Answer:
The magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
Explanation:
Given;
current in the eastern wire, [tex]I_e[/tex] = 15 A
current in the western wire, [tex]I_w[/tex] = 6 A
distance between the wires, d = 30 cm = 0.3 m
The magnetic field at a distance R from a line current I, is given as;
[tex]B = \frac{\mu_o I }{2 \pi R}[/tex]
The magnetic field between the wires, are in opposite directions, and since the currents are also in opposite directions, the magnetic fields of the wires will be added.
The total field = magnetic field (east) + magnetic field (west);
[tex]B = \frac{\mu_o I_e}{2 \pi R_e} + \frac{\mu_0 I_w}{2 \pi R_w} \\\\B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})[/tex]
where;
[tex]R_w[/tex] is the distance of the field from west = 10cm = 0.1 m
[tex]R_e[/tex] is the distance of the field on east from west = d - 10cm = 30cm - 10cm = 20cm = 0.2 m
The total magnetic field is;
[tex]B = \frac{\mu_o}{2\pi} (\frac{I_e}{R_e} + \frac{I_w}{R_w})\\\\B = \frac{4\pi *10^{-7}}{2\pi} (\frac{15}{0.2} + \frac{6}{0.1})\\\\B = 2*10^{-7}(75 + 60)\\\\B = 2*10^{-7}(135)\\\\B = 2.7*10^{-5} \ T[/tex]
Since total magnetic field is positive, the direction of the field is upwards (positive y direction)
Therefore, the magnitude and direction of the magnetic field is 2.7 x 10⁻⁵ T upwards
What is dark energy?
Explanation:
Dark Energy. Dark Energy is a hypothetical form of energy that exerts a negative, repulsive pressure, behaving like the opposite of gravity. It has been hypothesised to account for the observational properties of distant type Ia supernovae, which show the universe going through an accelerated period of expansion
The angle between the axes of two polarizing filters is 41.0°. By how much does the second filter reduce the intensity of the light coming through the first?
Answer:
The amount by which the second filter reduces the intensity of light emerging from the first filter is
z = 0.60
Explanation:
From the question we are told that
The angle between the axes is [tex]\theta = 41^o[/tex]
The intensity of polarized light that emerges from the second filter is mathematically represented as
[tex]I= I_o cos^2 \theta[/tex]
Where [tex]I_o[/tex] is the intensity of light emerging from the first filter
[tex]I = I_o [cos(41.0)]^2[/tex]
[tex]I =0.60 I_o[/tex]
This means that the second filter reduced the intensity by z = 0.60
The moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is: . If the axis of rotation passes through the center of the rod, then the moment of inertia is: . Give a physical explanation for this difference in terms of the way the mass of the rod is distributed with respect to the axis in the two cases.
Answer:
Explanation:
he moment of inertia for a rod that rotates about the axis perpendicular to the rod and passing through one end is: m L²/ 3 where m is mass and L is length of rod
If the axis of rotation passes through the center of the rod, then the moment of inertia is: m L² / 12
So for the former case , moment of inertia is higher that that in the later case .
In the former case , the axis is at one extreme end . Hence range of distance of any point on the rod from axis is from zero to L .
In the second case , as axis passes through middle point , this range of distance of any point on the rod from axis is from zero to L / 2 .
Since range of distance from axis is less , moment of inertia too will be less because
Moment of inertia = Σ m r² where r is distance of mass m from axis .
A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.
Answer:
Explanation:
The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.
In compound microscopes, the distance between the two lenses is expressed as L = v0+ue
v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.
Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.
Given L = 18.0cm
Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm
1/f0 = 1/u0+1/v0
f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.
1/v0 = 1/7.4-1/12
1/v0 = 0.1351-0.0833
1/v0 = 0.0518
v0 = 1/0.2184
v0 = 19.31cm
Note that v0 = ue = 19.31cm
To get ve, we will use the lens formula 1/fe = 1/ue+1/ve
1/ve = 1/fe-1/ue
Given ue = 19.31cm and fe = -7.00cm
1/ve = -1/7.0-1/19.31
1/ve = -0.1429-0.0518
1/ve = -0.1947
ve = 1/-0.1947
ve = -5.14cm
Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens
b) Magnification of the final image M = ve/ue
M = 5.14/19.31
M = 0.27
Magnification of the final image is 0.27
The first step to merging is entering the ramp and _____.
