Answer:
a) 4.17*10^-4 nm
b) 0.044 nm
c) 4.35*10^-17 joules
d) 271.875 eV
Explanation:
Given that
angle of scatter = θ = 34°
To find the change in wavelength of the photon, we use the relation
Δλ = λ(f) - λ(i) = [h/me*c] (1 - cos θ)
Δλ = [6.67*10^-34/9.1*10^-31*3*10^8] (1 - cos 34)
Δλ = 2.44*10^-12 * 0.171
Δλ = 4.17*10^-13 m = 4.17*10^-4 nm
B
To find the wavelength of scattered light, we have
Δλ = λ(f) - λ(i) = 4 17*10^-4 nm
λ(f) = λ(i) + 4.17*10^-4 nm, now we substitute the value if λ(i), to get
λ(f) = 4.36*10^-2 nm + 4.17*10^-4 nm
λ(f) = 0.044 nm
C
To find the change in energy, we use our wavelengths in m, instead of nm. The formula is thus
hc/λ(i) - hc/λ(f) = hc[λ(f) - λ(i)]/λ(f) *λ(i)
ΔE = 20.01*10^-26 [4.17*10^-13] / (0.044*10^-9 * 0.0436*10^-9)
ΔE = 8,344*10^-38 ÷ 1.918*10^-21
ΔE = 4.35*10^-17 joules
D
On converting to eV, we have
ΔE = 4.35*10^-17 /1.6*10^-19
ΔE = 271.875 eV (elevtron volt)
Can someone tell me what a free fall is
Answer:
free fall is any motion of a body where gravity is the only force acting upon it. In the context of general relativity, where gravitation is reduced to a space-time curvature, a body in free fall has no force acting on it
Explanation:
hope this helped
Answer:A freefall is what you experience when you go skydiving and that is because you are falling out of the sky
Explanation:
Two tectonic plates moving toward one another are at a
ANSWER CHOICES
convergent boundary.
divergent boundary.
subduction boundary.
transform boundary.
Answer:
A. cause i just took the test
Explanation:
Answer:
its A
Explanation:
no explanation is needed, just trust me.
How much energy is required to move a 1000 kg object from the Earth's surface to an altitude twice the Earth's radius?
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
Since the object must be moved away to a distance greater than the radius of the Earth, then change in gravitational potential energy must be based on Newton's Law of Gravitation.
By the Work-Energy Theorem, the work ([tex]W[/tex]), in joules, done on the object is equal to the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]W = U_{g}[/tex] (1)
[tex]W = -G\cdot m\cdot M\cdot \left(\frac{1}{r_{f}}-\frac{1}{r_{o}} \right)[/tex] (1b)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.[tex]m[/tex] - Mass of the object, in kilograms.[tex]M[/tex] - Mass of the Earth, in kilograms.[tex]r_{o}[/tex] - Initial distance, in meters.[tex]r_{f}[/tex] - Final distance, in meters.If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]m = 1000\,kg[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex], [tex]r_{o} = 6.371\times 10^{6}\,m[/tex] and [tex]r_{f} = 19.113\times 10^{6}\,m[/tex], then the energy required to move the object from the Earth's surface is:
[tex]W = -\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1000\,kg)\cdot (5.972\times 10^{24}\,kg)\cdot \left[\frac{1}{19.113\times 10^{6}\,m} - \frac{1}{6.371\times 10^{6}\,m} \right][/tex][tex]W = 4.171\times 10^{10}\,J[/tex]
An energy of [tex]4.171\times 10^{10}[/tex] joules to move a 1000-kg object from the Earth's surface to an altitude twice the Earth's radius.
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To remove a stain using a solvent the stain has to become dissolved in the solvent
True
False
Answer:
True
Explanation:
have a good day:)
Answer: This statement is True
A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what isthe current in the primary?and WHy?
Answer:
The correct solution will be "12.0 A".
