Answer:
a. 600 nm
b. 300 nm
Explanation:
According to the question it is given that there is two flat glass surface that is separated by 150nm air gap
Based on the above information
The computation is shown below
a. The bright thickness is
[tex]t = \frac{\lambda}{2} (m - \frac{1}{2} )\\\\\lambda = \frac{2t}{(m - \frac{1}{2} )} \\\\\frac{2(150)}{(1 - \frac{1}{2} )} \\\\[/tex]
= 600 nm
b. The dark thickness is
[tex]t = \frac{m\lambda }{2} \\\\\lambda = \frac{2t}{m} \\\\\frac{2(150)}{1}[/tex]
= 300 nm
Degree day calculations are based on a temperature of?
Answer:
65F
Explanation:
Consider two balls on a horizontal, frictionless surface. Ball A (with mass ma) is moving toward ball B (with mass mb = 1 kg). Ball B is initially at rest.a. (5 pts) If their collision is perfectly elastic, what should be the mass of ball A such that it stops upon collision if ball A's initial velocity is +3.1416 m/s?b. (5 pts) Using conditions in question (a), what is the speed of ball B after collision? Is it moving in the initial direction of ball A or opposite it?c. (5 pts) Consider this 2nd scenario: If the collision is perfectly inelastic, what is the speed of the balls after collision? [Use the mass that you calculated from question (a) for ball A.]d. (5 pts) Now, consider a 3rd scenario: Ball A (with arbitrary speed va) collides with a stationary ball B (mass is mb = 1 kg). After collision, you observed that ball A and ball B are now moving at the same speed va but toward opposite directions: Ball A is heading opposite its original direction, while B is moving along the initial direction of A. What is the mass of ball A?
Answer:
Explanation:
When two ball of identical mass collides perfectly elastically , there is exchange of velocity between the two balls .
Here ball be was at rest initially . After collision ball A comes to rest , so there is complete exchange of velocity . Hence ball A must have same mass as that of B . mass of ball A = 1 kg .
b ) Due to complete exchange of velocity , velocity of ball A will be picked up by ball B . Hence velocity of ball B = 3.1416 m /s . Yes it will be moving in the direction of ball A .
c )
In case of perfectly inelastic collision they will become single mass
total mass = 2 kg
applying conservation of momentum law
their common velocity after collision = 1 x 3.1416 / 2 = 1.57 m /s
d )
Applying conservation of momentum law
initial momentum = Ma x va
they move in opposite direction after collision
their total momentum after collision
1 x va - Ma va
applying law of conservation of momentum
1 x va - Ma va = Ma va
va = 2Ma va
Ma = .5 kg .
Something that has many particles in a small space would have a ————- density
Fill in blank
Answer:
a high density i believe
Explanation:
Answer:
it may have a higher density
At the equator, the earth spins a distance of 25,992 miles every day. What speed does the earth spin at in mph?
At the equator, the earth spins a distance of 25,992 miles every day, which is equal to 1083 mph.
What is equator?
An equator is a hypothetical line that circles the center of a planet or other celestial body. It is located at 0 degrees latitude, midway between the north and south poles. The globe is divided into northern and southern hemispheres by an equator.
The diameter of the Earth is also larger at the Equator, resulting in a phenomenon known as an equatorial bulge. A circle's diameter is measured by a straight line that goes through its center and has its ends on the circle's edge. The diameter of latitudes such as the Equator and the Arctic Circle can be calculated by scientists.
At the equator, the earth spins a distance of 25,992 miles every day
This is equal to (25992 ÷ 24) miles per hour
= 1083 mph.
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A bat moving at 3.7 m/s is chasing a ying insect. The bat emits a 36 kHz chirp and receives back an echo at 36.79 kHz. At what speed is the bat gaining on its prey? Take the speed of sound in air to be 340 m/s.
Answer:
The speed the bat is gaining on its prey is 0.03m/s
Explanation:
Given;
speed of the bat, v₀ = 3.7 m/s
frequency of the bat, F₀ = 36 kHz
frequency of the source, Fs = 36.79
This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.
