Answer:
3.77x10^-5T
Explanation:
Magnetic field at center of the loop is given as
B=uo*I/2r =(4pi*10-7)*6/2*0.1
B=3.77*10-5Tor 37.7 uTi
A certain shade of blue has a frequency of 7.06×1014 Hz. What is the energy E of exactly one photon of this light? Planck's constant h=6.626×10−34 J⋅s.
Answer:
Energy, [tex]E=4.67\times 10^{-19}\ J[/tex]
Explanation:
It is given that,
Frequency of blue shade is, [tex]f=7.06\times 10^{14}\ Hz[/tex]
We need to find the energy of exactly one photon of this light. The formula that is used to find the energy of photon is given by :
[tex]E=nhf[/tex]
Here, n is number of photon, n = 1
h is Planck's constant
So,
[tex]E=1\times 6.626\times 10^{-34}\times 7.06\times 10^{14}\\\\E=4.67\times 10^{-19}\ J[/tex]
So, the energy of this light is [tex]4.67\times 10^{-19}\ J[/tex].
a box container both cube and five side pyramid the total number of objects is 17 the total number of side for the cube and pyramid is 95 how many cubes are in the box
Answer:
There are 10 cubes in the box
Explanation:
Let the total number of cubes be x and the total number of pyramids be y
Since there are 17 total objects;
Then;
x + y = 17 •••••••••(i)
Also, the total number of sides of cubes is 6 * x = 6x ( a single cube has 6 sides)
For the pyramid we have 5 * y = 5y
adding both gives the total number of sides
6x + 5y = 95 •••••• (ii)
From i, we cabs say y = 17-x
plug this in ii
6x + 5(17-x) = 95
6x + 85 -5x = 95
6x-5x = 95-85
x = 10
At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light to (a) (1/3), (b) (1/10)
Answer:
35.3°
18.4°
Explanation:
a.
The first polariser polarises the unpolarised light reducing its intensity from I0 to I0/2. We have to reduce the intensity from I0/2 to I0/3.
Using to Law of Malus, I=I0cos²θ
cos²θ=I/I0=(I0/3)/I0/2 ,
cosθ=√2/3−−√=0.6667−−−−−√=0.8165
θ=cos−1(0.8165)=35.3∘
B.
Cos²θ=I/Io =Io/10/Io9
Cosθ= √9/10= 0.9487
= cos−10.9487
=18.4°
(a) The angle of polaroid such that intensity reduces by 1/3 is 35.26°
(b) The angle of polaroid such that intensity reduces by 1/10 is 63.43°
Angle of polarisation:According to the Malus Law: The intensity of light when passing through a polarizer is given by:
I = I₀cos²θ
where θ is the angle of the polarizer axis with the direction of polarization of the light
I₀ is the initial intensity
When an unpolarised light passes through a polarizer, θ varies from 0 to 2π, so the intensity after passing the first polarizer is :
I = I₀<cos²θ> { average of cos²θ, for 0<θ<2π}
I = I₀/2
Now, this emerging light passes through a second polarizer such that:
(a) the intensity is I' = I₀/3
From Malus Law:
I' = Icos²θ
I₀/3 = (I₀/2)cos²θ
cos²θ = 2/3
θ = 35.26°
(b) the intensity is I' = I₀/10
From Malus Law:
I' = Icos²θ
I₀/10 = (I₀/2)cos²θ
cos²θ = 1/5
θ = 63.43°
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For this study, the researcher is analyzing data by using __________ measures. A. experimental B. qualitative C. quantitative D. naturalistic
Answer:
B. qualitative
Explanation:
Since in the question it is mentioned that it takes the teacher opinion with respect to the student lunchroom that arranged the opinion in four classifications and according to that the report is also written
So this research represents the quality of the data collected as it does not specify the rating in terms of liking the student lunchroom
Therefore the correct option is B. qualitative
Three fish of equal volume are swimming in a fish tank. They all have the same volume, but A has the greatest mass, B has less, and fish C has even less mass. (The differences are significant.)
The fish are swimming as shown. At that instant, how do the buoyant forces exerted by the water on the three fish rank?
A. BFB > BFA = BFC
B. BFB = BFA > BFC
C. BFB > BFA > BFC
D. BFA = BFB = BFC
E. Some other ranking
If the fish stopped swimming, fish B would remain right where it is shown.
What would happen to fish A and C? Put all the correct answers. If none of these would happen, put N.