Explanation:
The given values are:
[tex]N_p= 500 \ turn[/tex]
[tex]N_s= 200 \ turn[/tex]
[tex]I_s= 3.0 \ A[/tex]
By using the transformer formula, we get
⇒ [tex]\frac{N_p}{N_s} =\frac{I_s}{I_p}[/tex]
⇒ [tex]I_p = I_s\times \frac{N_s}{N_p}[/tex]
On substituting the given values, we get
⇒ [tex]=3.0 \ A\times \frac{2000}{500}[/tex]
⇒ [tex]=12.0 \ A[/tex]
Someone help me pls
Explanation:
The dog was stationary at segment c
Answer:
I don't know if I'm wrong but I'm pretty sure stationary means that the thing is still. I would go with C And maybe D????
Which best describes a characteristic of an adiabatic process?
Answer:
Can you please provide the choices? For now this is all I can give you.Explanation:
In thermodynamics, an adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings.Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work. It also conceptually supports the theory used to explain the first law of thermodynamics and is therefore a key thermodynamic concept.
The system takes in heat but does not release it. If an adiabatic process is reversible, there is no change in entropy; otherwise, there is a rise in entropy, or degree of disorder.
What is adiabatic process?
A thermodynamic process known as an adiabatic process takes place when there is no exchange of mass or heat between the system and its surroundings.
An adiabatic process only transfers energy to the environment as work, in contrast to an isothermal process.
The system's overall heat remains constant since there is no heat exchange with the environment.
It is a fundamental idea in thermodynamics and theoretically supports the theory that underlies the first law of thermodynamics.
Therefore, no heat flow in and out of system, which is a characteristic of an adiabatic process.
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A boy on a bicycle rides in a circle of radius ro at speed vo. If the boy now rides at a radius equal to half the initial radius ro, by what approximate factor must he change his speed in order to have the same radial acceleration
Answer:
The speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
Explanation:
The radial or centripetal acceleration is given by:
[tex] a_{c} = \frac{v^{2}}{r} [/tex]
Where:
v: is the speed = v₀
r: is the radius = r₀
[tex] a_{c} = \frac{v_{0}^{2}}{r_{0}} [/tex] (1)
If the radius is now equal to half the initial radius the speed must be:
[tex]a_{c} = \frac{v^{2}}{r_{0}/2}[/tex] (2)
By equating equation (1) and (2):
[tex] \frac{v_{0}^{2}}{r_{0}} = \frac{v^{2}}{r_{0}/2} [/tex]
[tex]v^{2} = \frac{v_{0}^{2}}{2}[/tex]
[tex] v = \frac{v_{0}}{\sqrt{2}} [/tex]
Therefore, the speed must change by [tex]\frac{1}{\sqrt{2}}[/tex] factor in order to have the same radial acceleration.
I hope it helps you!
Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time interval?
Answer:
answer is 11.76 meter
Explanation:
use 2nd equation of motion
S=ut+1/2at^2
please help i’ll mark u branliest
Answer:
62
Explanation:
it doesn't need explanation
A jeweler's grinding wheel slows down at a constant rate from 185 rad/s to 105 rad/s while it rotates through 16.0 revolutions. How much time does this take?
Answer:
t = 0.6933 s
Explanation:
This is a rotational kinematics exercise, let's find the angular acceleration of the wheel
w² = w₀² + 2 α θ
α = (w² - w₀²) / 2 θ
Let's reduce the angles to the SI system
θ = 16 rev (2π rad / 1 rev) = 32π rad
let's calculate
α = (105² - 185²) / (2 32π)
α = -115.39 rad / s²
now let's use the relation
w = w₀ + α t
t = (w- w₀) /α
t = (105 - 185) / (- 115.39)
t = 0.6933 s
please answer this question
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
Required:
Find the minimum uncertainty in the momentum of the object. Find the minimum uncertainty in the object's velocity, assuming that the object is (b) a golf ball (mass=0.045 kg) and (c) an electron.
