }{\sqrt{1-\frac{v^{2} }{c^{2} } } }[/tex]
Here we are given,
[tex]m_0[/tex]=The rest mass of an electron = [tex]9. 11 \times 10^{-31}[/tex] kg.
v= The velocity of electron = 0. 85c.
c= The velocity of light = c. (c=[tex]3\times 10^8[/tex] m/s)
We have to found the mass of an electron = m Kg.
Now we substitute the known values, we found that,
[tex]m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-\frac{(0.85c)^{2} }{c^{2} } } }[/tex]
Or,[tex]m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-(\frac{0.85c}{c})^2 } }[/tex]
Or,[tex]m=\frac{9. 11 \times 10^{-31} }{\sqrt{1-({0.85})^2 } }[/tex]
Or, [tex]m=1.72\times 10^{-30}[/tex] Kg.
From the above calculation we can conclude that, The mass of the electron is [tex]1.72\times 10^{-30}[/tex] Kg.
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When starting a foot race, a 89 kilogram sprinter exerts an average force of 518 newtons backward on the ground for 0.75 seconds. what is his final speed in meters/second at the end of this action?
The final speed of the sprinter is 4.8 m/s.
What is newtons second law of motion?According to the second law, the mass of the item or the net force acting on it both affect how quickly an object accelerates.
An object's acceleration is directly proportional to the net force applied on it and inversely proportional to its mass.
An object's acceleration increases as the amount of force exerted on it does.
A decreasing acceleration is caused by a rise in an object's mass.
Calculation of final velocity;
The sprinter exerts an average force of 518 newtons backward on the ground.
The taken by sprinter is 0.75 sec.
The wight of the sprinter is 89Kg.
According to Newton's second law
Force = mass×acceleration
F = m×a
a = F/m
a = 518/89
a = 6.4 m/sec²
Now, apply the equation of motion of straight line;
final velocity = initial velocity + acceleration×time
As, the sprinter starts from rest; initial velocity = 0.
Substitute all the values;
final velocity = 0 + 6.4×0.75
= 4.8 m/sec
Therefore, the final velocity of sprinter comes out to be 4.8 m/sec.
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how is that velocity got like this? pls explain
Answer:
if a car is increasing it's acceleration uniformly in a unit time, the graph will be moving away from it's origin. that's how you get this kind of graph.
On a frictionless, horizontal air track. the glider has a head-on collision with a 0. 300-kg glider that is moving to the left with a speed of 2. 20 m>s. Find the final velocity?
The final velocity is v[tex]_{2}[/tex] =-0.2m/s , v[tex]_{1}[/tex] = -3.2 m/s
Given data,
Mass of glider A (M[tex]_{1}[/tex]) = 0.15 kg
Mass of glider B (M[tex]_{2}[/tex]) = 0.3 kg
Initial velocity of A (u[tex]_{1}[/tex]) = 0.80ms-1
Initial velocity of B ( u[tex]_{2}[/tex]) = -2.2m/s
Momentum and kinetic energy are conserved in elastic collision . So,
M[tex]_{1}[/tex]u[tex]_{1}[/tex]+M[tex]_{2}[/tex]u[tex]_{2}[/tex] = M[tex]_{1}[/tex]v[tex]_{1}[/tex]+M[tex]_{2}[/tex]v[tex]_{2}[/tex]
0.15×0.8+0.3x(-2.2) = 0.15v[tex]_{1}[/tex]+0.3v[tex]_{2}[/tex]
-0.54 = 0.15v[tex]_{1}[/tex]+0.3v[tex]_{2}[/tex]
Again if,
u[tex]_{1}[/tex]+v[tex]_{1}[/tex] =u[tex]_{2}[/tex]+v[tex]_{2}[/tex]
0.8 +v[tex]_{1}[/tex] =-2.2+v[tex]_{2}[/tex]
v[tex]_{1}[/tex]-v[tex]_{2}[/tex] = -3
Solving for -0.54 = 0.15v[tex]_{1}[/tex]+0.3v[tex]_{2}[/tex] and v[tex]_{1}[/tex]-v[tex]_{2}[/tex] = -3 , we get
v[tex]_{2}[/tex] =-0.2m/s , v[tex]_{1}[/tex] = -3.2 m/s
Therefore,The final velocity is v[tex]_{2}[/tex] =-0.2m/s , v[tex]_{1}[/tex] = -3.2 m/s
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Calculate the wavelength of an electron (m = 9. 11 × 10^-28 g) moving at 3. 66 × 10^6 m/s
The wavelength of an electron (m = 9. 11 × 10^-28 g) moving at 3. 66 × 10^6 m/s will be 0.197 * [tex]10^{-12}[/tex] m
De Broglie wavelength is an important concept while studying quantum mechanics. The wavelength (λ) that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle's de Broglie wavelength is usually inversely proportional to its force.
