Answer:
Time, [tex]t=1.07\times 10^{-23}\ s[/tex]
Explanation:
Given that,
When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel [tex]3.2\times 10^{-15}\ m[/tex] before interacting.
Let t is the time interval required for the strong interaction to occur. It will move with the speed of light. So,
[tex]t=\dfrac{d}{c}\\\\t=\dfrac{3.2\times 10^{-15}}{3\times 10^8}\\\\t=1.07\times 10^{-23}\ s[/tex]
So, the time interval is [tex]1.07\times 10^{-23}\ s[/tex]
An electron experiences a force of magnitude F when it is 5 cm from a very long, charged wire with linear charge density, lambda. If the charge density is doubled, at what distance from the wire will a proton experience a force of the same magnitude F?
Answer:
The distance of the proton is [tex]r_p =10 \ cm[/tex]
Explanation:
Generally the force experience by the electron is mathematically represented as
[tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]
Where [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled
Also the force experience by the proton is mathematically represented as
[tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]
Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are equal then
[tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]
[tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]
[tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]
[tex]r_p =2 r_e[/tex]
Now given from the question that [tex]r_e[/tex] the distance of the electron from the charged wire is 5 cm
Then
[tex]r_p =2 (5)[/tex]
[tex]r_p =10 \ cm[/tex]
A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy
Answer: her rotational kinetic energy increases
3. Is it possible for a scientific theory to become a law? Why or why not?
A theory does not change into a scientific law with the accumulation of new or better evidence. A theory will always remain a theory; a law will always remain a law. Both theories and laws could potentially be falsified by countervailing evidence. Theories and laws are also distinct from hypotheses.
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
A wire along the z axis carries a current of 4.9 A in the z direction Find the magnitude and direction of the force exerted on a 3.3 cm long length of this wire by a uniform magnetic field pointing in the x direction having a magnitude 0.43T
Answer:
0.069 N, in the X directionExplanation:
According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.
Mathematically the law is stated as
F= BIL
given data
Magnetic field B= 0.43T
Current I= 4.9 A
length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m
Applying the formula the force is calculated as
F= 0.43*4.9* 0.033= 0.069 N
According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction
Calculate the work performed by an ideal Carnot engine as a cold brick warms from 150 K to the temperature of the environment, which is 300 K. (Use 300 K as the temperature of the hot reservoir of the engine). The heat capacity of the brick is C
Answer
Work done is 57.9KJ
Explanation
First solve the problem according to work done due to variation in temperature
So W= intergral Cu( 1-Tu/T). at Tu and T
So Given that
C = Heat capacity of the Brick
TEPc= Cold Temperature
TEPh = Hot Temperature
W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)
So
W= (1)-(300-150)-300 (1) ln 2
W= -57.9KJ
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?
Answer: 0.0617
Explanation:
Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15
Let X be a binomial variable that represents the number of days having wet weather.
Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.
Here, n= 7 ( 1 week = 7 days)
The probability that it will take a week for it three wet weather on 3 separate days:
[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]
Hence, the required probability =0.0617
You plan to take your hair blower to Europe, where the electrical outlets put out 240 V instead of the 120 V seen in the United States. The blower puts out 1700 W at 120 V.Required:a. What could you do to operate your blower via the 240V line in Europe? which one is it?b. What current will your blower draw from a European outlet?c. What resistance will your blower appear to have when operated at 240 ?
Answer:
a) Connect a series resistance of 8,47 ohms
b)14,16 [A]
c) r = 10,96 ohms
Explanation:
My blower requires 120 (v) then, I have to connect a series resistor to make the nominal 240 (v) of the European voltage outlet drop to 120 (V) but at the same time keep the level of current to operate my blower
In America
P = V*I
1700 (w) = 120*I
I = 1700/120 [A]
I = 14,16 [A] current needed for the blower
In Europe
120 (v) (the drop of voltage I need) when a current of 14,16 passes through to series resistor is
V = I*R 120 = 14,16* R R = 8,47 ohms
c) P = I*r²
1700 (w) = 14,16 (A) * r²
r² = 120,06
r = 10,96 ohms
g A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75 C/s . Part A How many electrons are delivered to the
Answer:
Approximately 5 x 10^18 electrons are delivered to the smart phone charger.