A. Fish C would be pushed down to the bottom by the water above it.
B. Fish C would float to the top.
C. Fish A would be pushed to the top by the water below it.
D. Fish A would sink to the bottom.
Answer:
Fish C would float to the top & Fish A would sink to the bottom
Explanation:
A would sink to the bottom.Since B just remains where it is shown, its mass must balance the buoyant force and its weight must be equal to the weight of the water displaced. Since A is heavier, its weight will be greater than its buoyant force, it will sink. Since C is lighter, its buoyant force will be greater thanits weight and it will float to the top.
For a particle executing SHM with an amplitude ‘r’, the kinetic energy will be equal to the potential energy when the displacement is equal to-
Answer:
x = A sin w t displacement in SHM
v = A w cos w t velocity in SHM
PE = 1/2 k x^2 = 1/2 k A^2 sin^2 w t
KE = 1/2 m v^2 = 1/2 m w^2 A^2 cos^2 w t
If KE = PE then
k sin^2 w t = m w^2 cos^2 w t
sin^2 wt / cos^2 w t = tan^2 w t = m w^2 / k
but k / m = w^2
So tan^2 w t = 1 and tan w t = 1 or w t = pi / 4 or theta = 45 deg
Then x = r sin w t = r sin 45 = .707 r
To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?
Answer:
Vi = 2 m/s
Explanation:
First we find the force applied to the car by wall to stop it. We use Hooke's Law:
F = kx
where,
F = Force = ?
k = spring constant = 4 x 10⁶ N/m
x = compression = 3.18 cm = 0.0318 m
Therefore,
F = (4 x 10⁶ N/m)(0.0318 m)
F = 127200 N
but, from Newton's Second Law:
F = ma
a = F/m
where,
m = mass of car = 2010 kg
a = deceleration = ?
Therefore,
a = 127200 N/2010 kg
a = 63.28 m/s²
a = - 63.28 m/s²
negative sign due to deceleration.
Now, we use 3rd equation of motion:
2as = Vf² - Vi²
where,
s = distance traveled = 3.18 cm = 0.0318 m
Vf = Final Speed = 0 m/s
Vi = Initial Speed = ?
Therefore,
2(- 63.28 m/s²)(0.0318 m) = (0 m/s)² - Vi²
Vi = √4.02 m²/s²
Vi = 2 m/s
Two stars of masses M and 6M are separated by a distance D. Determine the distance (measured from M) to a point at which the net gravitational force on a third mass would be zero.
Answer:
0.29D
Explanation:
Given that
F = G M m / r2
F = GM(6m) / (D-r)2
G Mm/r2 = GM(6m) / (D-r)2
1/r2 = 6 / (D-r)2
r = D / (Ö6 + 1)
r = 0.29 D
See diagram in attached file
What would you predict would occur if you were able to
place a rotating magnet near a coil of wire?
Will occur if you were able to place a rotating magnet near a coil of wire is continually rotating magnetic field would continuously induce current.
When is the operator faster approaching or moving away from the magnetic flux?
When the magnet moves close to the coil or rapidly increasing until the magnet is within the flux of the coil. As it passes through the coil, the magnetic flux through the coil starts to decrease. Consequently, an induced EMF is inverted.
Whenever there is a relative movement between the loop and the magnet, regardless of who moves, an electric current, called induced current, appears in the loop.
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A car starts from rest and accelerates at a constant rate after the car has gone 50 m it has a speed of 21 m/s what is the acceleration of the car
Answer:
4.41 m/s^2
Explanation:
(v_f)^2 - (v_i)^2 = 2a * change in distance
(21)^2 - (0)^2 = 2a * 50
a = (21^2)/(2*50)
a = 4.41 m/s^2
An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is
Answer:
The magnetic field's direction is towards the north
Explanation:
The force on a positive charge in a uniform magnetic field is represented by the right hand rule. To determine the direction of the force, place your right hand with your palm up, and your thumb at 90° to the other fingers. If the fingers represent the magnetic field, and the thumb the direction of the positive charge, then the palm will push up in the direction of the force. But a negative charge like an electron pushes in exactly the opposite direction. So if you follow this rule, you will find that the magnetic field points towards the north.
The direction of the magnetic field is towards the North. This can be
determined using the right hand rule by Fleming.
The right hand rule states that to determine the direction of the magnetic
force, the right thumb should be pointed in in the direction of the velocity,
index finger in the direction of the magnetic field and middle finger in the
direction of magnetic force.
When this is applied, we will discover that the index finger will point towards
the north region.
