A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from left to right, through a uniform magnetic field, with the plane of the loop remaining perpendicular to the plane of the paper at all times. Determine the direction of the current induced in the loop as it swings past the locations labeled (a) I and (b) II. Specify the direction of the current in terms of the points x, y, and z on the loop (e.g., x→y→z or z→y→x). The points x, y, and z lie behind the plane of the paper. What is the direction of the induced current at the locations (c) I and (d) II when the loop swings back, from right to left?
Answers
Complete Question
The complete question iws shown on the first uploaded image
Answer:
a
[tex]y \to z \to x[/tex]
b
[tex]x \to z \to y[/tex]
Explanation:
Now looking at the diagram let take that the magnetic field is moving in the x-axis
Now the magnetic force is mathematically represented as
[tex]F = I L[/tex] x B
Note (The x is showing cross product )
Note the force(y-axis) is perpendicular to the field direction (x-axis)
Now when the loop is swinging forward
The motion of the loop is from y to z to to x to y
Now since the force is perpendicular to the motion(velocity) of the loop
Hence the force would be from z to y and back to z
and from lenze law the induce current opposes the force so the direction will be from y to z to x
Now when the loop is swinging backward
The motion of the induced current will now be x to z to y
Related Questions
Consider the interference/diffraction pattern from a double-slit arrangement of slit separation d = 6.60 um and slit width a. The wavelength of the monochromatic light incident normally upon the slits is 2n = d/10 (in air). There is a filter (of negligible thickness) placed on slit 2, so that the magnitude of the EM wave emitted from it is half of that emitted from slit 1. The space between the slits and the screen is filled with water, whose index of refraction is n = 1.33 (you can take noir as 1.00).
(a) What is the wavelength 2 of the light in water?
(b) If a << 1, What is the phase difference between the waves from slits 1 and 2?
(c) For a << 2, Derive an expression for the intensity / as a function of O and other relevant parameters, including the intensity at the center of the screen (where 0 = 0).
(d) Now suppose a = d/3. Redo part (b) above. How many interference maxima are present within the central diffraction peak? (Do not count the "clipped" maxima, if any.) (4) — E -2 d E ) в /Б/ = 1/5 | TT
Answers
Answer:
(a) λ = 0.496 um (b) S =2π Δ d sinθ/ λ (c) I =gI₀ (d) For the central diffraction peak, a total of 5 interference maxima are present or available.
Note: find an attached copy of a part of the solution to the given question below.
Explanation:
Solution
Recall that:
d = 6.6 um
λ₀ =d/10
λ₀ = 6.6 um
Now,
(a) We find the wavelength λ of the light in water.
Thus,
λ water = (λ₀ )/n
= 0.66/1.33
So,
λ water = λ = 0.496 um
(b) We find the phase difference between the waves from slit 1 and 2
Now,
if a <<d and a<<λ
Then the path difference between the rays will be
Δ S₂N = Δ d sinθ
Thus, the phase difference becomes,
S = 2π Δ/λ is S= 2π Δ d sinθ/ λ
(c) The next step is to derive an expression for the intensity I as function of O and other relevant parameters.
Now,
Let p be the point where these two rays interfere with each other.
Thus,
The electric field vector coming out from slot and and slot 2 is
E₁= E₀₁ cos (ks₁ p - wt) i
E₂ = E₀₂ cos (ks₂ p - wt) i
Note: Kindly find an attached copy of a part of the solution to the given question below.
Two vectors are being added, one at an angle of 20.0 , and the other at 80.0. The only thing you know about the magnitudes is they are both positive. Will the equilibrant vector be in the:
a. first quadrant
b. second quadrant
c. third quadrant
d. fourth quadrant
e. you cannot tell which quadrant from the available information
Answers
Answer:a
Explanation: they are all positive
The equivalent vectors of both the vectors lie in the first quadrant. So, the right choice is a. first quadrant.
How the resultant of the vectors can be calculated?The resultant of the vectors can be calculated using the Parallelogram theorem of vector addition. In this theorem, a parallelogram is formed using the vectors, having the vectors as the adjacent side and the diagonal of this parallelogram is the resultant of both the vectors.
How the theorem is implemented to find out the resultant?The vectors A and B are drawn at the given angle, angles of 20° and 80° respectively. Using the vectors, A and B, the dotted parallelogram is drawn, and the vector C is the resultant.