Explanation:
The electric current in a circuit is the flow of charges through a circuit with time.
The charges through the circuit are due to the electrons that flow through the circuit.
An individual electrons has a charge of -1.60 x 10^-19 C on it.
If the current through the circuit is -0.75 C/s, then the number of electrons that are delivered is gotten by dividing the charge per second by the charge on an electron.
==> -0.75/(-1.60 x 10^-19) = 4.67 x 10^18 electrons ≅ 5 x 10^18 electrons are delivered to the smart phone charger.
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.0 m/s. How much time does it take for the glove to return to the pitcher
Answer:
The glove takes 1.02s to return to the pitchers hand.
Explanation:
Given;
initial velocity the pitcher's glove, u = 5 m/s
Apply kinematic equation
s = ut - ¹/₂gt²
where;
g is acceleration due to gravity = 9.8 m/s²
t is the time takes the glove to return to the pitchers hand
s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)
0 = ut - ¹/₂gt²
ut = ¹/₂gt²
u = ¹/₂gt
gt = 2u
t = (2u) / g
t = (2 x 5) / 9.8
t = 1.02 s
Therefore, the glove takes 1.02s to return to the pitchers hand.
In an adiabatic process:
a. the energy absorbed as heat equals the work done by the systemon its environment
b. the energy absorbed as heat equals the work done by theenvironment on the system
c. the work done by the environment on the system equals the changein internal energy
Answer:
c. the work done by the environment on the system equals the changein internal energy.
Explanation:
Adiabatic process:
When the boundary of a system is perfectly insulated, it means that the energy can not flow from the system and into the system ,these system is known as adiabatic system.
When the energy transfer in the system is zero ,then these type of process is known as adiabatic process.
From the first law of thermodynamics
Q= ΔU + W
Q=Heat transfer
ΔU=Change in internal energy
W=Work transfer
In adiabatic process , Q= 0
Therefore
0=ΔU +W
W=- ΔU
Negative sign indicates that ,the work done by the environment.
Therefore the correct option will be (c).
A concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.
Answer:
D.
Inverted and reduced
If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
What is a concave mirror?When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.
A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
Hence option B is correct.
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If a train travels at a constant 18.0 m/s, how far would it move in one hour? In 1.00 minute? In 1.00 second?
Explanation:
Distance = speed × time
d = (18.0 m/s) (1 hr × 3600 s/hr)
d = 64,800 m
d = (18.0 m/s) (1 min × 60 s/min)
d = 1080 m
d = (18.0 m/s) (1 s)
d = 18.0 m
according to the big bang theory, what did protons and neutrons form from?
Answer:
Explanation:
In the first moments after the Big Bang, the universe was extremely hot and dense. As the universe cooled, conditions became just right to give rise to the building blocks of matter – the quarks and electrons of which we are all made. A few millionths of a second later, quarks aggregated to produce protons and neutrons. Within minutes, these protons and neutrons combined into nuclei. As the universe continued to expand and cool, things began to happen more slowly. It took 380,000 years for electrons to be trapped in orbits around nuclei, forming the first atoms. These were mainly helium and hydrogen, which are still by far the most abundant elements in the universe. Present observations suggest that the first stars formed from clouds of gas around 150–200 million years after the Big Bang. Heavier atoms such as carbon, oxygen and iron, have since been continuously produced in the hearts of stars and catapulted throughout the universe in spectacular stellar explosions called supernovae.
But stars and galaxies do not tell the whole story. Astronomical and physical calculations suggest that the visible universe is only a tiny amount (4%) of what the universe is actually made of. A very large fraction of the universe, in fact 26%, is made of an unknown type of matter called "dark matter". Unlike stars and galaxies, dark matter does not emit any light or electromagnetic radiation of any kind, so that we can detect it only through its gravitational effects.