momentum (p) = mv
DE Broglie wavelength = lambda = h / p
where
h = plank's constant
m = mass of electron
v = velocity of electron
lambda = h / mv
= 6.633 * [tex]10^{-34}[/tex] / 9. 11 × [tex]10^{-28}[/tex] * 3. 66 × [tex]10^{6}[/tex]
= 0.197 * [tex]10^{-12}[/tex] m
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Two point charges of 20. 0 μc and -8. 00 μc are separated by a distance of 20. 0 cm. what is the intensity of electric field e midway between these two charges? group of answer choices
The intensity of electric field E midway between these two charges is 2.52*10^7N/C.
To find the answer, we have to know more about the electric field.
How to find the electric field?We have the expression for net electric field as,[tex]E=\frac{kQ}{r^2}[/tex]
where, k is the constant equal to 8.99*10^9.
Thus, we can write the electric field created by 20micro coulomb charge at the midway between two charges is,[tex]E_1=\frac{8.99*10^9*20*10^{-6}}{10*10^{-2}} =1.79*10^7N/C[/tex]
We can write the electric field created by -8micro coulomb charge at the midway between two charges is,[tex]E_2=\frac{8.99*10^9*8*10^{-6}}{(10*10^{-2})^2} =7.19*10^6N/C[/tex]
Thus, the net field at the midway due to two charges will be,[tex]E=1.79*10^7+7.19*10^6=2.52*10^7N/C[/tex]
Thus, we can conclude that, the intensity of electric field e midway between these two charges is 2.52*10^7N/C.
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What is the work generated by a healthy adult who circulates 9 l of blood through the brachial artery in 10 min?
120kJ work is generated by a healthy adult who circulates 9L of blood through the brachial artery in 10 min.
What is mean by work?Work is defined as the energy that is applied to or removed from an object by applying force along a displacement. It is frequently described in its most basic form as the result of force and displacement.Calculation of Work generatedA flow rate of 900 mL per minute is equal to a flow of 9 L in 10 minutes.
Additionally, you have 200W of total power generated.
9L Equals 9000 mL
900mL/min x 9,000mL/10min
By multiplying the power (watt) by the time, we can derive the work performed by the adult.
10 min = 600s
200W x 600s = 120 kJ
Hence, the work generated is 120kJ.
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differemces between liquid pressure and air pressure . give this answer is box
Answer:
The key difference between air pressure and liquid pressure is that air pressure allows the gaseous state of matter to be compressible, whereas liquid pressure makes a liquid incompressible.
Liquid pressure is the pressure that we can observe in a liquid. Air pressure is also known as atmospheric pressure, and it is the pressure as the force exerted by the collisions of particles in the air.
Give one example of a thermodynamic cycle that does not account for the carnot efficiency.
Thermo-Electrochemical converter (UTEC) is a thermodynamic cycle that does not account for the Carnot Efficiency.
What is the Carnot cycle in thermodynamics?
An ideal closed thermodynamic cycle that is reversible and consists of the four steps of isothermal expansion to a desired point, adiabatic expansion to a desired point, isothermal compression, and adiabatic compression back to the starting state.
What is the purpose of Carnot cycle?
The Carnot cycle, first proposed by the French engineer Sadi Carnot in the early 19th century, is the optimum cyclical sequence of changes in pressure and temperature of a fluid, such as a gas utilized in an engine. It serves as a benchmark for all heat engines operating in the range of high and low temperatures.Learn more about Carnot Efficiency.
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What was the major problem with the hubble space telescope when it was first launched into orbit?
Answer:
Apparently, a measuring rod had been used incorrectly and the telescope was not focusing properly - changes were later made to the telescope to correct this problem.
The colors of a soap bubble or oil slick are due to __________. two-slit interference thin-film interference huygens' principle diffraction
Answer:
Thin film interference
A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg
initially at rest. The trains collide. After the collision the green train moves with a speed of 3 m/s. What is the final momentum of the blue train?
A. 90 kgm/s
B. 200 kgm/s
C. 110 kgm/s
D. 20 kgm/s
Answer:
C
Explanation:
Use conservation of momentum
Momentum before collision = mv = 50 (4) = 200 kg-m/s
AFTER collision
30 * 3 + Blue momentum = 200
blue momentum = 110 kg-m/s
What angle (in degrees) is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity to one-fifth of its initial value?
The 60 degrees is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity to one-fifth of its initial value.
It is given that axis of a polarizing filter to cut its intensity to one-fifth of its initial value.
It is required to find the angle between the direction of polarized light and the axis of a polarizing filter to cut its intensity to one-fifth of its initial value.
What is the angle between the direction of polarized light and the axis of a polarizing filter?Suppose the angle between the polarizer and the axis of filter is θ.
The intensity of light that is passing after the filter is 0.2 l₀.
From the law of Malus, we have
I = I₀ [tex]cos^{2}[/tex]θ
0.2I₀= I₀ [tex]cos^{2}[/tex]θ
0.2 = [tex]cos^{2}[/tex]θ
[tex]cos\\[/tex]θ = 0.447
θ = 60°
Thus the angle between the direction of polarized light and the axis of a polarizing filter is 60 degree.
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(Repost) I need help with this physics question. Thanks in advance! Answer ASAP.
(a) The gravitational force received by each 1 kg mass is 8.66 N.
(b) The magnitude of gravitational acceleration is 8.66 m/s².
(c) The orbital speed of the ISS is 7,663.6 m/s.
(d) The time take for the ISS to orbit round the Earth is 5,558.75 seconds which is equal to 1.54 hours.
Gravitational force received by each 1 kg massThe gravitational force received by each 1 kg mass is calculated as follows;
F = Gm₁m₂/r²
where;
m₁ is mass of Earthm₂ is mass of ISSr is the distance between the ISS and center of EarthF = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 1) / (6780,000)²
F = 8.66 N
Magnitude of gravitational accelerationmg = GMm/r²
g = GM/r²
where;
M is mass of Earthg = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ ) / (6780,000)²
g = 8.66 m/s²
Orbital Speed of the ISSv = √GM/r
v = √(6.67 x 10⁻¹¹ x 5.97 x 10²⁴ / 6780,000)
v = 7,663.6 m/s
Time of motion of the ISS round the EarthT = 2πr/v
T = (2π x 6780,000) / (7663.6)
T = 5,558.75 seconds
1 hour = 3600 seconds
= 5,558.75/3600
= 1.54 hours
Thus, the gravitational force received by each 1 kg mass is 8.66 N.
The magnitude of gravitational acceleration is 8.66 m/s².
The orbital speed of the ISS is 7,663.6 m/s.
The time take for the ISS to orbit round the Earth is 5,558.75 s = 1.54 hours.
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A crowbar of 2m is used to lift an object of 800N if the efforts arm is 160cm clalculate the efforts applied
The applied effort represents the time devoted to the analysis of theory, attendance at theory classes, and the authentic performance of all practical skills.
How to calculate efforts applied?The applied effort represents the time devoted to the analysis of theory, attendance at theory classes, and the authentic performance of all practical skills.
The effort, application, endeavor, and exertion imply activities directed or force expended toward a definite end. Effort exists as an expenditure of energy to perform some objective: He created an effort to control himself. The application exists continuous struggle plus careful attention: constant application to duties.
Length of crow bar = 2m
length of load arm = 0.1M
length effort arm = length of crow bar -length of load arm
=2−0.1=1.9m
so, velocity ratio = length of effort arm/length of load arm = 1.9/0.1 = 19
use formula, velocity ratio = load/effort
19 = 100kgf/effort
effort = 100/19= 5.26 kgf
hence,5.26 effort is required.
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What is the total mass of the earth's atmosphere? (the radius of the earth is 6.4 ´ 106 m, and atmospheric pressure at the surface is 105 n/m2.)
The total mass of the earth's atmosphere is 5×[tex]10^{8}[/tex] kg.
How do you calculate the total mass of the earth's atmosphere ?To calculate the total mass of the earth's atmosphere we use the expression
[tex]P=\frac{F}{A}[/tex]
or, [tex]F=P\times A[/tex]
or,[tex]M\times g=P\times A[/tex]
or,[tex]M=\frac{P\times 4\times \pi\times r^{2} }{g}[/tex]
Here, P= Atmospheric pressure at the surface = [tex]10^{5}[/tex] N/m2
F= Force at the earth surface.
A= Area of the earth.
r= Radius of the earth=6.4 × [tex]10^{6}[/tex] m
g= Acceleration due to gravity. = 9.8 m/s2 ≈ 10 m/s2
Let, M be the total mass of the earth's atmosphere.
Now, [tex]M=\frac{10^{5}\times4\times\pi\times(6.4\times10^{6} )^{2} }{10}[/tex]
M= 5×[tex]10^{8}[/tex] kg.
Thus from the above calculation we can show that, the total mass of the earth's atmosphere is 5×[tex]10^{8}[/tex] kg.
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An fm radio station broadcasts at 98. 6 mhz. What is the wavelength of the radiowaves?.
The wavelength of radiowaves is 3.042 m.
What is radiowaves?The electromagnetic spectrum's longest wavelengths, which are found in radio waves, are normally found at frequencies of 300 gigahertz and below.
Frequency given, f = 98.6 MHz = 98.6 x [tex]10^{6}[/tex] cycles/second
Electromagnetic waves including radiowaves also travel at the speed of light.
Therefore, c = 3.0 x [tex]10^{8}[/tex] m/sec
Wavelength = speed/frequency
wavelength of 98.6 MHz = 3.0 x [tex]10^{8}[/tex]/98.6 x [tex]10^{6}[/tex] meters
=3.042 meters
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Please help, I don't understand lol
The resulting velocity of marble B after the collision is 2.4 m/s.
The object with the faster velocity, and thus larger momentum, will impart more energy to the slower object during the collision than vice versa. Following the collision, the object with the lower beginning velocity will travel away from the other object with a higher speed and momentum.
Marble A's mass, marble B's mass, marble B's initial velocity, marble A's final velocity, marble B's final velocity, and marbles A and B's combined mass and initial velocity
Applying the mass-conservation principle
Momentum is conserved.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.08 kg)(0.5 m/s) + (0.05 kg)(0 m/s) = (0.08 kg)(-0.1 m/s) + (0.05 kg) v
0.04 kg m/s = -0.08 kg m/s + (0.05 kg) v
0.12 kg m/s = (0.05 kg) v
v = 2.4 m/s
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The first time that astronomers observed both gravitational waves and electromagnetic waves from the same event, what they were observing was:_____
Higher frequencies are present in more dramatic events and have thus been the first to be noticed, but the frequencies of ordinary gravitational waves are relatively low and considerably more difficult to detect.
A gamma-ray burst (GRB), which was discovered by the orbiting Fermi gamma-ray burst monitor on 2017 August 17 at 12:41:06 UTC, triggered an automatic notice throughout the world in addition to a merger of black holes. Six minutes later, a gravitational-wave observatory in Hanford, Washington, detected a gravitational-wave candidate that occurred 2 seconds before the gamma-ray explosion.
This collection of data supports the merger of two neutron stars, as shown by a multi-messenger transient event that was detected by gravitational waves as well as electromagnetic (gamma-ray burst, optical, and infrared) spectrum observations.
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