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Helppppp please thanks
Answers
Answer:
It represents battery
Answer: I think #1 represent lights and #2 represent the battery
Beginning from rest when = 20°, a 35-kg child slides with negligible friction down the sliding board which is in the shape of a 2.5-m circular arc. Determine the tangential acceleration and speed of the child, and the normal force exerted on her (a) when = 30° and (b) when = 90°
Answers
Answer:
a) [tex]a_{T}=8.50m/s^{2}[/tex], [tex]v_{T}=2.78 m/s[/tex] and [tex]N=279.83N[/tex]
b) [tex]a_{T}=0m/s^{2}[/tex], [tex]v_{T}=5.68 m/s[/tex] and [tex]N=795.2N[/tex]
Explanation:
a)
In order to solve this problem we need to start by drawing a diagram of what the problem looks like: See attached picture. Next, we can start by finding the initial height of the child which will happen at an angle of 20°. We can find this by subtracting the distance from the highest point and the initial point from the radius of the circle so we get:
[tex]h_{0}=2.5m-h_{1}[/tex]
so we get:
[tex]h_{1}=2.5m(sin (20^{o}))[/tex]
[tex]h_{1}=0.855m[/tex]
so
[tex]h_{0}=2.5m-h_{1}[/tex]
[tex]h_{0}=2.5m-0.855m[/tex]
[tex]h_{0}=1.645m[/tex]
once we got this value, we can find the final height the same way. This time the angle is 30° so we get:
[tex]h_{f}=2.5m-h_{2}[/tex]
[tex]h_{f}=2.5m-2.5m(sin 30^{o}))[/tex]
[tex]h_{f}=1.25m[/tex]
Once we have these heights, we can go ahead and use an energy balance equation to find the velocity at 30° so we get:
[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]
the initial kinetic energy is zero because its initial velocity is zero too, so the equation simplifies to:
[tex]U_{0}=U_{f}+K_{f}[/tex]
so now we substitute with the corresponding formulas:
[tex]mgh_{0}=mgh_{f}+\frac{1}{2}mv_{f}^{2}[/tex]
if we divided both sides of the equation by the mass, then the equation simplifies to:
[tex]gh_{0}=gh_{f}+\frac{1}{2}v_{f}^{2}[/tex]
and now we can solve for the final velocity so we get:
[tex]v_{f}=\sqrt{2g(h_{0}-h_{f}}[/tex]
and we can now substitute values:
[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m-1.25m)}[/tex]
which solves to:
[tex]v_{f}=2.78m/s[/tex]
which is our first answer. Once we got the velocity at 30° we can find the other data the problem is asking us for:
We can build a free body diagram (see attached picture) and do a balance of forces so we get:
[tex]\sum{F_{x}}=ma_{T}[/tex]
in this case we only have one x-force which is the x-component of the weight, so we get that:
[tex]w_{x}=ma_{T}[/tex]
so we get:
[tex]mgcos\theta=ma_{T}[/tex]
and solve for the tangential acceleration so we get:
[tex]a_{T}=gcos\theta[/tex]
[tex]a_{T}=9.81m/s^{2}*cos(30^{o})[/tex]
[tex]a_{t}=8.50m/s^{2}[/tex]
Now we can find the centripetal acceleration by using the formula:
[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]
[tex]a_{c}=\frac{(2.78m/s)^{2}}{2.5m}[/tex]
so we get:
[tex]a_{c}=3.09m/s^{2}[/tex]
Next, we can find the normal force which is found by doing a sum of forces on y, so we get:
[tex]\sum{F_{y}}=ma_{c}[/tex]
so we get:
[tex]N-w_{y}=ma_{c}[/tex]
and we solve for the normal force so we get:
[tex]N=ma_{c}+w_{y}[/tex]
and substitute:
[tex]N=ma_{c}+mg sin\theta[/tex]
when factoring we get:
[tex]N=m(a_{c}+g sin\theta)[/tex]
and we substitute:
[tex]N=(35kg)(3.09m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]
which yields:
N=279.83N
b)
The procedure for part b is mostly the same with some differences due to the angle. First:
[tex]h_{0}=1.645m[/tex]
[tex]h_{f}=0m[/tex]
so
[tex]U_{0}+K_{0}=U_{f}+K_{f}[/tex]
in this case the initial kinetic energy is zero because the initial velocity is zero and the final potential energy is zero because the final height is zero as well, so the equation simplifies to:
[tex]U_{0}=K_{f}[/tex]
so we get:
[tex]mgh_{0}=\frac{1}{2}mv_{f}^{2}[/tex]
so we solve for the final velocity so we get:
[tex]v_{f}=\sqrt{2gh_{0}}[/tex]
and we substitute:
[tex]v_{f}=\sqrt{2(9.81m/s^{2})(1.645m)}[/tex]
[tex]v_{f}=5.68m/s[/tex]
according to the free body diagram we get that:
[tex]a_{T}=gcos\theta[/tex]
[tex]a_{T}=9.81m/s^{2}(cos 90^{o})[/tex]
which yields:
[tex]a_{T}=0[/tex]
we can also find the centripetal acceleration, so we get:
[tex]a_{c}=\frac{V_{T}^{2}}{R}[/tex]
[tex]a_{c}=\frac{(5.68m/s)^{2}}{2.5m}[/tex]
so we get:
[tex]a_{c}=12.91m/s^{2}[/tex]
and we can do a sum of forces on y to find the normal force:
[tex]\sum{F_{y}}=ma_{c}[/tex]
so we get:
[tex]N-w_{y}=ma_{c}[/tex]
and we solve for the normal force so we get:
[tex]N=ma_{c}+w_{y}[/tex]
and substitute:
[tex]N=ma_{c}+mg [/tex]
when factoring we get:
[tex]N=m(a_{c}+g)[/tex]
and we substitute:
[tex]N=(35kg)(12.91m/s^{2}+9.81m/s^{2} sin30^{o})[/tex]
which yields:
N=795.2N
Following are the calculation to the angles:
For angle 30°:
consider the forces in tangential in direction:
[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 30^{\circ} \\\\a_t = 8.496 \ \frac{m}{s^2}\\\\[/tex]
calculate the speed of the child:
[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]
Integrating the equation:
[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{30} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 30-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 30-\sin 20) = \frac{V^2}{2}\\\\v=2.79 \ \frac{m}{s}\\\\[/tex]
consider the forces in the normal declaration:
[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 30) = \frac{35 \times 2.78^2}{2.5}\\\\N= 279.87\ N[/tex]
For angle 90°:
consider the forces in tangential in direction:
[tex]\Sigma F_t= ma_t\\\\ W \cos \theta= m a_t \\\\m g \cos \theta = m a_t\\\\ a_t = g \cos \theta = 9.81 \cos 90^{\circ} \\\\a_t = 0\ \frac{m}{s^2}\\\\[/tex]
calculate the speed of the child:
[tex]g\cos \theta = a_t\\\\ g\cos \theta = \frac{v.dv}{R d\theta} \\\\g R \cos \theta d \theta= V \ dv\\\\[/tex]
Integrating the equation:
[tex]\int_{20}^{30} g R \cos \theta d \theta= \int_{0}^{v} V \ dv\\\\g R (\sin \theta)_{20}^{90} = [\frac{V^2}{2}]_{0}^{v} \ dv\\\\g R (\sin 90-\sin 20) = \frac{V^2}{2}\\\\9.81 \times 2.5 (\sin 90-\sin 20) = \frac{V^2}{2}\\\\V=\sqrt{\frac{9.81 \times 2.5 (\sin 90-\sin 20) }{2}}[/tex]
[tex]=\sqrt{\frac{9.81 \times 2.5 (1 -0.342)}{2}}\\\\=\sqrt{8.06}\\\\=2.83[/tex]
consider the forces in the normal declaration:
[tex]\Sigma F_n = ma_n \\\\N-mg \sin \theta= \frac{mv^2}{R} \\\\N- (35\times 9.81\times \sin 90) = \frac{35 \times 2.78^2}{2.5}\\\\N= \frac{35 \times 2.78^2}{2.5 \times 35\times 9.81\times 1}\\[/tex]
[tex]= \frac{270.494}{858.375}\\\\=0.315[/tex]
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Suppose the electric field in problems 2 was caused by a point charge. The test charge is moved to a distance twice as far from the charge. What is the magnitude of the force that the field exerts on the test charge now ?
Answers
Answer:
it is reduced four times.
Explanation:
By definition, the electric field is the force per unit charge created by a charge distribution.
If the charge creating the field is a point charge, the force exerted by it on a test charge, must obey Coulomb´s Law, so, it must be inversely proportional to the square of the distance between the charges.
So, if the distance increases twice, as the force is inversely proportional to the square of the distance, and the square of 2 is 4, this means that the magnitude of the force exerted on the test charge must be 4 times smaller.
When we say that an object wants to maintain its state of motion, we’re talking about inertia. Which term determines the quantity of inertia for an object?
A.
mass
B.
velocity
C.
acceleration
D.
weight
Answers
Answer:
A. mass
Explanation:
Mass determines the quantity of inertia for an object. Mass is the quantity that depends upon the inertia of an object. The inertia that an object has is directly proportional to the mass of the object.
An object that has more mass has a greater tendency as compared to the object that has less mass to resist changes in its state of motion.
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the central plane 0.9 m above the ground and 1.7 m behind the front axle. When moving on the level at 90 km/h the brakes applied and it comes to a rest in a distance of 50 m.
Calculate the normal reactions at the front and rear wheels during the braking period and the least coefficient of friction required between the tyres and the road. (Assume g = 10 m/s2)
Answers
Answer:
normal reaction, front: 11,175 Nnormal reaction, rear: 6,825 Nminimum coefficient of friction: 0.625Explanation:
The speed in meters per second is ...
(90 km/h)(1000 m/km)(1 h/(3600 s)) = 25 m/s
The braking acceleration can be found from ...
a = v²/(2d) = (25 m/s)²/(2×50 m) = 6.25 m/s²
Then the braking force is ...
F = Ma = (1800 kg)(6.25 m/s²) = 11,250 N
The torque on the center of gravity is ...
