Answer:
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
Explanation:
The expression for the angle of refraction is
n₁ sin θ₁ = n₂ sin θ₂
refractive index n₁ is for incident light and n₂ is for transmitted light.
We have two cases
a) the light is close to normal therefore the reference incidence of medium 1 is less than medium n2 where the ray is transmitted.
b) The ray is far from normal in this case the refractive index of medium 1 is greater than index of medium 2
A liquid is poured into a vessel to a depth of 16cm when viewed from the top, the bottom appears to be raised 4cm. What is the refractive index of the liquid?
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
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SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
What happens to the deflection of the galvanometer needle (due to moving the magnet) when you increase the number of loops
Answer:
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil which will increase total induced electromotive force
Explanation:
galvanometer is an instrument that can detect and measure small current in an electrical circuit.
If the magnet is moved, the galvanometer needle will deflect, showing that current is flowing through the coil. If it is move in a way into the coil,the needle deflect in that way and if it move in another way, it will deflect in the other way.
The total induced emf is equal to the emf induced in each loop by the changing magnetic flux, then multiplied by the number of loops and an increase in the number of loops will cause increase in the total induced emf.
A sinewave has a period (duration of one cycle) of 645 μs (microseconds). What is the corresponding frequency of this sinewave, in kHz
The corresponding frequency of this sinewave, in kHz, expressed to 3 significant figures is: 155 kHz.
Given the following data:
Period = 645 μsNote: μs represents microseconds.
Conversion:
1 μs = [tex]1[/tex] × [tex]10^-6[/tex] seconds
645 μs = [tex]645[/tex] × [tex]10^-6[/tex] seconds
To find corresponding frequency of this sinewave, in kHz;
Mathematically, the frequency of a waveform is calculated by using the formula;
[tex]Frequency = \frac{1}{Period}[/tex]
Substituting the value into the formula, we have;
[tex]Frequency = \frac{1}{645 * 10^-6}[/tex]
Frequency = 1550.39 Hz
Next, we would convert the value of frequency in hertz (Hz) to Kilohertz (kHz);
Conversion:
1 hertz = 0.001 kilohertz
1550.39 hertz = X kilohertz
Cross-multiplying, we have;
X = [tex]0.001[/tex] × [tex]1550.39[/tex]
X = 155039 kHz
To 3 significant figures;
Frequency = 155 kHz
Therefore, the corresponding frequency of this sinewave, in kHz is 155.
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An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 11.2 eV and with its velocity directed toward the negative plate. How close to the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.6 x 10-19 J)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
The atomic number of a nucleus increases during which nuclear reactions?
Answer:
Option (A) : Nuclear Fusion and Beta Decay (electron emission)
Answer:
A : Fusion followed by beta decay (electron emission)
Explanation:
Ap3x
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero and increasing toward a maximum. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -I max, where Imax is the current amplitude. What is the unknown element?
a. a resistor
b. an inductor or a capacitor
c. an inductor
d. a capacitor
Object A, with heat capacity CA and initially at temperature TA, is placed in thermal contact with object B, with heat capacity CB and initially at temperature TB. The combination is thermally isolated. If the heat capacities are independent of the temperature and no phase changes occur, the final temperature of both objects is
Answer:
d) (CATA + CBTB) / (CA + CB)
Explanation:
According to the given situation, the final temperature of both objects is shown below:-
We assume T be the final temperature
while m be the mass
So it will be represent
m CA (TA - T) = m CB (T - TB)
or we can say that
CATA - CA T = CB T - CBTB
or
(CA + CB) T = CATA + CBTB
or
T = (CA TA + CBTB) ÷ (CA + CB)
Therefore the right answer is d
The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
The given parameters;
heat capacity of object A = CAinitial temperature of object A = TAheat capacity of object B = CBinitial temperature of object B = TBThe final temperature of both objects is calculated as follows;
heat lost by object A is equal to heat gained by object B
[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]
Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].
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g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?
Answer:
constructive interferencia 0, 1 , 2 m
destructive inteferencia 1/4, 3/4. 5/4, 7/4 m
Explanation:
This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression
Δr / λ = Φ / 2π
Δr = Φ/2π λ
let's apply this expression to our case
λ = 1 m
Δr = Φ 1 / 2π
We have constructive interference for angle of Φ = 0, 2π, ...
let's find the values where they occur
Φ Δr
0 0
2π 1
4π 2
Destructive interference occurs by Φ = π /2, 3π / 2, ...