Answer:
[tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
[tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
[tex]0.31\times 10^{-4}\ \text{m/s}[/tex]
Explanation:
[tex]\Delta x[/tex] = Uncertainty in position = 1.9 m
[tex]\Delta p[/tex] = Uncertainty in momentum
h = Planck's constant = [tex]6.626\times 10^{-34}\ \text{Js}[/tex]
m = Mass of object
From Heisenberg's uncertainty principle we know
[tex]\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}[/tex]
The minimum uncertainty in the momentum of the object is [tex]2.78\times 10^{-35}\ \text{kg m/s}[/tex]
Golf ball minimum uncertainty in the momentum of the object
[tex]m=0.045\ \text{kg}[/tex]
Uncertainty in velocity is given by
[tex]\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]6.178\times 10^{-34}\ \text{m/s}[/tex]
Electron
[tex]m=9.11\times 10^{-31}\ \text{kg}[/tex]
[tex]\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}[/tex]
The minimum uncertainty in the object's velocity is [tex]0.31\times 10^{-4}\ \text{m/s}[/tex].
where does a solid material go when a solution is made
Answer:
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Explanation:
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answer this plzzzzzzzzzzzzz
As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65 degrees above the horizontal 45 N at an angle of 65 degrees above the horizontal.
500 kg
The horizontal component of the force acting on the crate is?
Answer:
19.01 N
Explanation:
F = Force being applied to the crate = 45 N
[tex]\theta[/tex] = Angle at which the force is being applied = [tex]65^{\circ}[/tex]
Horizontal component of force is given by
[tex]F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}[/tex]
The horizontal component of the force acting on the crate is 19.01 N.
Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface
1)-96 cm
2)-24 cm
3)-48 cm
4)-72 cm
Answer:
Option 3: -48 cm
Explanation:
We are given:
refractive index; n = 1.5
radius of curvature; r2 = 24 cm
Formula for the focal length is given as;
1/f = (n - 1) × [(1/r1) - (1/r2)]
As r1 tends to infinity, 1/r1 = 0
Thus,we now have;
1/f = (n - 1) × (-1/r2)
Plugging in the relevant values;
1/f = (1.5 - 1) × (-1/24)
1/f = -0.02083333333
f = -1/0.02083333333
f = -48 cm
Jerry runs 60 meters east and then 20 meters west in 10
seconds. His average velocity is
m/s.
Answer: 8 meters per second
Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s
Helicopters rotor blades, could spin at high speed of 510 rpm. Find the angular displacement in radian for 3 hour(s) operation
Answer:
The angular displacement of the blade is 576,871.2 radians
Explanation:
Given;
angular speed of the Helicopters rotor blades, ω = 510 rpm (revolution per minute)
time of motion, t = 3 hours
The angular speed of the Helicopters rotor blades in radian per second is given as;
[tex]\omega = \frac{510 \ rev}{mins} *\frac{2 \pi \ rad}{1 \ rev} *\frac{1 \ min}{60 \ s}\\\\\omega = 53.414 \ rad/s[/tex]
The angular displacement in radian is given as;
θ = ωt
where;
t is time in seconds
θ = (53.414)(3 x 60 x 60)\\
θ = 576,871.2 radians
Therefore, the angular displacement of the blade is 576,871.2 radians
A 2 kg toy car moves at a speed of 5 m/s.
What is the kinetic energy of the car?
Answer:
The answer is 25 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]KE = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]KE = \frac{1}{2} \times 2 \times {5}^{2} \\ = 1 \times {5}^{2} [/tex]
We have the final answer as
25 JHope this helps you
A hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250 m/s, by how much will it miss the target
Answer:
The distance the bullet will miss the target is 1.13 m.
Explanation:
Given;
Initial velocity of the bullet = 250 m/s
Distance of the target = 120 m
Time of motion;
t = 120 / 250
t = 0.48 s
During this time the bullet is under the gravitational pull and the distance it will miss the target is given by;
Y = V₀y + ¹/₂gt²
where;
V₀y is the initial vertical velocity = 0
Y = 0+ ¹/₂gt²
Y = ¹/₂(9.8)(0.48)²
Y = 1.13 m
Therefore, the distance the bullet will miss the target is 1.13 m.
The asteroid 234 Ida has a mass of about 4 × 1016 kg and an average radius of about 16 km. What is the acceleration due to gravity on 234 Ida? Assume that the asteroid is spherical; use G = 6.67 × 10–11 Nm2/kg2.
A. 1 cm/s2
B. 2 cm/s2
C. 5 cm/s2
D. 6 cm/s2
Answer:
1 cm/s²
Explanation:
I just took the quiz
The asteroid 234 Ida has a mass of about 4×1016 kg and an average radius of about 16 km. The acceleration due to gravity will be 1.04 cm/s². Hence, option A is correct.
What is the acceleration due to gravity?The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s² is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on the earth's surface is 9.8 m/s².
The formula for the acceleration due to gravity is g=GM/r².
According to the question, the given values are :
Mass, M = 4 × 1016 kg or
M = 4 × 10¹⁶.
Radius, r = 16 km or,
r = 16000 meter.
G = 6.67 × 10⁻¹¹ Nm²/kg²
g = (6.67 × 10⁻¹¹ ) (4 × 10¹⁶) / 16000²
g = 0.0104 m/s² or,
g = 1.04 cm/s².
Hence, the acceleration due to gravity will be 1.04 m/s²
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You compress a spring by x, and then release it. Next you compress the spring by 2x. How much more work did you do the second time than the first
Answer:
Work done is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Explanation:
The work done by the spring is the same as the potential energy stored in the spring.
So that,
work done = potential energy = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
where k is the spring constant of the material of the spring, and x is the compression.
When the spring is compressed by x;
work done = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
When the spring is compressed by 2x;
work done = [tex]\frac{1}{2}[/tex] k[tex](2x)^{2}[/tex]
= [tex]\frac{1}{2}[/tex] k(4[tex]x^{2}[/tex])
= 2k[tex]x^{2}[/tex]
Therefore,
The work done the second time more than the first = 2k[tex]x^{2}[/tex] - [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
= [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex]
The work done the second time more than the first is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Spring work is equivalent to the effort done to extend the spring, Work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
What is spring work?Spring work is equivalent to the effort done to extend the spring, which is dependent on both the spring constant k and the distance stretched.
The potential energy stored as a result of the deformation of an elastic item, such as spring stretching, is referred to as elastic potential energy.
Work done by spring = potential energy
[tex](PE)_{spring }= \frac{1}{2} Kx^2[/tex]
Case 1
spring is compressed by x
[tex](PE)_1{spring }= \frac{1}{2} Kx^2[/tex]
Case 2
spring is compressed by 2x
[tex]\rm (PE)_2{spring }= \frac{1}{2} K(2x)^2\\\\\rm (PE)_2{spring }= \frac{1}{2}\times 4K(x)^2\\\\\rm (PE)_2{spring }= 2K(x)^2[/tex]
The difference in the potential energy is found by;
[tex](PE)_2-(PE)_1=2Kx^2-\frac{1}{2} Kx^2=\frac{3}{2} Kx^2[/tex]
Hence spring work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
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I need help with this science work
what happens to the grvatational fore between two objects as they move fartger apart
Answer:
♡ madeline here ♡
heres your answer
This force of gravitational attraction
is depending on the masses of both
objects and contrarily corresponding
to the square of the distance that divided
their middle parts. This means that as you
move away from an object the gravitational
force lowers. i hope i could help you! ☆
have a great day!
- madeline/madi ✧･ﾟ: *✧･ﾟ:･ﾟ✧*:･ﾟ✧･ﾟ
Explanation:
An object that is oscillating on a spring is given by the equation x = (10.0 cm) cos[(6.00 s-1)t]. At what value of t after t = 0.00 s is the cart first located at x = 8.00 cm?
Answer:
[tex]t=0.0107\ \text{s}[/tex]
Explanation:
[tex]x=10\cos(6t)[/tex]
Now [tex]x=8\ \text{cm}[/tex]
Substituting the value of [tex]x[/tex] in the equation we get
[tex]8=10\cos6t[/tex]
[tex]\Rightarrow 0.8=\cos6t[/tex]
[tex]\Rightarrow \cos^{-1}0.8=6t[/tex]
[tex]\Rightarrow t=\dfrac{\cos^{-1}0.8}{6}[/tex] the values here are used found in radians
[tex]\Rightarrow t=0.0107\ \text{s}[/tex]
So, at [tex]t=0.0107\ \text{s}[/tex] the value of [tex]x=8\ \text{cm}[/tex].
A 150 kg boy and his bike are traveling 12 m/s when he slams on his breaks and stop at his friend’s house. How much impulse is required to produce this change in momentum?
Please someone help me with this I’ll give brainliest
Answer:
J = 1800 kg-m/s
Explanation:
Given that,
Mass of a boy, m = 150 kg
Initial velocity of a boy, u = 12 m/s
Finally, it stops, v = 0
We need to find the impulse is required to produce this change in momentum. We know that impulse is equal to the change in momentum. So,
[tex]J=m(v-u)\\\\=150\times (0-12)\\\\=-1800\ kg-m/s\\\\|J|=1800\ kg-m/s[/tex]
So, the impulse is equal to 1800 kg-m/s
Help!!
A 30-N force is applied to a 4-kg object to move it with a constant
velocity of 2 m/s across a level surface. The coefficient of friction
between the object and the surface is approximately (Use the
approximation: g - 10 m/s/s.)
A 0.20
B O 0.50
C 0.55
D 0.75
Answer:
[tex]\mu=0.75[/tex]
Explanation:
Given that,
Force acting on an object, F = 30 N
Mass of the object, m = 4 kg
It is moving with a constant velocity of 2 m/s across a level surface.
We need to find the coefficient of friction between the object and the surface. Let it is μ. Force in terms of coefficient of friction is given by :
F = μ N, Where N is normal force, N = mg
[tex]\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{30}{4\times 10}\\\\\mu=0.75[/tex]
So, the coefficient of friction between the object and the surface is 0.75.
Which is one of Edwin Hubble’s findings that supports the big bang theory?
Answer:
Edwin Hubble found that galaxies are constantly moving away from us. According to his observations with the Hubble Space Telescope, galaxies are moving at different speeds. This shows that the universe is expanding. The farther away a galaxy is, the faster it is moving away. Found this on google hope this helps.
Answer:
A) the universe started at a central point
Explanation:
taking the quiz on eg. :))
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4. Determine the magnitude of force at point and determine if the ladder will slip.
This question is incomplete, the complete question;
The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.
Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb
Answer:
- the magnitude of force at point A is 79.1033 lb
- since FA < FA_max; Ladder WILL NOT slip
Explanation:
Given that;
∑'MA = 0
⇒ NB [Lsin∅] - W[L/2.cos∅] = 0
NB = W / 2tan∅ -------let this be equation 1
∑Fx = 0
⇒ FA - NB = 0
FA = NB
therefore from equation 1
FA = NB = W / 2tan∅
we substitute in our values
FA = NB = 76 / 2tan(60°) = 21.9393 lb
Now ∑Fy = 0
NA - W = 0
NA = W = 76 lb
Net force at A will be
FA' = √( NA² + FA²)
= √( (W)² + (W / 2tan∅)²)
we substitute in our values
FA' = √( (76)² + (21.9393)²)
= √( 5776 + 481.3328)
= √ 6257.3328
FA' = 79.1033 lb
Therefore the magnitude of force at point A is 79.1033 lb
Now maximum possible frictional force at A
FA_max = μ × NA
so, FA_max = 0.4 × 76
FA_max = 30.4 lb
So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.
Therefore since FA < FA_max; Ladder WILL NOT slip