T = (11,250 N)(0.90 m) = 10,125 N·m
__
If we let x and y represent the normal forces on the front and rear wheels, respectively, then we have ...
x + y = (10 m/s²)(1800 kg) = 18000 . . . . . newtons
1.7x -1.3y = 10,125 . . . . . . . . . . . . . . . . . . . newton-meters
The latter equation balances the torque due to the wheel normal forces with the torque due to braking forces.
Multiplying the first equation by 1.3 and adding that to the second, we have ...
3.0x = (1.3)(18,000) + 10,125
x = 33,525/3 = 11,175 . . . . . . . . . . . newtons normal force on front tyres
y = 18000 -11175 = 6,825 . . . . . . . .newtons normal force on back tyres
The least coefficient of friction is the ratio of horizontal to vertical acceleration, 6.25/10 = 0.625.
Consider an atom. Which contributes most to the size of the atom?
Answers
Answer:
Protons contribute the most to the size of the atom
Answer:
Protons
Explanation:
Protons contribute more to both the mass and size of an atom. Protons contribute more to an atom's mass while electrons contribute more to its size.
Two fledglings leave a nest which is 2.50 m off the ground. One drops from rest and then 0.10 s later the second charges out of the nest with a velocity with horizontal and
downward components of 1.00 and 1.50 m/s, respectively.
1. Which fledgling hits the ground first (first or second)?
2. With what speed, in m/s, does the first fledgling hit the ground?
3. With what speed, in m/s, does the second fledgling hit the ground?
4. Which fledgling travels the greater displacement?
5. How far does a parent bird have to travel on the ground between the two fledglings, in m, to make sure they are alright?
Answers
t Answer:
1) the time of the pigeon 1 is less, so it comes first
2) v = - 6,997 m / s , 3) v = 10.15 m / s ,
4) the displacement of the second point in greater
5) x = 0.883 m
Explanation:
For this exercise we will use the kinematics equations
1) ask which chick reaches the ground first
we calculate for the first chick that has zero initial velocity
y = y₀ + v₀ t - ½ g t²
0 = yo - ½ g t²
t = √ 2 y₀ / g
let's calculate
t = √ (2 2.50 / 9.8)
t = 0.714 s
We calculate the time it takes for the second chick that has velocity v = (1 i ^ - 1.5 j⁾ m / s
y = y₀ + v₀t - ½ g t²
0 = 2.5 - 1.5 t - ½ 9.8 t²
4.9 t² + 1.5 t - 2.5 = 0
t² + 0.306 t - 0.510 = 0
we solve the quadratic equation
t = [0.306 ± √ (0.306² - 4 (-0.510))] / 2
t = [0.306 ± 1.46] / 2
The results are
t₁ = -0.577 s
t₂ = 0.883 m / s
we take positive time as correct
the time of the pigeon 1 is less, so it comes first
2) the speed of the first chick is
v = v₀ - g t
we can see that
v = -gt
v = - 9.8 0.714
v = - 6,997 m / s
the negative sign indicates that the speed is down
3) the speed of the other bird is
v = -1.5 - 9.8 0.883
v = 10.15 m / s
4) which chick has the greatest displacement. The first point falls vertically and its displacement is y₀
The second point describes a parabola and its displacement is
d = √ (x² + y₀²)
therefore we see that the displacement of the second point in greater
5) calculate the horizontal displacement of the second point
x = vx t
x = 1 0.883
x = 0.883 m
Three Small Identical Balls Have Charges -3 Times 10^-12 C, 8 Times 10^12 C And 4 Times 10^-12 C Respectively. They Are Brought In Contact And Then Separated. Calculate Charge On Each Ball.
Answers
Answer:
The charge in each ball will be 3 * 10^-12 C
Explanation:
(Assuming the correct charge of the second ball is 8 * 10^-12)
When the balls are brought in contact, all the charges are split evenly among then.
So first we need to find the total charge combined:
(-3 * 10^-12) + (8 * 10^-12) + (4 * 10^-12) = 9 * 10^-12 C
Then, when the balls are separated, each ball will have one third of the total charge, so in the end they will have the same charge:
(9 * 10^-12) / 3 = 3 * 10^-12 C
So the charge in each ball will be 3 * 10^-12 C
The motion of a free falling body is an example of __________ motion
Answers
Answer:
uniformly accelerated motion
Explanation:
The motion of the body where the acceleration is constant is known as uniformly accelerated motion. The value of the acceleration does not change with the function of time.
On his way to deliver presents Santa has a minor accident. If the sleigh (1200 kg) was traveling at 322 m/s and the jet(4800 kg) was traveling at 680 m/s and they collided head on, what was the final velocity of the two objects after the collision
Answers
Answer:
608.4m/s
Explanation:
We are given that
Mass of Sleigh,M=1200 kg
Speed of Sleigh,u=322 m/s
Speed of jet,u'=680 m/s
Mass of jet,m=4800 kg
Total mass=M+m=1200+4800=6000 kg
We have to find the final velocity of the two objects after the collision.
