Answer:
the points are closer to each other
Explanation:
The expression for the diffraction of a grating is
d sin θ = m λ
sin θ = m λ / d (1)
where d is the distance between slits and m is the order of diffraction, the most general is to work in the order m = 1, the angle te is the angle of diffraction
When we immerse the apparatus in a medium with refractive index n = 1.33, the light emitted by the laser must comply
v = λ f
where v is the speed of light in the medium, the frequency remains constant
velocity and refractive index are related
n = c / v
v = c / n
we substitute
c / n = λf
λ = [tex]\frac{c}{f} \ \frac{1}{n}[/tex]
λ = λ₀ / m
where λ₀ is the wavelength in vacuum
we substitute is equation 1
d sin θ = m λ₀ / n
sin θ = λ₀/ n d
sin θ = [tex]\frac{1}{n}[/tex] sin θ₀
we can see that the value of the sine is redueced since the refractive index is greater than 1,
consequently the points are closer to each other
A linear accelerator can be used to accelerate which of the following?
Question 3 options:
protons and electrons
protons and neutrons
protons only
protons, electrons, and neutrons
According to Newton's law of universal gravitation, the force F between two bodies of constant mass m and M is given by the formula F = G m M d 2 , where G is the gravitational constant and d is the distance between the bodies. a. Suppose that are constants. Find the rate of change of force F with respect to distance d .
Answer:
One can write F = K d^-2 where K = G M m
So dF/dd = -2 K d^-3 = -2 K / d^3 (As d increases F decreases - it is opposite to the direction of F)
Which of the following would change mass as it accelerated? a bullet being shot out of a gun a roller skater pushing off a jet plane taking off a bowling ball slowing down
Answer:
Explanation:
A bullet being shot out of a gun tends to leave tiny amounts of the bullet behind due to friction between the bullet and the gun barrel.
A roller skater pushing requires the conversion of food chemical energy to muscle contraction energy. This conversion increases the body temperature and sweat is excreted to counteract the heat increase. The evaporation of the sweat causes a slight decrease in body mass.
A jet plane taking off consumes some of the fuel carried onboard to provide thrust. The products of combustion become part of the exhaust stream leaving the airplane rearward providing forward thrust.
An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.
Answer:
the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Explanation:
For an object to have a static equilibrium, it must meet two relationships
∑ F = 0
∑ τ =0
force acting on a body fulfills the relation of
sum F = F - F = 0
when two forces do not move from position.
To find the torque we assume that the counterclockwise rotations are positive
Σ τ = - F r - F r
Στ = -2 Fr <> 0
consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium
The area around a charged object that can exert a force on other charged objects is an electric ___
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.
Answer:
The speed of proton is 2.1 x 10^5 m/s .
Explanation:
potential difference, V = 234 V
let the initial speed of the proton is v.
The kinetic energy of proton is
KE = q V
[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]
State the law of conservation of momentum
Explanation:
Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant
A car whose tire have radii 50cm travels at 20km/h. What is the angular velocity of the tires?
Convert to m/s
[tex]\\ \sf\longmapsto v=20\times 5/18=5.5m/s[/tex]
We know
[tex]\boxed{\sf \omega=\dfrac{rv}{|r|^2}}[/tex]
[tex]\\ \sf\longmapsto \omega=\dfrac{0.5(5.5)}{|0.5|^2}[/tex]
[tex]\\ \sf\longmapsto \omega=\dfrac{2.75}{0.25}[/tex]
[tex]\\ \sf\longmapsto \omega=11rad/s[/tex]
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices
Answer:
check photo for solve
Explanation:
A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
the wave frequency is a measure of
Answer: hertz
Explanation:
Wavelength is also measured in metres ( ) - it is a length after all. The frequency ( ) of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz ( ), since one hertz is equal to one wave per second.
Answer:
The number of waves that pass a particular point in a unit of time.
Explanation:
The frequency of a wave describes the number of waves that pass a particular point in a unit of time. The frequency is inversely proportional to the period of the wave, which is the time required for one full wave cycle to complete.
PF
A transparent. dielectric coating is applied to glass (εr = 4.μr=1, σ= 0) to eliminate the reflection of red light (wavelength in air of 750 nm).
a. What is the required dielectric constant and minimum thickness of the coating?
b. If violet light (wavelength in air of 420 nm) is shone onto this glass coating (6-0). what percentage of the incident power will be reflected?