A. honking to indicate your location
B. matching your speed
C. signaling your intent
D. telling your passengers where you're going
Answer:
B. matching your speed
Explanation:
To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*
_____
* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.
Answer:
It's C "signaling your intent"
Explanation:
The key thing to look at is they are asking the rest of the first step and that;s C
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
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If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to:
5.19 x 10³Hz
Explanation:The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;
[tex]X_C = \frac{1}{2\pi fC }[/tex]
Where;
f = frequency of the signal through the capacitor
C = capacitance of the capacitor.
Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;
V = I x [tex]X_{C}[/tex] [Substitute the value of
=> V = I x [tex]\frac{1}{2\pi fC}[/tex] [Make f the subject of the formula]
=> f = [tex]\frac{I}{2\pi VC}[/tex] ---------------------(i)
From the question;
I = 3.33mA = 0.00333A
C = 8.50nF = 8.50 x 10⁻⁹F
V = 12.0V
Substitute these values into equation (i) as follows;
f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex] [Taking [tex]\pi[/tex] = 3.142]
f = 5.19 x 10³Hz
Therefore, the frequency is closest to f = 5.19 x 10³Hz
A 25 kg box sliding to the left across a horizontal surface is brought to a halt in a distance of 15 cm by a horizontal rope pulling to the right with 15 N tension.
Required:
a. How much work is done by the tension?
b. How much work is done by gravity?
The work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
The given parameters;
mass of the box, m = 25 kgdistance traveled by the box, d = 15 cm = 0.15 mtension on the rope, T = 15 NThe work done by the tension is calculated as follows;
W = Fd
W = 15 x 0.15
W = 2.25 J
The work done by gravity is calculated as;
W = (25 x 9.8) x 0.15
W = 36.75 J
Thus, the work done by tensional force of the rope is 2.25 J and the work done by gravity is 36.75 J.
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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 are the potential obstacles preventing you from completing your exercises as scheduled? How can you overcome those obstacles?
Answer:
Sleep, behavior patterns, mental state, and job
Explanation:
Answer:
If I exercise right after school, I might be low on energy. I suppose that I could eat a snack and drink something before my workout. If I exercise before school, I might be tired. But, as long as I keep getting eight to nine hours of sleep, I think that my body will adjust to the new schedule after a while. The trick will be getting to bed on time and eating a little breakfast before I work out. I’m kind of worried that the gym won’t be open early in the morning. On the weekends, my friends might keep me from exercising. I suppose I can try to get them to do it with me. We can pick things that we all like to do. I know they like to play tennis sometimes.
Explanation:
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
Two long parallel wires are separated by 11 cm. One of the wires carries a current of 54 A and the other carries a current of 45 A. Determine the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current.
Explanation:
It is given that,
The separation between two parallel wires, r = 11 cm = 0.11 m
Current in wire 1, [tex]q_1=54\ A[/tex]
Current in wire 2, [tex]q_2=45\ A[/tex]
Length of wires, l = 4.3 m
We need to find the magnitude of the magnetic force on a 4.3 m length of the wire carrying the greater current. The magnetic force per unit length is given by :
[tex]\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\F=\dfrac{\mu_o I_1I_2l}{2\pi r}\\\\F=\dfrac{4\pi \times 10^{-7}\times 54\times 45\times 4.3}{2\pi \times 0.11}\\\\F=0.0189\ N[/tex]
So, the magnetic force on a 4.3 m length of the wire on both of currents is F=0.0189 N.
Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you
Answer:
I = E/R e^{-t/RC}
Explanation:
In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,
E -q / c-IR = 0
we replace the current by its expression and divide by the resistance
I = dq / dt
dq / dt = E / R -q / RC
dq / dt = (CE -q) / RC
we solve the equation
dq / (Ce-q) = -dt / RC
we integrate and evaluate for the charge between 0 and q and for the time 0 and t
ln (q-CE / -CE) = -1 /RC (t -0)
eliminate the logarithm
q - CE = CE [tex]e^{-t/RC}[/tex]
q = CE (1 + 1/RC e^{-t/RC} )
In general the teams measure the current therefore we take the derivative to find the current
i = CE (e^{-t/RC} / RC)
I = E/R e^{-t/RC}
This expression is the one that describes the charge of a condensate in a DC circuit
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 26 m/s when a 60 kg skydiver drops out by releasing his grip on the glider.
What is the glider's speed just after the skydiver lets go?
Answer:
The glider’s speed after the skydiver lets go is 26 m/s
Explanation:
To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum
Mathematically;
mv = mv + mv
so 680 * 26 = (680-60)v + 60 * 26
17680 = 620v + 1560
17680-1560 = 620v
16120 = 620v
v = 16120/620
v = 26 m/s
An astronomy student, for her PhD, really needs to estimate the age of a cluster of stars. Which of the following would be part of the process she would follow?
A. plot an H-R diagram for the stars in the cluster
B. count the number of M type stars in the cluster
C. measure the Doppler shift of a number of the stars in the cluster
D. search for planets like Jupiter around the stars in the center of the cluster
E. search for x-rays coming from the center of the cluster
Answer:
A. plot an H-R diagram for the stars in the cluster.
Explanation:
A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.
The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.
A communications satellite orbiting the earth has solar panels that completely absorb all sunlight incident upon them. The total area A of the panels is 10m2.
1) The intensity of the sun's radiation incident upon the earth is about I=1.4kW/m2. Suppose this is the value for the intensity of sunlight incident upon the satellite's solar panels. What is the total solar power P absorbed by the panels?
Express your answer numerically in kilowatts to two significant figures.
2) What is the total force F on the panels exerted by radiation pressure from the sunlight?
Express the total force numerically, to two significant figures, in units of newtons.
Answer:
1) 14 kW
2) 4.67 x 10^-5 N
Explanation:
Area of solar panel = 10 m^2
Intensity of sun's radiation incident on earth = 1.4 kW/m^2
Solar power absorbed = ?
We know that the intensity of radiation on a given area is
[tex]I[/tex] = [tex]\frac{P}{A}[/tex]
where I is the intensity of the radiation
P is the power absorbed due to this intensity on a given area
A is the area on which this radiation is incident
From the equation, we have
P = IA
P = 1.4 kW/m^2 x 10 m^2 = 14 kW
b) For a perfect absorbing surface, the radiation pressure is given as
p = I/c
where p is the radiation pressure
I is the incident light intensity = 1.4 kW/m^2 = 1.4 x 10^3 kW/m^2
c is the speed of light = 3 x 10^8 m/s
substituting values, we have
p = (1.4 x 10^3)/(3 x 10^8) = 4.67 x 10^-6 Pa
we know that Force = pressure x area
therefore force on the solar panels is
F = 4.67 x 10^-6 x 10 = 4.67 x 10^-5 N
a football is kicked toward a goal keeper with an initial speed of 20m/s at an angle of 45 degrees with the horizontal .at the moment the ball is kicked the goal keeper is 50m from the player .at what speed and in what direction must the goalkeeper run in order to catch the ball at the same height at which it was kicked
Answer:
3.18 m/s
Explanation:
Given that
Initial speed of the ball, u = 20 m/s
Angle of inclination, θ = 45°
Distance from the ball, h = 50 m
Using equations of projectile to solve this, we have
We start by finding the time of flight, T
T = 2Usinθ/g
T = (2 * 20 * sin45)/9.8
T = (40 * 0.7071) / 9.8
T = 28.284/9.8
T = 2.89 s
Next we find the Range, R
R = u²sin2θ/g
R = (20² * sin 90) / 9.8
R = (400 * 1) / 9.8
R = 400/9.8 = 40.82 m
Distance the gk must cover
40.82 - 50 m
-9.18 m or 9.18 m in the opposite direction.