Φ Δr
π/2 ¼ m
3π /2 ¾ m
5π /2 5/4 m
7π /2 7/4 m
"Two waves of the same frequency have amplitudes 1.00 and 2.00. They interfere at a point where their phase difference is 60.0°. What is the resultant amplitude?"
Answer:
The resultant amplitude of the two waves is 2.65.
Explanation:
Given;
amplitude of the first wave, A₁ = 1
amplitude of the second wave, A₂ = 2
phase difference of the two amplitudes, θ = 60.0°.
The resultant amplitude of two waves after interference is given by;
[tex]A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2Cos \theta} \\\\A = \sqrt{1^2 + 2^2 + 2(1)(2)Cos 60} \\\\A= 2.65[/tex]
Therefore, the resultant amplitude of the two waves is 2.65.
Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.3 kg, down a theta= 79.6º slope at constant acceleration a=-4.3 m/s2, as shown in Figure (here we assume the positive direction is going down the slope. So the given acceleration is a negative value, it means its direction is going up the slope, slowing down as it moving downward). So, the coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 2.1 m along the hill? Use g= 10 m/s2.
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
In this case, we need to construct the Free Body Diagram of the sled-victim System in order to determine what Forces are doing Work. Then, we construct the respective Energy equation by Newton's Laws of Motion, Work-Energy Theorem and definition of Work.
Given that system experiments an uniform Acceleration, we must solve the resulting model for the work done by the Tension in the rope.
From the Free Body Diagram (see image attached), we see that both Weight of the sled and Friction between sled and snow are doing work in favor of gravity, whereas Tension forces is against gravity. Normal force is not doing work as its direction is perpendicular to the direction of motion. The energy equation of this system is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex] (1)
Where:
[tex]W_{T}[/tex] - Work done by tension, in joules.
[tex]m[/tex] - Mass of the sled-victim system, in kilograms.
[tex]\mu[/tex] - Coefficient of kinetic friction, no unit.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]s[/tex] - Travelled distance, in meters.
[tex]\theta[/tex] - Slope angle, in sexagesimal degrees.
[tex]a[/tex] - Net acceleration of the sled-victim system, in meters per square second.
If we know that [tex]\mu = 0.100[/tex], [tex]m = 55.3\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]s = 2.1\,m[/tex], [tex]\theta = 79.6^{\circ}[/tex] and [tex]a = -4.3\,\frac{m}{s^{2}}[/tex], then the work done by the tension in the rope is:
[tex]-W_{T} + \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta = m\cdot a\cdot s[/tex]
[tex]W_{T} = \mu\cdot m\cdot g \cdot s\cdot \cos \theta + m\cdot g\cdot s\cdot \sin \theta -m\cdot a\cdot s[/tex]
[tex]W_{T} = (0.100)\cdot \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \cos 79.6^{\circ} + \left(55.3\,kg\right)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot (2.1\,m)\cdot \sin 79.6^{\circ} - (55.3\,kg)\cdot \left(-4.3\,\frac{m}{s^{2}} \right) \cdot (2.1\,m)[/tex]
[tex]W_{T} = 1662.544\,J[/tex]
The tension in the rope is doing a work of 1662.544 joules as the sled moves 2.1 meters along the hill.
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A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall. Find the potential energy of a particle due to this force when it is at a distance x from the wall, assuming the potential energy at the wall to be zero.
Answer:
it will be 10x
Explanation:
workdone(potential energy before it hits the wall)= horizontal force × distance
=10× x = 10x joules
A mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall.The potential energy of a particle due to this force is 10x.
What is force?A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to describe force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.
Given in the question a mysterious constant force of 10 N acts horizontally on everything. The direction of the force is found to be always pointed toward a wall in a big hall the potential energy,
Work done (potential energy before it hits the wall)
= horizontal force × distance
=10× x = 10x joules
The potential energy of a particle due to this force is 10x.
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The filament in the bulb is moving back and forth, first pushed one way and then the other. What does this imply about the current in the filament
Answer:
energy carried by the current is given by the pointyng vector
Explanation:
The current is defined by
i = dQ / dt
this is the number of charges per unit area over time.
The movement of the charge carriers (electrons) is governed by the applied potential difference, when the filament has a movement the drag speed of these moving electrons should change slightly.
But the energy carried by the current is given by the pointyng vector of the electromagnetic wave
S = 1 / μ₀ EX B
It moves at the speed of light and its speed depends on the properties of the doctor and is not disturbed by small changes in speed, therefore the current in the circuit does not change due to this movement
6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?
Answer:
t = t₀ / 2
Explanation:
In this exercise we must use Newton's second law
F = m a
a = F / m
now we can use kinematics
as in object part of rest (v₀ = 0)
v =a t₀
t₀ = v / a
these results are with the first experiment
now repeat the experiment, but F = 2F₀
a = 2F₀ / m = 2 a₀
v = 2 a₀ t
t = v / 2a₀
t = t₀ / 2
The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].
Given the following data:
Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = VTo find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:
Mathematically, Newton's Second Law of Motion is given by this formula;
[tex]F = \frac{M(V-U)}{t}[/tex]
Where:
F is the force.V is the final velocity.U is the initial velocity.t is the time.Substituting the given parameters into the formula, we have;
[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]
When the experiment is repeated, the magnitude of the force is doubled:
[tex]F = 2F[/tex]
Now, we can find the time interval that is required to reach the same final speed (V):
[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]
Substituting the value of F, we have:
[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]
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A 1300-turn coil of wire 2.40 cm in diameter is in a magnetic field that increases from 0 T to 0.120 T in 9.00 ms . The axis of the coil is parallel to the field. What is the emf of the coil?
Answer:
The induced emf in the coil is 7.843 V
Explanation:
Given;
number of turns of the coil, N = 1300 turn
diameter of the coil, d = 2.4 cm = 0.024 m
initial magnetic field, B₁ = 0 T
final magnetic field, B₂ = 0.12 T
change in time, dt = 9.0 ms = 9 x 10⁻³ s
Area of the coil is given by;
A = πr²
radius of the coil, r = 0.024 / 2
radius of the coil, r = 0.012 m
A = π(0.012)²
A = 4.525 x 10⁻⁴ m²
The induced emf in the coil is given by;
E = NA(dB/dt)
E = NA [(B₂ - B₁) /dt]
E = 1300 x 4.525 x 10⁻⁴ (0.12 - 0) / (9 x 10⁻³)
E = 7.843 V
Therefore, the induced emf in the coil is 7.843 V
How do for-profit and nonprofit fitness centers compare with each other?
For-profit centers offer more luxury services than nonprofit centers.For-profit centers offer more luxury services than nonprofit centers. , ,
For-profit centers have lower membership fees than nonprofit centers.For-profit centers have lower membership fees than nonprofit centers. , ,
For-profit centers place more emphasis on weight lifting than nonprofit centers.For-profit centers place more emphasis on weight lifting than nonprofit centers. , ,
For-profit centers are more common in low-income communities than nonprofit centers.
Comparison of For-profit and Non-profit Fitness Centers
This is how they compare with each other: For-profit centers offer more luxury services than nonprofit centers.
For-profit fitness centers are fitness centers that offer their services to cover their costs and make some profits. They are business entities that generate income for their shareholders. Their fees are based on cost plus profit. This approach enables them to remain sustainable as business ventures.
They are not like non-profit fitness centers, which may offer their services for free or at cost. Sometimes, the nonprofit centers may not even cover their fixed and operational costs. They are usually located in low-income communities, unlike for-profit fitness centers, which are mainly located in rich-income communities.
Considering their clientele, for-profit centers provide modern and better facilities to satisfy their rich-income clients.
Thus, these for-profit centers offer more luxury services than their non-profit counterparts.
For more information about how for-profit centers base their charges, see https://brainly.com/question/21302488.
A locomotive is pulling three train cars along a level track with a force of 100,000N. The car next to the locomotive has a mass of 80,000kg, next one, 50,000kg, and the last one, 70,000 kg. you can neglect the friction on the cars being pulled.
A) what if the magnitude of the force between that the 80,000-kg car exerts on the 50,000-kg car?
B) what is the magnitude of the force that the 50,000-kg car exerts on the 70,000-kg car?
Answer:
a) 60000 N
b) 35000 N
Explanation:
Force from locomotive = 100000 N
mass of first car = 80000 kg
mass of second car = 50000 kg
mass of third car = 70000 kg
friction is neglected in this system
Total mass of the cars = 80000 + 50000 + 70000 = 200000 kg
All the car in the system will accelerate at the same rate since they are pulled by the same force
We know that force F = ma
where
a is the acceleration of the cars
m is the total mass in the system
from this we can say that
a = F/m
a = 100000/200000 = 0.5 m/s^2
a) The total mass involved in this case = mass of the last two cars after the 80000 kg car = 50000 + 70000 = 120000 kg
therefore force exerted F = ma
F = 0.5 x 120000 = 60000 N
b) The total mass in this case = mass of the third car only = 70000 kg
F = ma
F = 70000 x 0.5 = 35000 N
The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The average induced current in the coil is 140 A. What is the resistance of the wire
Answer:
Resistance is 1.75 ohms
Explanation:
Magnetic flux:
[tex]{ \phi{ = NBA}}[/tex]
N is number of turns, N = 8
B is magnetic flux
A is area of projection.
From faradays law:
[tex]E = - \frac{ \triangle \phi}{t} [/tex]
where E is the Electro motive force.
But E = IR
where I is current and R is resistance:
[tex]IR = \frac{( \phi_{1} - \phi _{2}) }{t} \\ \\ 140 \times R = \frac{8 \times (8 - 5)}{0.098} \\ \\ R = \frac{24}{0.098 \times 140} \\ \\ resistance = 1.75 \: ohms[/tex]
Figure (3) shows a car travelling along the route PQRST in 30 minutes. What is the average speed of the car in km/hour?
Answer:
60 km/hour.
Explanation:
We'll begin by calculating the total distance traveled by the car. This is illustrated below:
Total distance traveled = sum of distance between PQRST
Total distance = 10 + 5 + 10 + 5
Total distance = 30 km
Next, we shall convert 30 mins to hour. This can obtained as follow:
Recall:
60 mins = 1 hour
Therefore,
30 mins = 30/60 = 0.5 hour.
Finally, we shall determine the average speed of the car as follow:
Distance = 30 km
Time = 0.5 hour
Speed =?
Speed = distance /time
Speed = 30/0.5
Speed = 60 km/hour
Therefore, the speed of the car is 60 km/hour.
What is the displacement of the object after 3 seconds?
Answer:
3 meters
Explanation:
A radar pulse returns 3.0 x 10-4 seconds after it is sent out, having been reflected by an object. What is the distance between the radar antenna and the object
Answer:
The distance is [tex]D = 45000 \ m[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 3.0 *10^{-4 } \ s[/tex]
Generally the speed of the radar is equal to the speed of light and this has a value
[tex]c = 3.0*10^{8} \ m /s[/tex]
Now the distance covered by the to and fro movement of the radar is mathematically evaluated as
[tex]d = c * t[/tex]
=> [tex]d = 3.0*10^{8} * 3.0*10^{-4}[/tex]
=> [tex]d = 90000 \ m[/tex]
Therefore the distance between the radar antenna and the object is
[tex]D = \frac{d}{2}[/tex]
[tex]D = \frac{ 90000}{2}[/tex]
[tex]D = 45000 \ m[/tex]
The distance between the radar antenna and the object will be 45000 m.
What is a radar antenna?A radar antenna is a device that sends out radio waves and listens for their reflections. The ability of an antenna to identify the exact direction in which an item is placed determines its performance.
The given data in the problem is;
t is the time= 3.0 x 10⁻⁴
d is the distance between the radar antenna and the object=?
c is the peed of light=3×10⁸ m/sec
The radar's speed is usually equal to the speed of light, and this has a value. The distance covered by the radars to and fro movement is now calculated mathematically as
[tex]\rm d= c \times t \\\\ \rm d= 3.0 \times 10^8 \times 3.0 \times 10^{-4} \\\\ d=90000 \ m[/tex]
As a result, the radar antenna's distance from the target is
[tex]\rm D=\frac{d}{2} \\\\ \rm D=\frac{90000}{2} \\\\ \rm D=\ 45000 \ m[/tex]
Hence the distance between the radar antenna and the object will be 45000 m.
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An object on a level surface experiences a horizontal force of 12.7 N due to kinetic friction. The coefficient of kinetic friction is 0.42.
What is the mass of the object? (Express your answer to two significant figures)kg
Answer:
The mass of the object is 3.08 kg.
Explanation:
The horizontal force is12.7 N and the coefficient of the kinetic fraction are 0.42. Now we have to compute the mass of the object. Thus, use the below formula to find the mass of the object.
Let the mass of the object = m.
The coefficient of kinetic friction, n = 0.42
Therefore,
Force, F = n × mg
12.7 = 0.42 × 9.8 × m
m = 3.08 kg
The mass of the object is 3.08 kg.
Calculate the answers to the appropriate number of significant
12.21 x 9.19 =
Is the study of the moons places applied or pure science
Answer:
It is pure science
Explanation:
A basic knowledge for the discovery of unknown laws based on well controlled experiments and deductions from demonstrated facts or truths.
The primary difference between a barometer and a manometer is
A. a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
B. a barometer uses mercury, while a manometer can use any liquid. a barometer is used to measure atmospheric pressure, and a manometer is used to measure absolute pressure.
C a barometer reads in mm, while a manometer reads in Pa.
D a barometer can measure either positivee or negative pressure, while a manometer only
E positive pressure. measures
Answer:
a barometer is used to measure atmospheric pressure, and a manometer is used to measure gauge pressure.
Explanation:
A barometer measures air pressure at any locality with sea level as the reference.
However, a manometer is used to measure all pressures especially gauge pressures. Thus, if the aim is to measure the pressure at any point below a fluid surface, a barometer is used to determine the air pressure. The manometer may now be used to determine the gauge pressure
The algebraic sum of these two values gives the absolute pressure.
A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.
Answer:
82.2 mL
Explanation:
The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;
C1V1=C2V2
Where;
C1= concentration of stock solution
V1= volume of stock solution
C2= concentration of dilute solution
V2= volume of dilute solution
V2= C1V1/C2
V2= 1.48 × 55.6/ 1.0
V2= 82.2 mL
Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is
Answer:
The radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
Explanation:
Given;
intensity of light, I = 1 kW/m²
The radiation pressure of light is given as;
[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]
I kW = 1000 J/s
The energy flux density = 1000 J/m².s
The speed of light = 3 x 10⁸ m/s
Thus, the radiation pressure of the light is calculated as;
[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]
Therefore, the radiation pressure of the light is 3.33 x 10⁻⁶ Pa.
The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is
Answer:
In a distant galaxy, whose light is just arriving from 10 billion light years away, our spectroscope should reveal that the most common element is HELIUM
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is located 1.03 m from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen
Answer:
1.696 nm
Explanation:
For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,
dsinθ = mλ = (1)λ = λ
dsinθ = λ
sinθ = λ/d.
Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m
From trig ratios 1 + cot²θ = cosec²θ
1 + (1/tan²θ) = 1/(sin²θ)
substituting the values of sinθ and tanθ we have
1 + (D/w)² = (d/λ)²
(D/w)² = (d/λ)² - 1
(w/D)² = 1/[(d/λ)² - 1]
(w/D) = 1/√[(d/λ)² - 1]
w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹ = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.
w is also the distance from the center to the other principal maximum on the other side.
So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm
So, the minimum width of the screen must be 1.696 nm
Experiment to find ways to make rainbows.
a) Insert at least one setup where light passing through a prism gives a rainbow and describe why a rainbow is formed.
b) Explain why only some types of light will yield rainbows.
Answer:
Explanation:
a) To get a rainbow from a prism arrangement, we will need
A triangular prismA black cardboard boxA source of white light (light from the window will suffice)A pocket knifeFirst, you cut a slit in one end of the cardboard with the pen knife.
Next you open up a space on top of the cardboard through which you can observe the experiment and its result.
Next, you place the triangular prism with its slant face facing the the cut slit.
Finally, position the slit to face the light from the open window, and adjust the prism till the projected bands of colored light (rainbow) is very much obvious on the other end of the box, opposite the slit.
b) For a light to yield rainbow, it most be composed of different component colors of light. The colors of light is due to the difference in wavelength, and dispersion is due to the different in the wavelengths of the component light. So to get rainbow from a light source, the light must not be monochromatic. This means that only light composed of component light of different colors can produce rainbow. Light from the sun for example is composed of 7 distinct colors of light, and white light can be created with just three colors; blue, green, and red light.