Answer:
a) Dielectric constant ( λ ) = 750 * 10^-9 m
minimum thickness of coating ( d ) = 187.5 nm
b) 3.6%
Explanation:
Given data:
wavelength of red light in air = 750 nm
εr = 4
μr = 1, σ = 0
a) Determine the required dielectric constant and min thickness of coating used
Refractive index of coating ( n ) = √εr * μr = √4*1 = 2
the refractive index of glass( ng) = 1.5 which is < 2
λ = 750 * 10^-9 m
Dielectric constant ( λ ) = 750 * 10^-9 m
To determine the minimum thickness we will apply the formula below
d = m λ/2n ; where m = 1
∴ d = 750 nm / 2 ( 2 )
= 187.5 nm
minimum thickness of coating ( d ) = 187.5 nm
b) Determine the percentage of the incident power that will be reflected
R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2
= [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2
= 0.03628 = 3.6%
A capacitor is connected to an ac generator that has a frequency of 3.2 kHz and produces a rms voltage of 2.0 V. The rms current in the capacitor is 28 mA. When the same capacitor is connected to a second ac generator that has a frequency of 4.7 kHz, the rms current in the capacitor is 70 mA. What rms voltage does the second generator produce
Answer:
The rms voltage of new generator is 3.4 V.
Explanation:
f = 3200 Hz
rms voltage, V = 2 V
rms current, i = 28 mA
Now
f' = 4700 Hz
rms current, i' = 70 mA
let the new rms voltage is V'.
[tex]i = \frac{V}{Xc} = V \times 2\pi fC....(1)\\\\i' = V' \times 2 \pi f' C..... (2)\\\\\frac{i}{i'} =\frac{V f}{V' f'}\\\\\frac{28}{70}=\frac{2\times 3200}{V'\times 4700}\\\\V' = 3.4 V[/tex]
Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 21 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 61 cm. a. What is the wavelength of the sound
Answer:
The answer is "80 cm".
Explanation:
The distance of 21 cm between the speaker's effect of high strength but a spacing of 61 cm corresponds to a zero to zero intensity, that also is, the waves are all in phase with others [tex]\Delta \ x_1 = 21 \ cm[/tex] this is out of phase [tex]\Delta\ x_2 = 61\ cm[/tex]
[tex]\therefore\\\\\Delta\ x_2 -\Delta\ x_1 = \frac{\lambda}{2}\\\\\lambda= 2( \Delta\ x_2 -\Delta\ x_1)[/tex]
[tex]= 2 ( 61\ cm - 21\ cm)\\\\ = 2(40\ cm)\\\\= 80\ cm[/tex]
The spectral lines of two stars in a particular eclipsing binary system shift back and forth with a period of 6 months. The lines of both stars shift by equal amounts, and the amount of the Doppler shift indicates that each star has an orbital speed of 64,000 m/s. What are the masses of the two stars
Answer:
the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
Explanation:
Given the data in the question;
Time period = 6 months = 1.577 × 10⁷ s
orbital speed v = 64000 m/s
since its a circular orbit,
v = 2πr / T
we solve for r
r = vT/ 2π
r = ( 64000 × 1.577 × 10⁷ ) / 2π
r = 1.6063 × 10¹¹ m = ( (1.6063 × 10¹¹) / (1.496 × 10¹¹) )AU = 1.0737 AU
Now, from Kepler's law
T² = r³ / ( m₁ + m₂ )
T = 6 months = 0.5 years
we substitute
(0.5)² = (1.0737)³ / ( m₁ + m₂ )
0.25 = 1.2378 / ( m₁ + m₂ )
( m₁ + m₂ ) = 1.2378 / 0.25
( m₁ + m₂ ) = 4.9512
m₁ = m₂ = 4.9512 / 2 = 2.4756 solar mass
we know that solar mass = 1.989 × 10³⁰ kg
so
m₁ = m₂ = 2.4756 × 1.989 × 10³⁰ kg
m₁ = m₂ = 4.92 × 10³⁰ kg
Therefore, the masses of the two stars are; m₁ = m₂ = 4.92 × 10³⁰ kg
a student standing between two walls shouts once.he hears the first echo after 3 seconds and the next after 5 seconds. calculate the distance between the walls.