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The speed of sound in air is 340 m/s, and the density of air is 1.2 kg/m3. If the displacement amplitude of a 330-Hz sound wave is 14 µm, what is its pressure-variation amplitude?
I bel.ieve the answer is 279Ghz
The required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.
Given data:
The speed of sound in air is, v = 340 m/s.
The density of air is, [tex]\rho = 1.2 \;\rm kg/m^{3}[/tex].
The frequency of sound wave is, f = 330 Hz.
The displacement amplitude of sound wave is, [tex]A = 14 \;\rm \mu m= 14 \times 10^{-6} \;\rm m[/tex].
The standard expression for the pressure variation amplitude for the sound wave propagating in air medium is,
[tex]\Delta P= B \times A \times K[/tex]
Here,
B is the Bulk Modulus and its value is, [tex]B = \rho \times v^{2}[/tex].
K is the wave form constant and its value is, [tex]K = \dfrac{2 \pi f}{v}[/tex].
Solving as,
[tex]\Delta P= (\rho \times v^{2}) \times A \times \dfrac{2 \pi f}{v}\\\\\Delta P= (\rho \times v) \times A \times (2 \pi f)\\\\\Delta P= (1.2 \times 340) \times (14 \times 10^{-6}) \times (2 \pi \times 330)\\\\\Delta P= 11.84 \;\rm Pa[/tex]
Thus, we can conclude that the required value of pressure-variation amplitude of the given sound wave is 11.84 Pa.
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A 68.5kg astronaut floating motionless next to the space station throws a 2.25kg tool away from her at 3.20m/s. With what speed and direction will the astronaut begin to move?
Answer:
-0.105 m/s
Explanation:
Given that
Mass of the astronaut, m(a) = 68.5 kg
Mass of the tool, m(t) = 2.25 kg
Speed of the tool after it is thrown, v(t) = 3.20 m/s
We know that momentum of a particle,
p = mv
See the attachment for calculations
Therefore, the speed is 0.105 m/s and it moves in the opposite direction.
Astronauts increased in height by an average of approximately 40 mm (about an inch and a half) during the Apollo-Soyuz missions, due to the absence of gravity compressing their spines during their time in space. Does something similar happen here on Earth
Answer:
Yes. Something similar occurs here on Earth.
Explanation:
Gravity tends to pull objects perpendicularly to the ground. In space, the absence of this force means there is no compression on the spine due to gravity trying to pull it down. This means that astronauts undergo an increase in height in space.
Here on Earth, we experience gravity pull on our spine during the day. At night when we sleep, we lie down with our spine parallel to the ground, which means that our spine is no longer under compression from gravity force. The result is that we are a few centimetres taller in the morning when we wake up, than we are before going to bed at night. The increase is not much pronounced here on Earth because there is a repeated cycle of compression and decompression of our spine due to gravity, unlike when compared to that of astronauts that spend long duration in space, all the while without gravity forces on their spine
The process of star and planet formation begins with a large cloud of gas and dust called a solar nebula. Rank the formation events that occur within a cloud from earliest to latest.
Rank from earliest to latest. To rank items as equivalent, overlap them.
A. The cloud is large, cool, and slowly rotating
B. The cloud collapses into a disk
C. Competing rotational and gravitational forces begin to flatten the cloud
D. The cloud becomes denser, heats up, and rotates faster
E. The cloud starts to contract under the influence of gravity
A, B, E , C, D
What is Nebula?A nebula is an enormous cloud of dust and gas occupying the space between stars and acting as a nursery for new stars.
Nebulae are made up of dust, basic elements such as hydrogen and other ionized gases.
Nebula Formation:
In essence, a nebula is formed when portions of the interstellar medium undergo gravitational collapse.
Mutual gravitational attraction causes matter to clump together, forming regions of greater and greater density.
The formation events that occur within a cloud from earliest to latest are:
A. The cloud is large, cool, and slowly rotating
B. The cloud collapses into a disk.
E. The cloud starts to contract under the influence of gravity
C. Competing rotational and gravitational forces begin to flatten the cloud.
D. The cloud becomes denser, heats up, and rotates faster
Therefore , The rank from earliest to latest is A, B, E , C, D
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When the magnet falls toward the copper block, the changing flux in the copper creates eddy currents that oppose the change in flux. The resulting braking force between the magnet and the copper block always opposes the motion of the magnet, slowing it as it falls. The rate of the fall produces a rate of flux change sufficient to produce a current that provides the braking force. If the copper is cooled with liquid nitrogen, the resistivity of the copper drops dramatically. How will this affect the speed at which the magnet falls toward the copper
Answer:
The speed at which the magnet falls through the copper block will be reduced dramatically.