The collision is inelastic .
By using law of conservation of momentum
[tex]Mu+mu'=(m+M)v[/tex]
Using the formula
[tex]1200\times 322+4800\times 680=6000v[/tex]
[tex]6000v=3650400[/tex]
[tex]v=\frac{3650400}{6000}[/tex]
[tex]v=608.4m/s[/tex]
Hence, the final velocity of two objects after the collision=608.4m/s
There is an old physics joke involving cows, and you will need to use its punchline to solve this problem
A cow is standing in the middle of an open, flat field. A plumb bob with a mass of 1 kg is suspended via an unstretchable string 10 meters long so that it is hanging down roughly 2 meters away from the center of mass of the cow. Making any reasonable assumptions you like or need to, estimate the angle of deflection of the plumb bob from vertical due to the gravitational field of the cow.
Answers
Answer:
The angle of deflection will be "1.07 × 10⁻⁷°".
Explanation:
The given values are:
Mass of a cow,
m = 1100 kg
Mass of bob,
mb = 1 kg
The total distance between a cow and bob will be,
d = 2 m
Let,
The tension be "t".
The angle with the verticles be "[tex]\theta[/tex]".
Now,
Vertically equating forces
⇒ [tex]T\times Cos \theta =mb\times g[/tex] ...(equation 1)
Horizontally equating forces
⇒ [tex]T\times Sin \theta = G\times M\times \frac{mb}{d^2}[/tex] ...(equation 2)
From equation 1 and equation 2, we get
⇒ [tex]tan \thata=\frac{G\times M}{g\times d^2}[/tex]
O putting the estimated values, will be
⇒ [tex]\theta = tan(\frac{6.674\times 10^{-11}\times 1100}{9.8\times 2^2} )[/tex]
⇒ [tex]\theta = 1.07\times 10^{-7}^{\circ}[/tex]
Convert 45m/s to mph. Choose the best answer to two significant figures.
101mph
45mph
200mph
24mph
Answers
Explanation:
45 m/s × (3.28 ft/m) × (1 mi / 5280 ft) × (3600 s/h) = 101 mph
This is actually 3 significant figures. To write in 2 significant figures, you must use scientific notation:
1.0×10² mph
A 64.7 cm long straight section of wire is located entirely inside a uniform magnetic field of |B| = 0.370 T. The wire is perpendicular to the direction of the magnetic field. When a current runs through the wire a magnetic force of |F| = 0.110 N acts on that section of wire. Calculate the size of the current.
Answers
Answer:
The current is [tex]I = 0.4595 \ A[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 64.7 \ cm = 0.647 \ m[/tex]
The magnetic field is [tex]B = 0.370 \ T[/tex]
The magnetic force is [tex]F = 0.110 \ N[/tex]
Generally the magnetic force exerted by the field is mathematically represented as
[tex]F = IL B sin \theta[/tex]
Making I the subject we have
[tex]I = \frac{F}{LB \ sin\theta}[/tex]
The angle here is 90° since the wire is perpendicular to the direction of the magnetic field
Substituting values
[tex]I = \frac{0.110}{ (0.370) * 0.647 \ sin(90)}[/tex]
[tex]I = 0.4595 \ A[/tex]
A wall clock has a second hand 22.0 cmcm long. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Fast car, flat curve. Part A What is the radial acceleration of the tip of this hand
Answers
Answer:
Explanation:
The tip of the second hand moves on a circular path having radius equal to .22 m . Redial acceleration is given by the expression
ω²R where ω is angular velocity and R is radius of the circular path .
angular velocity of second hand = 2π / T where T is time period of circular motion . For second hand it is 60 s.
ω = 2π / T
= 2π / 60
= .1047
angular acceleration = .1047² x .22
= 2.41 x 10⁻³ rad / s² .
So, the required radial acceleration is [tex]a=0.21 cm/s[/tex]
Acceleration:The rate of change of velocity with respect to time.
Acceleration is a vector quantity.
The formula for the acceleration is,
[tex]a=\frac{V^2}{R}[/tex]
It is given that the radius is 22 cm
Now, substituting the given values into the above formula we get,
[tex]a=\frac{V^2}{22}[/tex]
Here, 1 second is equal to [tex]\frac{2\pi}{60}[/tex] then,
[tex]\\w=\frac{2\pi}{60}\\v=wR\\v=\frac{2\pi}{60}\times 19\\a=(\frac{2\pi}{60}\times 19)^2\times 19\\a=0.21 cm/s[/tex]
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The speed of light changes when it goes from ethyl alcohol (n = 1.361) to carbon tetrachloride (n = 1.461). The ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is
Answers
Answer:
The ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is 0.93
Explanation:
The formula for the refractive index of a medium is given as:
n = c/v
where,
n = refractive index of the medium
c = speed of light in vacuum
v = speed of light in that medium
FOR CARBON TETRACHLORIDE:
n = n₁ = 1.461
v = v₁
Therefore,
1.461 = c/v₁
v₁ = c/1.461 ----- equation (1)
FOR ETHYL ALCOHOL:
n = n₂ = 1.361
v = v₂
Therefore,
1.361 = c/v₂
v₂ = c/1.361 ----- equation (2)
Now, dividing equation (1) by equation (2), we get:
v₁/v₂ = (c/1.461) / (c/1.361)
v₁/v₂ = 1.361/1.461
v₁/v₂ = 0.93
Hence, the ratio of the speed of light in carbon tetrachloride to the speed in ethyl alcohol is 0.93
If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.