Explanation:
It took [tex]t_1 =1.5\:\text{s}[/tex] for the sound to reach the 1st wall and at the same time time, the same sound took [tex]t_2 = 2.5\:\text{s}[/tex] to reach the 2nd wall. Assuming that the sound travels at 343 m/s, then let [tex]x_1[/tex] be the distance of the person to the 1st wall and [tex]x_2[/tex] be the distance to the 2nd wall. So the distance between the walls X is
[tex]X = x_1 + x_2 = v_st_1 + v_st_2 = v_s(t_1 + t_2)[/tex]
[tex]\:\:\:\:\:= (343\:\text{m/s})(4.0\:\text{s}) = 1372\:\text{m}[/tex]
Present the ways to observe interference patterns on thin films. Why the thickness of thin films should be in scale of wavelength?
the president of The proposal ahr stay Sheba SBR Abba and and I send svrvs you svrvs
Which of the following groups is the largest ?
population
community
ecosystem
biome
Answer:
B. Community
Took science classes for 6 years now
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike the ground
Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Trong thực nghiệm nhiễu xạ qua 1 khe hẹp, bề rộng khe là 2.10-5 m, màn quan sát đặt cách khe 0,5 m. Chiếu ánh sáng có bước sóng 480 nm xuyên qua khe và thấy có nền nhiễu xạ trên màn quan sát. Tính bề rộng của vân sáng liền kề vân sáng trung tâm.
Answer:
tyygggghgtyhyrdfgyyyhjjillbxsrfvgygvnjj
Explanation:
cffczhxucuxoyitxohvojcdivbjv ohxc
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
A 15kg mass suspended from a ceiling is pulled aside with a horizontal force, F. Calculate the value of the tension.
Answer:
147 Newtons
Explanation:
To find tension, you can use the formula Tension = (mass)(gravity)
*Gravity's acceleration = 9.8 m/s^2 because of Newton's law of universal gravitation*
T = (15kg)(9.8m/s^2)
= 147 Newtons
Hope this helps! Best of luck <3
Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at a given distance:___________.
A) must be directed inward.
B) acts like it originated in a point charge Q at the center of the distribution.
C) must be directed outward.
D) must be greater than zero.
E) must be zero.
Answer:
Q at the center of the distribution.
Explanation:
The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!
Answer:
y = 1/2at^2
we could also write it as-
y = (at^2)/2
2y = at^2
2y/a = t^2
√2y/a = t
hope it helps
An observer on Earth sees rocket 1 leave Earth and travel toward Planet X at 0.3c. The observer on Earth also sees that Planet X is stationary. An observer on Planet X sees rocket 2 travel toward Earth at 0.4c. What is the speed of rocket 1 according to an observer on rocket 2?
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
[tex]$u'=\frac{u-v}{1-\frac{uv}{c^2}}$[/tex]
Speed of rocket 1 with respect to rocket 2 :
[tex]$u' = \frac{0.4 c- (-0.3 c)}{1-\frac{(0.4 c)(-0.3 c)}{c^2}}$[/tex]
[tex]$u' = \frac{0.7 c}{1.12}$[/tex]
[tex]u'=0.625 c[/tex]
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
Explanation:
Work cannot be increased by using a machine of some kind.
Work cannot be increased by using a machine of some kind.
A machine is any device in which the effort applied at one end overcomes a load at the other end.
Machines are generally used to perform different tasks faster.
However, a simple machine can not be used to increase the amount of work done at any time.
Force, speed and torque can all be increased using machines.
Learn more: https://brainly.com/question/15365822
a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force
Explanation:
Hey there!
According to the question;
A person can lift mass of 60 kg on earth.
mass(m1) = 60kg
acceleration due to gravity on earth (a) = 9.8m/s²
Now;
force (f) = m.a
= 60*9.8
= 588 N
Since, there is application of same magnitude of force on moon,
mass(m) =?
acceleration due to gravity on moon (a) = 1.67m/s²
Now;
force (f) = m.a
588 = m*1.67
m = 352.09 kg
Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.
Hope it helps!