Speed of the GK = d/t
9.18 / 2.89 = 3.18 m/s
A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?
a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.
Answer:
the greatest intensity is obtained from c
Explanation:
An electromagnetic wave stagnant by the expression
E = E₀ sin (kx -wt)
when two waves meet their electric fields add up
E_total = E₁ + E₂
the intensity is
I = E_total . E_total
I = E₁² + E₂² + 2E₁ E₂ cos θ
where θ is the phase angle between the two rays
Let's examine the two waves
in this case E₁ = E₂ = E₀
I = Eo2 + Eo2 + 2 E₀ E₀ coasts
I = E₀² (2 + 2 cos θ )
I = 2 I₀ (1 + cos θ )
let's apply this expression to different cases
a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀
b) cos π = -1 this implies that I_total = 0
c) the cosine is 1,
I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ
I = E₀² (5 +4 cos θ)
I = E₀² 9
I = 9 Io
d) in this case the cos pi = -1
I = E₀² (5 -4)
I = I₀
e) we rewrite the equation
I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ
I = Eo2 (10 + 6 cos θ)
cos π = -1
I = E₀² (10-6)
I = 4 I₀
the greatest intensity is obtained from c
The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.
What is an amplitude?An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.
In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.
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Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)
Answer:
1/4F
Explanation:
We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.
So F α Qq
But if it is now half the initial charges, then
F α (1/2)Q *(1/2)q
F α (1/4)Qq
Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.
Thus the answer will be 1/4F
A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration
Answer:
The average velocity for this trip is 0 km/hr
Explanation:
We know that average velocity = total displacement/total time.
Now, its displacement is d = final position - initial position.
Since the car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.
So, its displacement is d = 0 km - 0 km = 0 km.
Since the total time for the round trip is 2 hours, the average velocity is
total displacement/ total time = 0 km/2 hr = 0 km/hr.
So the average velocity for this trip is 0 km/hr
You are pushing a 60 kg block of ice across the ground. You exert a constant force of 9 N on the block of ice. You let go after pushing it across some distance d, and the block leaves your hand with a velocity of 0.85 m/s. While you are pushing, the work done by friction between the ice and the ground is 3 Nm (3 J). Assuming that the ice block was stationary before you push it, find d.
Answer: d = 33 cm or 0.33 m
Explanation: In physics, Work is the amount of energy transferred to an object to make it move. It can be expressed by:
W = F.d.cosθ
F is the force applied to the object, d is the displacement and θ is the angle formed between the force and the displacement.
For the ice block, the angle is 0, i.e., force and distance are at the same direction, so:
W = F.d.cos(0)
W = F.d
To determine d:
d = [tex]\frac{W}{F}[/tex]
d = [tex]\frac{3}{9}[/tex]
d = 0.33 m
The distance d the block ice moved is 33 cm.
You need to design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2
Answer:
Hello your question has some missing parts and the required diagram attached below is the missing part and the diagram
Digital circuits require actions to take place at precise times, so they are controlled by a clock that generates a steady sequence of rectangular voltage pulses. One of the most widely
used integrated circuits for creating clock pulses is called a 555 timer. shows how the timer’s output pulses, oscillating between 0 V and 5 V, are controlled with two resistors and a capacitor. The circuit manufacturer tells users that TH, the time the clock output spends in the high (5V) state, is TH =(R1 + R2)*C*ln(2). Similarly, the time spent in the low (0 V) state is TL = R2*C*ln(2). Design a clock that will oscillate at 10 MHz and will spend 75% of each cycle in the high state. You will be using a 500 pF capacitor. What values do you need to specify for R1 and R2?
ANSWER : R1 = 144.3Ω, R2 = 72.2 Ω
Explanation:
Frequency = 10 MHz
Time period = 1 / F = 0.1 u s
Duty cycle = 75% = 0.75
Duty cycle can be represented as : Ton / T
Also: Ton = Th = 0.75 * 0.1 u s = 75 n s
TL = T - Th = 100 ns - 75 n s = 25 n s
To find the value of R2 we use the equation for time spent in the low (0 V) state
TL = R2*C*ln(2)
hence R2 = TL / ( C * In 2 )
c = 500 pF
Hence R2 = 25 / ( 500 pF * 0.693 ) = 72.2 Ω
To find the value of R1 we use the equation for the time the clock output spends in the high (5V) state,
Th = (R1 + R2)*C*ln(2)
from the equation make R1 the subject of the formula
R1 = (Th - ( R2 * C * In2 )) / (C * In 2)
R1 = ( 75 ns - ( 72.2 * 500 pF * 0.693)) / ( 500 pF * 0.693 )
R1 = ( 75 ns - ( 25 ns ) / 500 pf * 0.693
= 144.3Ω
Why would physics be used to study light emitted by a star?
O A. Stars form interesting shapes in the sky.
B. Light is very pretty.
O C. The positions of stars control our lives.
O D. Light is a form of energy.
Answer:
O D.
Explanation:
Physics has an aspect that deals with the study of energy
Answer:
D. Light is a form of energy
Explanation:
what is transmission of heat?
Answer:
Heat transfer is the transmission of heat energy from a body at higher temperature to lower temperature. The three mechanisms of heat transfer are
Conduction ConvectionRadiation.Example of Conduction:
Heating a metal
Example of Convection:
Sea Breeze
Example of Radiation:
Sun
Hope this helps ;) ❤❤❤
Answer:
Transmission of heat is the movement of thermal energy from one thing to another thing of different temperature.
There are three(3) different ways heat can transfer and they are:
a) Conduction (through direct contact).
b) Convection (through fluid movement).
c) Radiation (through electromagnetic waves).
Examples: 1.Heating a saucepan of water using a coalpot.(conduction&convection).
2. Baking a pie in an oven(radiation).
Hope it helps!!Please mark me as the brainliest!!!Thanks!!!!❤❤❤
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If of the light passes through this combination, what is the angle between the transmission axes of the two filters
Answer:
The angle between the transmission axes of the filters is 65°
Explanation:
The complete question is
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 18% of the light passes through this combination, what is the angle between the transmission axes of the two filters.
From Malus law,
[tex]I = I_{0} cos^{2} \beta[/tex] ....1
where [tex]I[/tex] is the intensity of the polarized light,
[tex]I_{o}[/tex] is the intensity of the incident light
β the angle between the transmission axes of the two filters
Since the intensity is reduced to 18% or 0.18 of its initial value, this means that
[tex]cos^{2} \beta[/tex] = 0.18
substituting into the equation above, we have
[tex]I = 0.18I_{0}[/tex] ....2
equating the two equations, we have
[tex]I_{0}cos^{2} \beta[/tex] = [tex]0.18I_{0}[/tex]
[tex]cos^{2}\beta[/tex] = [tex]\frac{0.81I_{0} }{I_{0} }[/tex] = 0.18
[tex]cos \beta[/tex] = [tex]\sqrt{0.18}[/tex] = 0.424
[tex]\beta[/tex] = [tex]cos^{-1} 0.424[/tex] = 64.9 ≅ 65°
The index of refraction of a certain material is 1.5. If I send red light (700 nm) through the material, what will the frequency of the light be in the material
Answer: [tex]4.29\times10^{14}\text{ Hz}[/tex]
Explanation:
Given: Speed of red light = 700 nm
= [tex]700\times10^{-9}[/tex] m
[tex]= 7\times10^{-7}[/tex] m
Frequency of red light = [tex]\dfrac{\text{Speed of light}}{\text{Speed of red light}}[/tex]
Speed of light = [tex]3\times10^8[/tex] m
Then, Frequency of red light = [tex]\dfrac{3\times10^8}{7\times10^{-7}}[/tex]
[tex]=0.429\times10^{8-(-7)}=0.429\times10^{15}\\\\=4.29\times10^{14}\ Hz[/tex]
Hence, Frequency of red light = [tex]4.29\times10^{14}\text{ Hz}[/tex]
The frequency of the light be in the material is [tex]4.29\times10^{14}\text{ Hz}[/tex].
The planets how and block are near each other in the Dorgon system. the Dorgons have very advanced technology, and a Dorgon scientist wants to increase the pull of gravity between the two planets. Which proposals would the scientist make to accomplish this goal? check all that apply.
Answer:
Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.
Explanation:
The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.
Or
If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.
On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously.
Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:
"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."
Answer:
[tex]\mu_{sB}=0.126[/tex]
[tex]\mu_{sC}=0.168[/tex]
Explanation:
In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:
Sum of torques:
[tex]\sum \tau_{A}=0[/tex]
[tex]N(3m)-W(1.5m)=0[/tex]
When solving for N we get:
[tex]N=\frac{W(1.5m)}{3m}[/tex]
[tex]N=\frac{(1962N)(1.5m)}{3m}[/tex]
[tex]N=981N[/tex]
Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:
First, the forces in y.
[tex]\sum F_{y}=0[/tex]
[tex]-F_{By}+N_{c}=0[/tex]
[tex]F_{By}=N_{c}[/tex]
Next, the forces in x.
[tex]\sum F_{x}=0[/tex]
[tex]-f_{sB}-f_{sC}+P_{x}=0[/tex]
We can find the x-component of force P like this:
[tex]P_{x}=360N(\frac{4}{5})=288N[/tex]
and finally the torques about C.
[tex]\sum \tau_{C}=0[/tex]
[tex]f_{sB}(1.75m)-P_{x}(0.75m)=0[/tex]
[tex]f_{sB}=\frac{288N(0.75m)}{1.75m}[/tex]
[tex]f_{sB}=123.43N[/tex]
With the static friction force in point B we can find the coefficient of static friction in B:
[tex]\mu_{sB}=\frac{f_{sB}}{N}[/tex]
[tex]\mu_{sB}=\frac{123.43N}{981N}[/tex]
[tex]\mu_{sB}=0.126[/tex]
And now we can find the friction force in C.
[tex]f_{sC}=P_{x}-f_{xB}[/tex]
[tex]f_{sC}=288N-123.43N=164.57N[/tex]
[tex]f_{sC}=N_{c}\mu_{sC}[/tex]
and now we can use this to find static friction coefficient in point C.
[tex]\mu_{sC}=\frac{f_{sC}}{N}[/tex]
[tex]\mu_{sC}=\frac{164.57N}{981N}[/tex]
[tex]\mu_{sB}=0.168[/tex]
Science activity
Imagine that some settlers have left Earth and gone to the Moon, taking
their recipe books with them. The first cake they baked was a disaster. It had
far too little moisture and was about six times the size they had expected.
the cake recipe was:
1.25 N butter
1.50 N sugar
4 eggs
1.50 N flour
20 ml milk
ANALYTICAL THINKING
Q. Why was the cake so big? Why was it se
dry?
Answer:
Answer in explanation
Explanation:
The reason for the big size and less moisture of the cake is due to difference in weight of the ingredients on the surface of moon. So, the same has the lesser weight on the surface of the moon than it has on the surface of earth. Or in other words, The same weight of the ingredients will have greater mass and thus the greater quantity on the surface of earth than the surface of earth. For example, on earth 1.25 N butter will have a mass:
m = W/g = 1.25 N/(9.8 m/s²) = 0.13 kg
But, on moon:
m = W/g = 1.25 N/(1.625 m/s²) = 0.77 kg
Hence, it is clear that the mass of the same weight of the substance becomes 6 times greater on the surface of moon. This explains why the cake was so big.
Now, coming to the second part about the dryness of the cake. The main and only source of moisture in recipe is the eggs bu the eggs are taken in a quantity of numbers. So they are exactly the same on moon as well. While all the other ingredients are increased, the same amount of eggs are not sufficient to provide them with enough moisture. Hence the cake was dry.
A cylinder is closed by a piston connected to a spring of constant 2.20 10^3 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C. The piston has a cross sectional area of 0.0100 m^2 and negligible mass. What is the pressure of the gas at 250 °C?
Answer:
1.3515x10^5pa
Explanation:
Plss see attached file