Answers
Complete Question
The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ . The average rate of electric energy consumption in one house is 1.0 kW .
If light energy to electric energy conversion using solar cells is 12 % efficient, how many square miles of land must be covered with solar cells to supply the electrical energy for 350000 houses? Assume there is no cloud cover.
Answer:
The area is [tex]A = 1.26 *10^{7} m^2[/tex]
Explanation:
From the question we are told that
The efficiency is [tex]\eta =[/tex]12%
The number of houses is [tex]N = 350000[/tex]
The light energy per day is [tex]E = 20 \ MJ[/tex]
The average rating of electric energy for a house is [tex]E_h = 1.0 \ k W = 1000W[/tex]
Generally the electric energy which the solar cells covering [tex]1 \ m^2[/tex] produces in a day is
[tex]E_s = \eta * E[/tex]
[tex]E_s = 0.12 * 20*10^{6}[/tex]
[tex]E_s = 2.4 MJ m^{-2}[/tex]
Energy for required by one house for one day is
[tex]E_H = E_h * 1 \ day[/tex]
[tex]E_H = 1000 * 24 * 3600[/tex]
[tex]E_H = 86.4 MJ[/tex]
Energy needed for 350000 house is
[tex]E_z = 86.4 *10^{6} * 350000[/tex]
[tex]E_z = 3.02 *10^{7} MJ[/tex]
The area covered is mathematically represented as
[tex]A = \frac{3.02*10^{7} \ MJ}{2.4 \ MJ m^{-2}}[/tex]
[tex]A = 1.26 *10^{7} m^2[/tex]
A p-type Si sample is used in the Haynes-Shockley experiment. The length of the sample is 2 cm, and two probes are separated by 1.8 cm. Voltage applied at the two ends is 5 V. A pulse arrives at the collection point at 0.608 ms, and the separation of the pulse is 180 sec. Calculate mobility and diffusion coefficient for minority carriers. Verify it from the Einstein relation.
Answers
Answer:
Mobility of the minority carriers, [tex]\mu_{n} =1184.21 cm^{2} /V-sec[/tex]
Diffusion coefficient for minority carriers,[tex]D_{n} = 29.20 cm^2 /s[/tex]
Verified from Einstein relation as [tex]\frac{D_{n} }{\mu_{n} } = 25 mV[/tex]
Explanation:
Length of sample, [tex]l_{s} = 2 cm[/tex]
Separation between the two probes, L = 1.8 cm
Drift time, [tex]t_{d} = 0.608 ms[/tex]
Applied voltage, V = 5 V
Mobility of the minority carriers ( electrons), [tex]\mu_{n} = \frac{V_{d} }{E}[/tex]
Where the drift velocity, [tex]V_{d} = \frac{L}{t_{d} }[/tex]
[tex]V_{d} = \frac{1.8}{0.608 * 10^{-3} } \\V_{d} = 2960.53 cm/s[/tex]
and the Electric field strength, [tex]E = \frac{V}{l_{s} }[/tex]
E = 5/2
E = 2.5 V/cm
Mobility of the minority carriers:
[tex]\mu_{n} = 2960.53/2.5\\\mu_{n} =1184.21 cm^{2} /V-sec[/tex]
The electron diffusion coefficient, [tex]D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }[/tex]
[tex]\triangle x = (\triangle t )V_{d}[/tex], where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)
[tex]\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm[/tex]
[tex]D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s[/tex]
For the Einstein equation to be satisfied, [tex]\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V[/tex]
[tex]\frac{D_{n} }{\mu_{n} } = \frac{29.20}{1184.21} \\\frac{D_{n} }{\mu_{n} } = 0.025 = 25 mV[/tex]
Verified.
If the molar mass of helium is 4.0 g/mol and the molar mass of neon is 20.2 g/mol, then
a. All the atoms have exactly the same velocity.
b. All the atoms have the same average speed.
c.The average speed of the helium atoms is greater than the average speed of the neon atoms.
d. The average speed of the neon atoms is greater than the average speed of the helium atoms.
e. The atoms diffuse from high temperature to low temperature.
Answers
Hope it helps
Complete the sentence below correctly.