Explanation:
Eddy's current are loops of electrical current induced within conductors by a changing magnetic field in the conductor. Eddy's current is proportional to the the magnetic field strength, the rate of change of flux, the area of the loop, and is inversely proportional to the resistivity of the material. Eddy currents flows perpendicularly to the magnetic field, and in closed loops within conductors.
Reducing the resistivity of the copper will increase the Eddy current on the copper, which will in turn increase the opposition to the action producing the flux change (the falling magnet through the copper block). The result is that the speed at which the magnet falls through the copper block will be reduced dramatically.
Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy U from the light. The linear momentum these objects will receive is such that:
Answer:
absorbent p = S / c
reflective p = 2S/c
Explanation:
The moment of radiation on a surface is
p = U / c
where U is the energy and c is the speed of light.
In the case of a fully absorbent object, the energy is completely absorbed. The energy carried by the light is given by the Poynting vector.
p = S / c
in the case of a completely reflective surface the energy must be absorbed and remitted, therefore there is a 2-fold change in the process
p = 2S/c
define the term DNA
Answer:
It is the carrier of genetic information.
An automobile of mass 2500 kg moving at 49.4 m/s is braked suddenly with a constant braking force of 8,868 N. How far does the car travel before stopping
Answer:
344.68 m
Explanation:
The computation of the far does the car travel before stopping is
Data provided in the question
Force = F = 8,868 N
mass = m = 2,500 kg
So,
accleration = a is
[tex]= \frac{-F}{m}\\\\\= \frac{8868}{2500}[/tex]
a = -3.54 m/s^2
The initial speed = u = 49.4 m / s
final speed = v = 0
Based on the above information
Now applying the following formula
v^ 2- u^ 2= 2aS
Therefore
[tex]S = \frac{v^ 2- u^ 2}{2a}\\\\\ = \frac{0- 49.4^ 2}{2\times -3.54}[/tex]
= 344.68 m
what effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?
Answer:
Explanation:
When water droplet condenses on the outer wall of glass of ice , it releases heat equal to mass x latent heat of condensation of water . This heat reaches the ice melting inside glass . Due to this heat , the melting process is accelerated .
Hence the process of melting gets accelerated when water droplet condenses on the outer wall of glass containing mixture of ice and water .
6) Set the battery to a value between 0.0 V and 1.5 V. Now drag the voltage meter toward the capacitor and move the red and black leads to measure the voltage. Determine the potential difference between the two plates and whether the top plate is at higher or lower voltage than the bottom plate.
Answer:
A) 1.5 v
B) Top plate is at higher voltage than the bottom plate
Explanation:
Battery value set between 0.0 V and 1.5 V
a) The potential difference between the plates
Δ V = V1( potential at top plate) - V2( potential at lower plate )
potential at top plate = 1.5 V
potential at lower plate = 0.0 V
hence potential difference = 1.5 V
b ) The top plate is always connected to the positive terminal of the DC source ( which is at a higher potential )while the bottom plate is connected to the negative terminal of the DC source ( which is at a lower potential )
hence the Top plate is at higher voltage than the bottom plate
1 A diffraction grating has a spacing of 1.6 10-m.
A beam of light is incident normally on the
grating. The first order maximum makes an angle
of 20° with the undeviated beam.what is wavelength of the incident light
16. In single-slit diffraction, the central band gets thicker as the distance to the screen increases. True False
Answer:
the right answer is true
Answer:
True
Explanation:
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
In what directions is it possible that the wave is traveling?
A. The-z direction.
B. The ty direction
C. The +x direction.
D. The -y direction
E. The -x direction.
F. The +z direction.
Answer:
The wave will be travelling in the y-axis
Explanation:
An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.
The wave number (k) and the angular frequency of a wave are 6.2rad/m and 12rad/s respectively. Which of the followings could be the equation of the wave?
a. 8sin(12x-6.2t)
b. 8sin(6.2x-12t)
c. 12 sin (6.2x-8t)
d. all of the above
Answer:
B. [tex]8sin(6.2x-12t)[/tex]
Explanation:
The general equation of a wave is expressed as [tex]y = Asin(kx-\omega t)[/tex]
A is the amplitude of the wave
k is the wave number and it is expressed as [tex]k =\frac{2\pi}{\lambda}[/tex]
[tex]\omega[/tex] is the angular frequency expressed as [tex]\omega = 2\pi f[/tex]
[tex]\lambda[/tex] is the wavelength and f is the angular frequency
Given k = 6.2rad/m and [tex]\omega = 12rad/s[/tex]
On substituting this value into the general wave equation;
[tex]y = Asin(kx-\omega t)\\y = Asin(6.2x-12t)[/tex]
From the expression gotten, the only equation that could be the equation of the wave is [tex]y = 8sin(6.2x-12t)[/tex]
To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written
ΦE=∫E.dA =qencl/ϵ0
, where ϵ0=8.85×10−12C2/(N⋅m2) is the permittivity of vacuum.