An even more mysterious form of energy called “dark energy” accounts for about 70% of the mass-energy content of the universe. Even less is known about it than dark matter. This idea stems from the observation that all galaxies seems to be receding from each other at an accelerating pace, implying that some invisible extra energy is at work.
Protons and neutrons formed from quarks in the early universe, a process driven by extremely high temperatures and energies.
In the early cosmos, quarks generated protons and neutrons, which form atomic nuclei, according to the Big Bang hypothesis. Quarks and gluons, which bond quarks, existed freely in a quark-gluon plasma state in the early cosmos due to extreme heat and energy.
Hadronization formed protons and neutrons from quarks as the cosmos cooled and expanded. Eventually, protons and neutrons formed atomic nuclei as the cosmos evolved. This key stage in the early universe's existence formed the basic particles that make up matter today.
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A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides without friction down the inner surface toward the bottom of the bowl. At a certain point of its path the mass achieves a speed v = 3.57 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?
Answer:
The angle is [tex]\theta = 36.24 ^o[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.6 \ kg[/tex]
The radius is [tex]r = 1.1 \ m[/tex]
The speed is [tex]v = 3.57 \ m /s[/tex]
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
[tex]m * g * h = \frac{1}{2} * m * v^2[/tex]
=> [tex]h = \frac{1}{2 g } * v^2[/tex]
Here h is the vertical distance traveled by the mass which is also mathematically represented as
[tex]h = r * sin (\theta )[/tex]
So
[tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]
substituting values
[tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]
[tex]\theta = 36.24 ^o[/tex]
A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black c. White d. Green
The color that is reflected when a red card is illuminated by red light is white.
The color an object is perceived to have, depends on the frequency of light it reflects.
If white light incidents on a red filter, red is transmitted while blue and green are absorbed.
Consequently, when a red card is illuminated by red light, the red card will reflect back almost all the incident light on it, causing it to appear brighter which creates an illusion of white color to the eyes.
Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.
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A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.
Answer:
PlROCA
Explanation:
What did Edwin Hubble discover by observing absorption spectrum
blue light
red light
grey matter
ultra-violet light
Answer:
According to different sources Edwin Hubble observed red light of galaxies directly proportional to the distance of the galaxy from earth.
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OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular
Answer:
option A
Explanation:
simple harmonic motion
Answer:
random motion I think not sure
A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?
Answer:
Explanation:
Relation between flux and inductance is as follows
φ = Li
where φ is flux associated with induction of inductance L when a current i flows through it
putting the values
3.25 x 10⁻³ x 800 = L x 2.9
L = .9 H
for induced emf in an induction , the relation is
emf induced = L di / dt
Putting the values
7.5 x 10⁻³ = .9 x di / dt
di / dt = 8.33 x 10⁻³ A / s
(a) The self inductance of the solenoid is 0.897 H.
(b) The magnitude of the rate of change of the current is 0.00836 A/s.
The given parameters;
number of turns, N = 800 turnscurrent in the solenoid, I = 2.9 flux through the solenoid, Ф = 3.25 x 10⁻³ WbThe self inductance of the solenoid is calculated as follows;
[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]
The magnitude of the rate of change of the current is calculated as follows;
[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]
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A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What minimum energy is required to change the orbit to one for which the height of the satellite is 3R above the surface of the planet
Answer:
ΔE = GMm/24R
Explanation:
centripetal acceleration a = V^2 / R = 2T/mr
T= kinetic energy
m= mass of satellite, r= radius of earth
= gravitational acceleration = GM / r^2
Now, solving for the kinetic energy:
T = GMm / 2r = -1/2 U,
where U is the potential energy
So the total energy is:
E = T+U = -GMm / 2r
Now we want to find the energy difference as r goes from one orbital radius to another:
ΔE = GMm/2 (1/R_1 - 1/R_2)
So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R
ΔE = GMm/2R (1/3 - 1/4)
ΔE = GMm/24R
3. A very light bamboo fishing rod 3.0 m long is secured to a boat at the bottom end. It is
held in equilibrium by an 18 N horizontal force while a fish pulls on a fishing line
attached to the rod shown below. How much force F does the fishing line exert on the
rod? (3)
18 N
pivot
30°
1.8 m
3.0 in
The image in the attachment describes the situation of the fishing rod.