A compression of a longitudinal wave is like a
of a transverse wave.
OA) crest
OB) trough
OC) rarefaction
OD) compression
Answers
Answer:
A. Crest
Explanation:
Longitudinal wave is a type of wave that is characterized by the particles of the medium's movement in a parallel direction in comparison to the direction in which wave travels, such that, in compression of longitudinal wave, the density of the wave medium is at its highest due to its closeness together than natural state, while in rarefaction, the density is at its lowest due to wave medium spread apart than normal.
Similarly, in Transverse wave, the crest of a wave implies the medium has reached the highest point while the trough of the wave depicts the lowest point the wave medium has reached.
Therefore, longitudinal wave's compression and rarefaction equates accordingly to the crest and trough of a transverse wave.
Hence, a compression of a longitudinal wave is like a CREST of a transverse wave.
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho=rho0[1+α(T−T0)], where T0 is a reference temperature, usually 20∘C, and α is the temperature coefficient of resistivity. Part A First find an expression for the current I through a wire of length L, cross-section area A, and temperature T when connected across the terminals of an ideal battery with terminal voltage ΔV. Then, because the change in resistance is small, use the binomial approximation to simplify your expression. Your final expression should have the temperature coefficient α in the numerator. Express your answer in terms of L, A, T, T0, ΔV, rho0, and α.
Answers
Answer:
I = ΔVA[1 - α (T₀ - T)]/Lρ₀
Explanation:
We have the following data:
ΔV = Battery Terminal Voltage
I = Current through wire
L = Length of wire
A = Cross-sectional area of wire
T = Temperature of wire, when connected across battery
T₀ = Reference temperature
ρ = Resistivity of wire at temperature T
ρ₀ = Resistivity of wire at reference temperature
α = Temperature Coefficient of Resistance
From OHM'S LAW we know that;
ΔV = IR
I = ΔV/R
but, R = ρL/A (For Wire)
Therefore,
I = ΔV/(ρL/A)
I = ΔVA/ρL
but, ρ = ρ₀[1 + α (T₀ - T)]
Therefore,
I = ΔVA/Lρ₀[1 + α (T₀ - T)]
I = [ΔVA/Lρ₀] [1 + α (T₀ - T)]⁻¹
using Binomial Theorem:
(1 +x)⁻¹ = 1 - x + x² - x³ + ...
In case of [1 + α (T₀ - T)]⁻¹, x = α (T₀ - T).
Since, α generally has very low value. Thus, its higher powers can easily be neglected.
Therefore, using this Binomial Approximation, we can write:
[1 + α (T₀ - T)]⁻¹ = [1 - α (T₀ - T)]
Thus, the equation becomes:
I = ΔVA[1 - α (T₀ - T)]/Lρ₀
A 50.0 Watt stereo emits sound waves isotropically at a wavelength of 0.700 meters. This stereo is stationary, but a person in a car is moving away from this stereo at a speed of 40.0 m/s. The frequency of sound waves that the car receives is ________. In addition, when the car is 70.0 meters away from the speaker, the car will hear sound waves with a sound intensity level of _________ .
Answers
Answer:
a) f' = 432 Hz
b) I = 8.12*10^-4 W/m^2
Explanation:
a) To calculate the frequency of sound waves that car receives, you take into account the Doppler effect. In this case (observer moves away of the source) you have the following formula:
[tex]f'=f(\frac{v-v_o}{v+v_s})[/tex] (1)
where
f: frequency of the source = ?
v: speed of sound = 343 m/s
vo: speed of the observer = 40.0 m/s
vs: speed of the source = 0 m/s (stationary)
You replace the values of all parameters in the equation (1):
To calculate f' you first calculate the frequency of the sound wave, by using the following formula:
[tex]v=\lambda f\\\\[/tex]
v: speed of sound
λ: wavelength = 0.700 m
[tex]f=\frac{v}{\lambda}=\frac{343m/s}{0.700m}=480Hz[/tex]
Next, you replace the values of all parameters in the equation (1):
[tex]f'=(490Hz)(\frac{343m/s-40.0m/s}{343m/s})=432Hz[/tex]
hence, the frequency perceived by the car is 432 Hz
b) To calculate the power of the sound wave, when the car is 70.0 maway from the speaker, you use the following formula:
[tex]I=\frac{P}{4\pi r^2}[/tex]
P: power of the source = 50.0 W
r: distance to the source = 70.0 m
[tex]I=\frac{50.0 W}{4\pi(70.0m)^2}=8.12*10^{-4}\frac{W}{m^2}[/tex]
hence, the intensity is 8.12*10^⁻4 W/m^2
An early submersible craft for deep-sea exploration was raised and lowered by a cable from a ship. When the craft was stationary, the tension in the cable was 6000 N. When the craft was lowered or raised at a steady rate, the motion through the water added an 1800 N drag force. What was the tension in the cable when the craft was being lowered to the seafloor?