How should the integral in Gauss's law be evaluated?
a. around the perimeter of a closed loop
b. over the surface bounded by a closed loop
c. over a closed surface
Answer:
Explanation:
jjjjjjjjjjjjjjjj
Match the followings.
a. The current is induced when there is ________ magnetic flux through a closed loop of wire.
b. If the magnetic flux was constant then there __________ induced current regardless of the magnetic flux value.
c. If the magnetic flux was not constant then there ________ induced current regardless of the magnetic flux value.
1. a constant
2. a strictly decreasing
3. either decreasing or Increasing
4. will be
5. will be no
6. a strictly Increasing
Answer:
a. The current is induced when there is ____3. EITHER DECREASING OR INCREASING___ magnetic flux through a closed loop of wire.
b. If the magnetic flux was constant then there _______WILL BE NO___ induced current regardless of the magnetic flux value.
c. If the magnetic flux was not constant then there __WILL BE______ induced current regardless of the magnetic flux value.
Explanation:
THIS IS BECAUSE IF FARADAY'S LAW OF ELECTROMAGNETIC INDUCTION WHICH STATES THAT WHENEVER THERE IS A CHANGE IN MAGNETIC LINES OF FORCE LINKED WITH A CLOSED CIRCUIT AN EMF IS INDUCED
Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3, 2, 1, 4 D. 3, 1, 2, 4
Given that,
Rank in order from largest to smallest the magnitude of the electric field at block dot.
Electric field :
Electric field is proportional to the charge divided by square of distance.
In mathematically,
[tex]E\propto\dfrac{q}{r^2}[/tex]
Where, q = charge
r = distance
If the charge is greater then electric field will be greater.
If the distance is greater then electric field will be smaller.
We need to find the electric field at black dot
According to figure,
(I). The electric field at black dot due to positive charge point q to left. the distance is r.
The electric field will be
[tex]E=\dfrac{kq}{r^2}[/tex]
The electric field will be largest.
(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{2r^2}[/tex]
The electric field will be smallest.
(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(r)^2}[/tex]
The electric field will be very largest.
(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{kq}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{4r^2}[/tex]
The electric field will be very smallest.
So, The electric field from largest to smallest will be
[tex]E_{3}>E_{1}>E_{2}>E_{4}[/tex]
Hence, The ranking will be 3, 1, 2, 4.
(D) is correct option.
A conducting sphere with radius R is charged until the magnitude of the electric field just outside its surface is E. The electric potential of the sphere, relative to the potential for away, is: Group of answer choices 0 E/R E/R2 ER ER2
Answer:
he correct answer is V = ER
Explanation:
In this exercise they give us the electric field on the surface of the sphere and ask us about the electric potential, the two quantities are related
ΔV = ∫ E.ds
where E is the elective field and normal displacement vector.
Since E is radial in a spray the displacement vector is also radial, the dot product e reduces to the algebraic product.
ΔV = ∫ E ds
ΔV = E s
since s is in the direction of the radii its value on the surface of the spheres s = R
ΔV = E R
checking the correct answer is V = ER
A man hits a 50 grams golf ball such that it leaves the tee at an angle of 40o with the horizontal and strikes the ground at the same elevation 20 m away. Determine the impulse of the club C on the ball.
Answer:
Explanation:
Range of projectile R = 20 m
formula of range
R = u² sin2θ / g
u is initial velocity , θ is angle of projectile
putting the values
20 = u² sin2x 40 / 9.8
u² = 199
u = 14.10 m /s
At the initial point
vertical component of u
= u sin40 = 14.1 x sin 40
= 9.06 m/s
Horizontal component
= u cos 30
At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .
Horizontal component of velocity
u cos 30
Vertical component
= - u sin 30
= - 9.06 m /s
So its horizontal component remains unchanged .
change in vertical component = 9.06 - ( - 9.06 )
= 18.12 m /s
change in momentum
mass x change in velocity
= .050 x 18.12
= .906 N.s
Impulse = change in momentum
= .906 N.s .