Answer: F = 10.8 N
Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.
Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:
τ = F.r.sin(θ)
F is the force acting on the object;
r is distance between where the torque is measured to where the force is applied;
θ is the angle between F and r;
For the fishing rod:
[tex]\tau_{1} = \tau_{2}[/tex]
[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]
Assuming part (1) is related to unknown force:
[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]
Replacing the corresponding values:
[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]
[tex]F = \frac{18*1.8}{3}[/tex]
F = 10.8
The fishing line exert on the the rod a force of 10.8N.
PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!
Vf^2 = v0^2 + 2a (xf -x0)
Solve for v0
b. solve for Vo
[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]
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A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire
Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum uncertainty in the electron's momentum
Answer:
Uncertainty in position (∆x) = 24 × 10⁻¹⁵ mUncertainty in momentum (∆P) = ?Planck's constant (h) = 6.26 × 10⁻³⁴ Js[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]
The place you get your hair cut has two nearly parallel mirrors 6.5 m apart. As you sit in the chair, your head is
Complete question is;
The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?
Answer:
13 m
Explanation:
We are given;
Distance between two nearly parallel mirrors; d = 6.5 m
Distance between the face and the nearer mirror; x = 3 m
Thus, the distance between the back-head and the mirror = 6.5 - 3 = 3.5m
Now, From the given values above and using the law of reflection, we can find the distance of the first reflection of the back of the head of the person in the rear mirror.
Thus;
Distance of the first reflection of the back of the head in the rear mirror from the object head is;
y' = 2y
y' = 2 × 3.5
y' = 7
The total distance of this image from the front mirror would be calculated as;
z = y' + x
z = 7 + 3
z = 10
Finally, the second reflection of this image will be 10 meters inside in the front mirror.
Thus, the total distance of the image of the back of the head in the front mirror from the person will be:
T.D = x + z
T.D = 3 + 10
T.D = 13m
A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot a 20cm long arrow parallel to the axle through the wheel at a speed of 6m/s. The arrow and the spokes are supposed to be thin. Calculate the maximum value of w ( in rad/second and in rev/second) so that the arrow just goes through without hitting any of the spokes. Does it matter where between the axle and the rim of the wheel you aim? If so, what is the best location?
Explanation:
We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is
[tex]s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}[/tex]
The 20-cm arrow, traveling at 6 m/s, will travel its length in
[tex]t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}[/tex]
The fastest speed that the wheel can spin without clipping the arrow is
[tex]v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}[/tex]
The angular velocity [tex]\omega[/tex] of the wheel is given by
[tex]\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}[/tex]
In terms of rev/s, we can convert the answer above as follows:
[tex]23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}[/tex]
As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.
Answer:omega=23.56... rad/s
Explanation:for the arrow to pass without hitting the 8 spokes
taking the spokes thickness to be a point along the spokes linear motion path
lets calculate the time athe arrow pass point length of the wire 0.25m and it's speed(6m/s)
t=0.25m÷6m/s=0.0333...s
so every spoke around the wheel must move replacing each other's position so distribution of eight spokes=2π÷8=0.78... rad so using as angular displacement and t=0.033s then omega=∅/t=0.78/0.033=23.56rad/s
*yes
*space between spokes
A polarized laser beam of intensity 285 W/m2 shines on an ideal polarizer. The angle between the polarization direction of the laser beam and the polarizing axis of the polarizer is 16.0 ∘. What is the intensity of the light that emerges from the polarizer?
Answer:
The intensity is [tex]I_1 = 263.35 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the beam is [tex]I = 285\ W/m^2[/tex]
The angle is [tex]\theta = 16^o[/tex]
The intensity of the light that emerges from the polarizer is mathematically represented by Malus' law as
[tex]I_1 = I * cos^2 (\theta )[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
[tex]I_1 = 263.35 \ W/m^2[/tex]