Answers
Answer:
T = 4200N
Explanation:
When the submersible craft is at rest, the tension in the cable is 6000N.
With this information you can calculate the weight of the craft by summing the forces (the summation of the force is zero because the craft is at rest):
[tex]T-W=0\\\\W=T=6000N[/tex]
When the craft is going down with a constant speed, there is a drag force of 1800N. Then, by using the second Newton law you have:
[tex]T-W+F_d=0[/tex] (1)
Fd: drag force
The summation of the forces is zero because the craft moves with constant velocity, that is, there is no acceleration.
You calculate the new tension on the cable by solving the equation (1) for T:
[tex]T=W-Fd=6000N-1800N=4200N[/tex]
hence, the tension is 4200N
An advertisement for a new fish food claims that lab studies show that fish grew three inches in three weeks while eating the food. Travis wondered if the food was worth the extra cost. To evaluate the claim in the advertisement, Travis should
Answers
Answer:
look at the data for the control group to see how much the fish grew without the new food.
Explanation:
i got it right
Monochromatic light is incident on (and perpendicular to) two slits separated by 0.200 mm, which causes an interference pattern on a screen 613 cm away. The light has a wavelength of 656.3 nm. (a) What is the fraction of the maximum intensity at a distance of 0.600 cm from the central maximum of the interference pattern
Answers
Answer:
I = 0.636*Imax
Explanation:
(a) To find the fraction of the maximum intensity at a distance y from the central maximum you use the following formula:
[tex]I=I_{max}cos^2(\frac{\pi d}{\lambda L}y)[/tex] (1)
I: intensity of light
Imax: maximum intensity of light
d: separation between slits = 0.200mm = 0.200 *10^-3 m
L: distance from the screen = 613cm = 0.613 m
y: distance to the central peak of the interference pattern
λ: wavelength of light = 656.3 nm = 656.3 *10^-9 m
You replace the values of all variables in the equation (1):
[tex]I=I_{max}cos^2(\frac{\pi (0.200*10^{-3}m)}{(656.3*10^{-9}m)(0.613m)}0.600m)\\\\I=I_{max}cos^2(937.06)=0.636I_{max}[/tex]
Hence, the fraction of the maximum intensity is I = 0.636*Imax
-
is a side effect of tobacco use.
A. Lightheadedness
B. Irritability
C. Dizziness
D. All of the above
Please select the best answer from the choices provided.
Answers
Answer:
D. all of the above
Explanation:
What is a limitation of the electron cloud model theory that
a law about electrons would not have?
Answers
Answer:
Electron cloud model describes the region in an atom which negatively charged and which has probability to find an electron. according to this model, an electron can move closer or away from the nucleus that is it can be inside the nucleus but according to Bohr's model, an electron is always at a fixed distance from the nucleus. Thus, it is the limitation of the electron cloud model but still it is a widely accepted model.
A flat, rectangular coil consisting of 60 turns measures 23.0 cmcm by 34.0 cmcm . It is in a uniform, 1.40-TT, magnetic field, with the plane of the coil parallel to the field. In 0.230 ss , it is rotated so that the plane of the coil is perpendicular to the field.
(a) What is the change in the magnetic flux through the coil due to the rotation?
(b) What is the magnitude of the average emf induced in the coil during the rotation?
(c) What is the average current induced in the coil during the rotation?
Answers
Answer:
(a) ΔФ = -0.109W
(b) emf = 28.43V
(c) Iin = emf/R
Explanation:
(a) In order to calculate the magnetic flux you use the following formula:
[tex]\Delta\Phi_B=\Phi-\Phi_o=BAcos(90\°)-BAcos(0\°)[/tex] (1)
B: magnitude of the magnetic field = 1.40T
A: area of the rectangular coil = (0.23m)(0.34m)=0.078m^2
Where it has been taken into account that at the beginning the normal vector to the cross sectional area of the coil, and the magnetic field vector are parallel. When the coil is rotated the vectors are perpendicular.
Then, you obtain:
[tex]\Delta\Phi_B=(1.40T)(0.078m^2)=-0.109W[/tex]
The change in the magnetic flux is -0.109 W
(b) During the rotation of the coil the emf induced is given by:
[tex]emf=-N\frac{\Delta \Phi}{\Delta t}[/tex] (2)
N: turns of the coil = 60
ΔФ: change in the magnetic flux = 0.109W
Δt: lapse time of the rotation = 0.230s
You replace the values of the parameters in the equation (2):
[tex]emf=-(60)(\frac{-0.109W}{0.230s})=28.43V[/tex]
The induced emf is 28.43V
(c) The induced current in the coil is given by:
[tex]I_{in}=\frac{emf}{R}[/tex] (3)
R: resistance of the coil (it is necessary to have this value)
emf :induced emf = 28.43V
⦁ Consider an atom. Which contributes most to the mass of the atom?
Answers
Answer:
protons
Explanation: