Answer:
this point we have a destructive inference m=1
Explanation:
This is an interference exercise, where the light has two possible states one of constrictive interference
d sin θ = m λ
and another of destructive interference
d sin θ = (m + ½) λ
the term on the left is the difference in trajectory that tells us that it is worth
d sin θ = 300 nm = 300 10⁻⁹ m
We can find what type of interference we have by finding m, which must be an integer
constructive interference
m = d sin θ /λ
let's calculate
m = 300 10⁻⁹ / 500 10⁻⁹
m = 0.6
this value is not possible because m must be an integer
destructive interference
m + ½ = d sin θ /λ
m = d sin θ /λ + ½
m = 0.6 + 0.5
m = 1.1
that can be approximated to m = 1
this value if possible, so at this point we have a destructive inference
A school bus's velocity decreases from 15 m/s to 5 m/s forwards over a period of 10 s. What is the bus's average acceleration? [Vf=v,+at]
Answer:-1 m/s/s
Explanation:
The acceleration of the bus when its velocity changes from 15 m/s to 5 m/s will be a=1 m/s^2
What is acceleration?Acceleration is defined as the change of the velocity with the time. Acceleration is a vector quantity and is defined by both the magnitude and the direction.
Now from the first equation of motion we have,
[tex]a= \dfrac{v-u}{t}[/tex]
we have V=15 m/s, u=5 m/s , t=10 seconds
[tex]a=\dfrac{15-5}{10}=1\ \frac{m}{s^2}[/tex]
Hence the acceleration of the bus will be a=1 m\s^2
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A model airplane with mass 1.0 kg is held by a wire so that it flies in a horizontal circle with radius 20.0 m. The airplane engine provides a net thrust of 1.0 N perpendicular to the wire. (a) Find the torque the net thrust produces about the center of the circle. (b) Find the angular acceleration of the airplane when it is in this horizontal flight.
Answer:
330
Explanation:
(a) The torque the net thrust produces about the center of the circle is of 20 N-m.
(b) The angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
Given data:
The mass of model airplane is, m = 1.0 kg.
The radius of horizontal circle is, r = 20.0 m.
The magnitude of net thrust by engine is, F = 1.0 N.
(a)
The effort made to turn any object is known as the torque. The mathematical expression for the torque is given as,
T = F × r
Solving as,
T = 1.0 × 20.0
T = 20 N-m
Thus, we can conclude that the torque the net thrust produces about the center of the circle is of 20 N-m.
(b)
The expression for the angular acceleration of airplane during the horizontal flight is given as,
[tex]T = I \times \alpha[/tex]
Here, I is the moment of inertia of airplane and its value is,
[tex]I = \dfrac{1}{2}mr^{2}\\\\\\I = \dfrac{1}{2} \times 1.0 \times 20^{2}\\\\\\I =200 \;\rm kg.m^{2}[/tex]
So, the angular acceleration is,
20 = 200 × α
α = 20/200
α = 0.1 rad/s²
Thus, we can conclude that the angular acceleration of the airplane when it is in this horizontal flight is 0.1 rad/s².
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in what condition do acceleration and deaccelration
Im not sure what are you asking about. But in physic/ mechanic usually when the object against gravity, its deceleration but when the object follow the gravity its acceleration
Also usually when the object move to the left side the speed will be negative (-) this is probably because of deceleration? since deceleration = -acceleration.
6. a. find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm. B. the electron makes a transition from the n
Complete Question
(A) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 nm .
(B) The electron makes a transition from the n=1 to n= 4 level by absorbing a photon. Calculate the wavelength of this photon.
Answer:
A
[tex]\Delta E = 337 \ eV[/tex]
B
[tex]\lambda = 3.439 *10^{-9} \ m[/tex]
Explanation:
Considering question a
From the question we are told that
The width of the box is [tex]w = 0.125 \ nm = 0.125 *10^{-9} \ m[/tex]
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 3\ (Third \ excited \ level ) \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [3^2 - 1^2 ][/tex]
=> [tex]\Delta E = 539 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{539 *10^{-19}}{1.60 *10^{-19}} \ J[/tex]
=> [tex]\Delta E = 337 \ eV[/tex]
Considering question b
Generally the energy level of a particle confined to a box is mathematically represented as
[tex]E_n = \frac{n^2 h^2}{8 m L^2 }[/tex]
Generally the excitation energy is mathematically represented as
[tex]\Delta E = \frac{h^2 }{ 8 m L^2 } [n_2^2 - n_1 ^2 ][/tex]
From the question [tex]n_2 = 4 \ \ and \ \ n_1 = 1[/tex]
Here h is the Planck's constant with a value of [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
m is the mass of electron with value [tex]m = 9.11 *10^{-31} \ kg[/tex]
So
[tex]\Delta E = \frac{[6.62607015 * 10^{-34} ]^2 }{ 8 * (9.11 *10^{-31}) (0.125 *10^{-9})^2 } [4^2 - 1^2 ][/tex]
=> [tex]\Delta E = 578 *10^{-19} \ J[/tex]
=> [tex]\Delta E = \frac{ 578 *10^{-19}}{1.60 *10^{-19 }}[/tex]
=> [tex]\Delta E = 361.45 \ eV[/tex]
Gnerally the wavelength is mathematically represented as
[tex]\lambda = \frac{hc}{\Delta E }[/tex]
=> [tex]\lambda = \frac{ 6.626 *10^{-34} * (3.0 *10^{8})}{578 *10^{-19} }[/tex]
=> [tex]\lambda = 3.439 *10^{-9} \ m[/tex]
a girl weighing 350N runs up hill upto 5m in 20sec. calculate her work done and power
Work done against gravity becomes potential energy of the object-Earth system. Its given by:
W = Ep
W = mgh but weigh(w) = mg
W = wh
W = 350 N × 5 m
W = 1750 JPower is the amount of energy transferred or converted per unit time. This is:
P = W/t
P = 1750 J / 20 s
P = 87.5 Wwith a mass if 74 kg, is a moving with a velocity of 11m/s, what is the kinetic energy ?
Answer:
4477 JExplanation:
The kinetic energy of an object can be found by using the formula
[tex]k = \frac{1}{2} m {v}^{2} \\ [/tex]
m is the mass
v is the velocity
From the question we have
[tex]k = \frac{1}{2} \times 74 \times {11}^{2} \\ = 37 \times 121[/tex]
We have the final answer as
4477 JHope this helps you
A typical donation of whole blood is equal to a pint (473 ml). If the average density of blood rhoBLOOD = 1060 kg/m^3, what is the mass of donated blood?
a. 5.01 kg
b. 0.501 lb
c. 0.501 kg
d. 0.501 g
e. 5.01 g
Answer:
c. 0.501 kg
Explanation:
Given;
volume of the denoted blood, V = 473 ml = 0.000473 m³
density of the denoted blood, ρ = 1060 kg/m³
The mass of donated blood is given as;
mass of the donated blood = density of the denoted blood x volume of the denoted blood
m = 1060 x 0.000473
m = 0.501 kg
Therefore, the mass of donated blood is 0.501 kg
Pls Help physics 8th class question
Answer:
angle of reflection and angle of incident is always equal
Strontium chloride, SrCl2
Is ionic or covalent
Answer:
it is a ionic bond because electrons are transferred from chlorine to strontium
Strontium chloride, SrCl₂, is an ionic compound because Ionic compounds are formed through the transfer of electrons from one atom to another, resulting in the formation of positively charged ions (cations) and negatively charged ions (anions).
In the case of strontium chloride, strontium (Sr) is a metal, and chlorine (Cl) is a non-metal. Strontium readily donates two valence electrons to chlorine, which has a strong affinity for gaining one electron to achieve a stable electron configuration. The electron transfer between strontium and chlorine leads to the formation of S[tex]r^2^+[/tex] cations and C[tex]l^-[/tex] anions. The resulting ionic bond holds the cations and anions together in a crystal lattice structure. The attraction between the oppositely charged ions creates the ionic bond, making strontium chloride an ionic compound.
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______ Is the distance and direction of an object's change in position from its starting point.
(5 Points)
Distance
Displacement
Velocity
Rate
Answer:
Displacement
Explanation:
I took the test.
HELP NOW PLS !!
If you exert 950 N*s of impulse on a 12 kg frictionless cart over the course of 5 seconds, how far will it travel during those seconds?
Answer:
The value is [tex]s =197.88 \ m[/tex]
Explanation:
From the question we are told that
The impulse is [tex]I = 950 \ N\cdot s[/tex]
The mass is [tex]m = 12 \ kg[/tex]
The time t = 5 s
Generally impulse is mathematically represented as
[tex]I = F * t[/tex]
=> [tex]F = \frac{I}{t}[/tex]
=> [tex]F = \frac{950 }{ 5 }[/tex]
=> [tex]F = 190 \ N[/tex]
Generally force is mathematically represented as
[tex]F = m * a[/tex]
=> [tex]a = \frac{F }{ m }[/tex]
=> [tex]a = \frac{190 }{ 12 }[/tex]
=> [tex]a = 15.83 \ m/s^2[/tex]
Generally from kinematic equation , the distance covered is
[tex]s = ut + \frac{1}{2} * at^2[/tex]
Here u is the initial velocity of the cart and the value is [tex]u = m/s[/tex]
=> [tex]s = 0 * 5 + \frac{1}{2} * 15.83 * 5^2[/tex]
=> [tex]s =197.88 \ m[/tex]
Suppose a shrimp has been put on the ground that has just been taken out of water.Now touch the shrimp from a distance by a stick.The shrimp will jump straight upward.Will the shrimp do any work in this case?
Answer:
yes
Explanation:
I would say yes because it's exerting some type of force as it reacts to the stick to make it jump straight in an upwards direction
Select the correct answer.
A perlodic wave travels from one medlum to another. Which palr of varlables are likely to change In the process?
OA velocity and wavelength
OB. velocity and frequency
OC. frequency and wavelength
OD. frequency and phase
OE. wavelength and phase
Answer:
velocity and wavelength likely to change in the process.
Explanation:
When an object passes from one medium to another, refraction occurs. Due to refraction, it changes iits direction. Some of the properties is changed when it undergoes refraction. The velocity of a wave and wavelength change in this process.
Hence, the correct option is (a) "velocity and wavelength".
A 0.7 kg block attached to a spring with force constant 160 Nm is free to move on a frictionless, horizontal surface. The block is released from rest when the
spring is stretched + 0.15 m as shown in figure below. At the instant the block is released the force vector on the block is.
(Consider Right direction is positive direction and Left direction is negative direction)
- 34.29 N
+ 34.29 N
+ 24 N
Оe
h Oa-241
Answer:
+ 24 N
Explanation:
the computation is shown below:
Given that
Mass of the block = m = 0.7 kg
Sprint constant = k = 160 N / m
x = 0.15m
Now the force on the block is
F = kx
= (160) (0.15)
= 24 N
As the instant block is released so the acting of the force on the block is positive and it would be in a positive direction i.e. right direction
Therefore the third option is correct
Which is not true about waves?
O Compression and rarefaction are found in longitudinal waves.
O Sound waves are transverse waves.
O Amplitude, wavelength, and troughs all describe parts of a wave.
O Electromagnetic waves are transverse waves.
Answer:
sound waves are transverse waves
We now have two tappers that are tapping the surface of the water together. They are tapping in sync (both are hitting the surface at the same time). A small cork is sitting in the tank at the position of the small brown circle. As the stars continue to tap and the ripples from each star continue to move outward, what will happen to the cork
Answer:
The cork moves up and down.
Explanation:
The cork moves up and down because of the waves produce due to tapping the surface of water. The cork experience movement due to placed on the water surface so when the wave passes through the medium by tapping of water, movement of cork also occur. Cork has high volume and lower mass so it floats on the water surface and experience movement due to waves.
Find the resultant of the following displacement:
A = 20 Km 30° south of east;
B = 50 Km due west;
C = 40 Km north east;
D = 30 Km 60° south of west.
Answer:
Explanation:
Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees
charge of one of electron is 1.6022×10*-19 coulomb. what is the total charge on 1 mole of electrons
There are
[tex]6.023 \times {10}^{23} [/tex]
electrons in a mole. So charge of 1 mole electron is
[tex]6 \times {10}^{23} \times ( - 1.6022) \times {10 }^{ - 19} [/tex]
Determine the weight of a 5.1 kg scooter that moves with a constant acceleration of 3.0 ms2. (Make sure you use the weight equation: W or Fg = mg)
Answer:
15.3 sorry if i got it wrong
Explanation:
An ambulance is traveling east at 62.4 m/s. Behind it a car travels along the same direction at 34.5 m/s. The ambulance driver hears his siren with a wavelength of 0.47 m. What wavelength would a stationary observer behind the ambulance measure for the sound? The velocity of sound in air is 343 m/s.
Answer:
The answer is "0.5555 m"
Explanation:
Where the reference leaves the list and the viewer is at rest:
[tex]\lambda'=\frac{v-v_s}{v} \times \lambda\\\\[/tex]
[tex]=\frac{343 \frac{m}{s} - (-62.4 \frac{m}{s})}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{343 \frac{m}{s} + 62.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m\\\\ =\frac{405.4 \frac{m}{s}}{343 \frac{m}{s}} \times 0.47 \ m[/tex]
[tex]=0.5555 \ m[/tex]
In 1940, it was discovered that uranium could be artificially changed to plutonium, a process called transmutation. What combination of decay processes would convert U-239 (92 protons) into Pu-239 (94 protons)?
A. Beta-decay followed by beta decay.
B. Positron emission followed by electron capture.
C. Electron capture followed by alpha decay.
D. Alpha decay followed by beta decay.
The combination of decay processes that would convert U-239 into PU- 239 is beta decay followed by another beta decay. There are two different types of beta decay (positive and negative beta decays) they both occur simultaneously in order to move the atoms towards stability.
Which is an example of a crime against a person?
A. Cheating on a test
B. Leaving military service without approval
C. Injuring a stranger
D. Refusing to pay taxes
The example of a crime against a person is Injuring a stranger.
What is the Injuring ?Injury is any physical damage or harm caused to the body, including cuts, scrapes, bruises, broken bones, and burns. It can be caused by an accident, medical negligence, or intentional harm. Injuries can range from mild to severe and can be temporary or permanent. Treatment for an injury often involves rest, medications, physical therapy, and sometimes surgery. In more severe cases, an injury can cause lasting disability or even death. Prevention of injuries is important, and can be achieved through education, wearing protective gear, and avoiding risky activities.
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What is true of the bottom layer of the ocean
Answer:
well i tough it was infinte but its not an the layer is a d i c k
Explanation:
What is the dog's kinetic energy?
16.3 J
12.1 J
19.2 J
It’s 12.1
Answer:12.1 j
Explanation:
A brick is moving at a speed of 3 m/s and a pebble is moving at a speed of 5 m/s. If both objects have the same kinetic energy, what is the ratio of the brick's mass to the rock's mass
Answer:
let M be the mass of brick
let m be the mass of the pebble
.5M(3)^2 =KE
.5m(5)^2= KE
.5M9 = .5m25
9M = 25m
(9/25)M = m
Answer:
A
Explanation:
e=1/2M[tex]V^{2}[/tex]
1/2[tex]M_{1}[/tex]9=1/2[tex]M_{2}[/tex]25
[tex]\frac{M_{1} }{M_{2} }[/tex] =[tex]\frac{25}{9}[/tex]
Which has more mass Helium or mass
given that there is no second choice i have to go with helium. if you get the second choice i will revise my answer
Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalpost and experiences a backward force of 17600 N for 55 ms.
Answer:
0.8 m/s
Explanation:
F(avg) * Δt = I, where
F(avg) = Average force
Δt = change in time
I = impulse
From the question, we know that the
average force is given as -17600 N
time change is given as 55 milliseconds
speed of the player, v = 8 m/s
Mass of the player is given as 110 kg
Impulse, I = F(avg) * Δt
I = -17600 * 0.055
I = -968
Now, using the Impulse-Momentum theory, we have that
Δp = I and thus,
Δp = m [v(f) - v(i)], where
Δp = change in the momentum
v(f) = final speed of the player
v(i) = initial speed of the player
Substituting the values, we have
I = m [v(f) - v(i)]
-968 = m.v(f) - m.v(i)
m.v(f) = m.v(i) - 968
110v(f) = 110 * 8 - 968
110v(f) = 880 - 968
110v(f) = -88
v(f) = -88 / 110
v(f) = -0.8 m/s
A boat is moving due east in a region where the earth's magnetic field is 5.0×10⁻⁵ NA⁻¹ m⁻¹ due north and horizontal. The boat carries a vertical aerial 2 m long. If the speed of the boat is 1.50 ms⁻¹ , the magnitude of the induced emf in the wire of aerial is:
Answer: Induced emf is given by:
ε= Bvl
On putting the values we get
=5×10
−5
×1.50×2
=0.15mV
Explanation:
Hope these helped have a wonderful Christmas break ❄️
As a freely falling object speeds up, what is happening to its acceleration - does it increase, decrease, or stay the same? (a) Ignore air resistance. (b) Consider air resistance.
Essay ANSWER
Answer:
(a)When ignoring air resistance its accelerating increases steadily .
(b)When considering air resistance then its acceleration decreases this could either be uniformly or unevenly.
Hope this helped.
Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as much as Jack's power output. four times Jack's power output. twice Jack's power output. one-half as much as Jack's power output. the same as Jack's power output.
Answer:
Second Choice.
Explanation:
Jack's Power = W/t
Jill's Power = 2W/(0.5)*t
2/0.5 = 4
Jill's Power = 4*W/t
Jill's Power is 4 times greater than Jack's
Second Choice