Apply the following equation to determine the speed of the insect which is the source;
[tex]F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s[/tex]
The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s
Therefore, the speed the bat is gaining on its prey is 0.03m/s
The half-lives for the radioactive decay of various elements are listed below.
Parent Isotope
Daughter Isotope
Half Life (years)
Carbon-14
Nitrogen-14
5730
Uranium-235
Lead-207
710 million
Potassium-40
Argon-40
1.25 billion
a) Suppose you discovered a meteorite that contains a small amount of potassium-40 and its decay product argon-40. You determine that 1/8 of the original potassium-40 remains; the other 7/8 has decayed into argon-40. How old is the meteorite?
b) A piece of cloth was painted with organic dye 34380 years ago. What percentage of the original carbon-14 in the organic dye remains after all this time?
c) The oldest minerals found on Earth are zircons. One isotope used to measure their age is uranium-235. A single crystal contained 20 nanograms of uranium-235 when it first formed. How much uranium-235 does the crystal still contain if it formed 2.84 billion years ago?
Answer:
D ko alam pasensya ka na ha
An 4.0-kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting
from rest. Assuming the acceleration is constant, find the magnitude
and direction of the net force acting on the fish during this interval.
Answer:
2.64N
Explanation:
Force = mass * acceleration
Given
mass = 4kg
distance = 1.9m
Time t = 2.4s
Get the acceleration using the equation of motion
S = ut + 1/2at²
1.9 = 0 + 1/2a(2.4)²
1.9 = 5.76a/2
1.9 = 2.88a
a = 1.9/2.88
a = 0.66m/s²
Get the magnitude of the force
Force = 4 * 0.66
Force = 2.64N
Hence the net force acting on the fish is 2.64N
You are driving down the road heading north at 15 m/s. An ambulance is driving in the opposite direction (south) towards you at 25 m/s. If the ambulance's siren emits sound at a frequency of 2 kHz, what frequency do you hear? Assume that the speed of sound is 340 m/s. a. 1780 Hz b. 1945 Hz c. 2254 Hz d. 2063 Hz e. 2000 Hz
Answer:
d. 2063 Hz
Explanation:
Given that the source of the sound (the ambulance) is heading towards the observer, we have;
fL= (v ± vL/v ± vS) fS
Where;
v = speed of sound = 340 m/s
vL = velocity of the listener = 15 m/s
vS = velocity of the source = 25 m/s
fS = frequency of source = 2 kHz
Since the source is moving towards the observer we subtract;
Substituting values;
fL = (340 - 15/340 - 25)2 *10^3
fL = 2063Hz
The frequency of the siren heard by the driver is approximately 2254Hz.
Hence, Option C) 2254Hz is the correct answer.
Given the data in the question;
Velocity of observer; [tex]v_o = 15m/s[/tex]Velocity of source; [tex]v_s = 25m/s[/tex]Actual frequency (source); [tex]f = 2kHz = 2000Hz[/tex]Velocity of sound; [tex]v = 340m/s[/tex]Frequency heard by observer; [tex]f' = \ ?[/tex]
To find the frequency heard by the observer, we Doppler Effect Equation:
[tex]f' = [ \frac{v+v_o}{v - v_s} ] f[/tex]
Where [tex]f'[/tex] is the observed frequency, [tex]f[/tex] is the actual frequency, [tex]v[/tex] is the velocity of sound, [tex]v_o[/tex] is the velocity of the observer and [tex]v_s[/tex] is the velocity of the source.
We substitute our given values into the equation
[tex]f' = [ \frac{340m/s\ +\ 15m/s}{340m/s\ -\ 25m/s} ] 2000Hz\\\\f' = [ \frac{355m/s}{315m/s} ] 2000Hz\\\\f' = 2253.968Hz\\\\f' = 2254Hz[/tex]
The frequency of the siren heard by the driver is approximately 2254Hz.
Hence, Option C) 2254Hz is the correct answer.
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