Answer:
Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.
Explanation:
A scientific method can be defined as a research method that typically involves the use of experimental and mathematical techniques which comprises of a series of steps such as systematic observation, measurement, and analysis to formulate, test and modify a hypothesis.
Measurements refers to a process which typically involves identifying and determining the dimensions of a physical object.
Basically, the dimensions include important parameters such as width, height, length, area, volume, circumference, breadth, etc.
A 40-turn coil has a diameter of 11 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.40 T so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf induced in the coil (in V) if the magnetic field is reduced to zero uniformly in the following times.
(a) 0.30 s V
(b) 3.0 s V
(c) 65 s V
Answer:
(a) emf = 0.507 V
(b) emf = 0.0507 V
(c) emf = 0.00234 V
Explanation:
Given;
number of turns of the coil, N = 40 turns
diameter of the coil, d = 11 cm
radius of the coil, r = 5.5 cm = 0.055 m
magnitude of the magnetic field, B = 0.4 T
The magnitude of the induced emf is calculated as;
[tex]emf = - N\frac{d\phi}{dt} \\\\where;\\\\\phi \ is \ magnetic \ flux= BA \\\\A \ is the \ area \ of \ the \ coil = \pi r^2 = \pi (0.055)^2 = 0.0095 \ m^2\\\\emf = - N \frac{dB.A}{dt} = -NA\frac{dB}{dt} \\\\emf = -NA\frac{(B_2 - B_1)}{t} \\\\emf = NA \frac{(B_1 - B_2)}{t} \\\\the \ final \ magnetic \ field \ is \ reduced \ to \ zero;\ B_2 = 0\\\\emf = \frac{NAB_1}{t}[/tex]
(a) when the time, t = 0.3 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{0.3} = 0.507 \ V[/tex]
(b) when the time, t = 3.0 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{3} = 0.0507 \ V[/tex]
(c) when the time, t = 65 s
[tex]emf = \frac{NAB_1}{t} = \frac{40\times 0.0095\times 0.4}{65} = 0.00234 \ V[/tex]
The current in resistor Y is..?
(A)
Explanation:
We can see that the resistors are connected in parallel so all of them have the same voltage of 100 V. We also know that
[tex]P = VI[/tex]
Since resistor Y dissipates 100 W of power, we can solve for the current as
[tex]I = \dfrac{P}{V} = \dfrac{100\:\text{W}}{100\:\text{V}} = 1.0\:\text{A}[/tex]
The current in resistor Y is
a)1.0 AYou are working on a project to make a more efficient engine. Your team is investigating the possibility of making electrically controlled valves that open and close the input and exhaust openings for an internal combustion engine. Determine the stability of the valve by calculating the force on each of its sides and the net force on the valve.
The valve is made of a thin but strong rectangular piece of non-magnetic material that has a current-carrying wire along its edges. The rectangle is 0.35 cm x 1.83 cm. The valve is placed in a uniform magnetic field of 0.15 T such that the field lies in the plane of the valve and is parallel to the short sides of the rectangle. The region with the magnetic field is slightly larger than the valve. When a switch is closed, a 1.7 A current enters the short side of the rectangle on one side and leaves on the opposite short side of the rectangle. At the suggestion of a colleague, who is hoping to ensure different currents along the sides of the valve, resistors have been included along the wire on each of the short sides of the valve. The value of the resistor on one side is twice that on the other side.
Answer:
The answer is "0.00466 N".
Explanation:
[tex]F=(B \times i) L\\\\[/tex]
therefore the smaller side is parallel to magnetic field
[tex]\therefore \\\\F= B i L\ \sin\ 'o'=0 \ N[/tex]
calculating the force on the layer side:
[tex]\to F=0.15 \times 1.7 \times 0.0183 \times \sin 90^{\circ}=0.00466\ N\\\\[/tex]
Therefore [tex]F_o[/tex] the net force on the rectangular loop [tex]= 0.00466 \ N[/tex]
Use the DC Construction kit to build a simple circuit to perform the following task:
You are asked to use a single resistor and a 110 V DC battery for the purpose of boiling a litter of water (4,184 Joule/Kg*degree Celsius), with a starting temperature of 20 C, in exactly 4 minutes.
Answer:
The resistance is 8.7 ohm.
Explanation:
Voltage, V = 110 V
mass, m = 1 kg
change in temperature, T = 100 - 20 = 80 C
time, t = 4 min = 4 x 60 = 240 s
specific heat, c = 4184 J/kg C
let the resistance is R.
The heat generated by the heater is used to the heat the water.
[tex]\frac{V^2}{R} t = m c T \\\\\frac{110^2}{R}\times 240 = 1\times 4184\times 80\\\\R = 8.7 ohm[/tex]
Twin skaters approach each other with identical speeds. Then, the skaters lock hands and spin. Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate
Answer:
[tex]\omega=3.135rad/s[/tex]
Explanation:
From the question we are told that:
initial Speed [tex]V_1=2.50[/tex]
Mass [tex]m=70.0kg[/tex]
Center of mass [tex]d=0.0.800m\[/tex]
Generally the equation for angular velocity is is mathematically given by
[tex]\omega=\frac{v}{r}\\\\\omega=\frac{2.50}{0.0800}[/tex]
[tex]\omega=3.135rad/s[/tex]
A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Answer:
N = 107.94 N
Explanation:
For this exercise we must use Newton's second law.
Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical
X axis
Fₓ = ma
ej and
N -F_y - W = 0
let's use trigonometry to decompose the applied force
cos -35 = Fₓ / F
sin -35 = F_y / F
Fₓ = F cos -35
F_y = F sin -35
Fₓ = 40.0 cos -35 = 32.766 N
F_y = 40.0 sin -35 = -22.94 N
we substitute
N = Fy + W
N = 22.94 + 85
N = 107.94 N
Is there any absolute rest or motion? Describe the types of motion with one example of each type
A singly charged 7Li ion has a mass of 1.16 10-26 kg. It is accelerated through a potential difference of 523 V and subsequently enters a uniform magnetic field of magnitude 0.370 T perpendicular to the ion's velocity. Find the radius of its path.
Answer:
[tex]R=0.023m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.16*10^{-26}[/tex]
Potential difference [tex]V=523V[/tex]
Magnitude [tex]m=0.370 T[/tex]
Generally the equation for Velocity is mathematically given by
[tex]\frac{1}{2}mv^2=ev[/tex]
[tex]v=\frac{2ev}{m}[/tex]
[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]
[tex]v=12.22*10^4m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=qvBsin \theta[/tex]
Where
[tex]qVB=m\frac{v^2}{R}[/tex]
[tex]F=m\frac{v^2}{R}sin\theta[/tex]
Therefore
[tex]R=\frac{mv}{qB sin \theta}[/tex]
[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]
[tex]R=0.023m[/tex]
When tightening a bolt, you push perpendicularly on a wrench a with force of 165 N at a distance of 0.140m from the center of the bolt. (A) How much torque are you exerting in newton x meters (relative to the center of the bolt)
Answer:
Part a)
23.1 Nm
..........
How can I solve the following statement?
What is the magnitude of the electric field at a point midway between a −8.3μC and a +7.8μC charge 9.2cm apart? Assume no other charges are nearby.
Answer:
The net electric field at the midpoint is 6.85 x 10^7 N/C.
Explanation:
q = − 8.3 μC
q' = + 7.8 μC
d = 9.2 cm
d/2 = 4.6 cm
The electric field due to the charge q at midpoint is
[tex]E = \frac{k q}{r^2}\\\\E = \frac{9\times 10^9\times 8.3\times 10^{-6}}{0.046^2}\\\\E = 3.53\times 10^7 N/C[/tex] leftwards
The electric field due to the charge q' at midpoint is
[tex]E' = \frac{k q}{r^2}\\\\E' = \frac{9\times 10^9\times 7.8\times 10^{-6}}{0.046^2}\\\\E' = 3.32\times 10^7 N/C[/tex]
The resultant electric field at mid point is
E'' = E + E' = (3.53 + 3.32) x 10^7 = 6.85 x 10^7 N/C
. A ball of mass 0.50 kg is rolling across a table top with a speed of 5.0 m/s. When the ball reaches the edge of the table, it rolls down an incline onto the floor 1.0 meter below (without bouncing). What is the speed of the ball when it reaches the floor?
Answer:
4
Explanation:
a method of reducing friction
Answer:
Lubrication
Explanation:
People oil/lubricate bicycle chains because the chain turns around the cogs and rub together so this help with friction.
Hope this helps :)
Answer:
The method of reducing friction are :
i) In moving parts of machine friction can be reduced by using a ball bearing between the moving surfaces
ii) The bodies of aeroplane ,ship ,boat etc are made streamlined to reduce friction.
iii) Friction can be reduced by polishing rough surfaces. For example : carrom boards are highly polished to reduce friction.
I hope this help you:)
True or false : conservation of energy gives a relationship between the speed of a falling object and the height from which it was dropped
Answer:
truee
Explanation:
Dựa vào môi trường hoạt chất, laser được phân thành
Answer:
ok
Explanation:
An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 3.3 kg determines its density to be 8100 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube
Answer:
[tex]v=0.9833\ c[/tex]
Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,
[tex]$\text{Density} = \frac{m}{lwh}$[/tex]
Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 [tex]kg/m^3[/tex]
So,
[tex]$8100=\frac{3.3}{l \times 0.13 \times 0.13}$[/tex]
[tex]$l=\frac{3.3}{8100 \times 0.13 \times 0.13}$[/tex]
l = 0.024 m
Then for relativistic length contraction,
[tex]$l= l' \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.024= 0.13 \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.184= \sqrt{1-\frac{v^2}{c^2}}$[/tex]
[tex]$0.033= 1-\frac{v^2}{c^2}}$[/tex]
[tex]$\frac{v^2}{c^2}= 0.967$[/tex]
[tex]$\frac{v}{c}=0.9833$[/tex]
[tex]v=0.9833\ c[/tex]
Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 mmmm. The charge on the capacitor is 7.8 nCnC . A proton is released from rest next to the positive plate. Part A How long does it take for the proton to reach the negative plate
Answer:
t = 2.09 10⁻³ s
Explanation:
We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance
let's start with Newton's second law
∑ F = m a
the force is electric
F = q E
we substitute
q E = m a
a = [tex]\frac{q}{m} \ E[/tex]
a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]
a = 1.37 10³ m / s²
now we can use kinematics
x = v₀ t + ½ a t²
indicate that rest starts v₀ = 0
x = 0 + ½ a t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]
t = 2.09 10⁻³ s
To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 meters from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom
The magnitude of the forces acting at the top are;
[tex]\mathbf{F_{Top, \ x}}[/tex] = 132.95 N
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0
The magnitude of the forces acting at the bottom are;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]\mathbf{ F_f}[/tex] = -132.95 N
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = 784.8 N
The known parameters in the question are;
The mass of the person, m₁ = 70.0 kg
The length of the ladder, l = 6.00 m
The mass of the ladder, m₂ = 10.0 kg
The distance of the base of the ladder from the house, d = 2.00 m
The point on the roof the ladder rests = A frictionless plastic rain gutter
The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder
The location of the point the person is standing = 3 meters from the bottom
g = The acceleration due to gravity ≈ 9.81 m/s²
The required parameters are;
The magnitudes of the forces on the ladder at the top and bottom
The strategy to be used;
Find the angle of inclination of the ladder, θ
At equilibrium, the sum of the moments about a point is zero
The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C
Taking moment about the point of contact of the ladder with the ground, B gives;
[tex]\sum M_B[/tex] = 0
Therefore;
[tex]\sum M_{BCW}[/tex] = [tex]\sum M_{BCCW}[/tex]
Where;
[tex]\sum M_{BCW}[/tex] = The sum of clockwise moments about B
[tex]\sum M_{BCCW}[/tex] = The sum of counterclockwise moments about B
Therefore, we have;
[tex]\sum M_{BCW}[/tex] = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81
[tex]\sum M_{BCCW}[/tex] = [tex]F_R[/tex] × √(6² - 2²)
Therefore, we get;
2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = [tex]F_R[/tex] × √(6² - 2²)
[tex]F_R[/tex] = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95
The reaction force on the wall, [tex]F_R[/tex] ≈ 132.95 N
We note that the magnitude of the reaction force at the roof, [tex]F_R[/tex] = The magnitude of the frictional force of bottom of the ladder on the floor, [tex]F_f[/tex] but opposite in direction
Therefore;
[tex]F_R[/tex] = [tex]-F_f[/tex]
[tex]F_f[/tex] = - [tex]F_R[/tex] ≈ -132.95 N
Similarly, at equilibrium, we have;
∑Fₓ = [tex]\sum F_y[/tex] = 0
The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;
[tex]\sum F_y[/tex] = -70.0 × 9.81 - 10 × 9.81 + [tex]F_{By}[/tex]
∴ The upward force acting at the bottom, [tex]F_{By}[/tex] = 784.8 N
Therefore;
The magnitudes of the forces at the ladder top and bottom are;
At the top;
[tex]\mathbf{F_{Top, \ x}}[/tex] = [tex]F_R[/tex] ≈ 132.95 N←
[tex]\mathbf{F_{Top, \ y}}[/tex] = 0 (The surface upon which the ladder rest at the top is frictionless)
At the bottom;
[tex]\mathbf{F_{Bottom, \ x}}[/tex] = [tex]F_f[/tex] ≈ -132.95 N →
[tex]\mathbf{F_{Bottom, \ y}}[/tex] = [tex]F_{By}[/tex] = 784.8 N ↑
Learn more about equilibrium of forces here;
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Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?
Answer:
The charges are + 74.3 μC and - 74.3 μC
Explanation:
Let the charges be q and q'.
Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by
F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m
When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.
They now repel each other.
So, the magnitude of the force of repulsion is given by
F' = k[(q + q')/2][(q + q')/2]/r²
F' = k[(q + q')²/4r²
Since the magnitude of the force of attraction and repulsion are the same, we have that
F = F'
kqq'/r² = k[(q + q')²/4r²
qq' = (q + q')²/4
(q + q')² = 4qq'
q² + 2qq' + q'² = 4qq'
q² + 2qq' - 4qq' + q'² = 0
q² - 2qq' + q'² = 0
(q - q')² = 0
q - q' = 0
q = q'
Substituting q = q' into F, we have
F = kqq'/r²
F = kq²/r²
making q subject of the formula, we have
q² = Fr²/k
q = √(Fr²/k)
q = r√(F/k)
Substituting the values of the variables into the equation, we have
q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)
q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)
q = 0.189 m(0.3923 × 10⁻³ C/m)
q = 0.0743 × 10⁻³ C
q = 74.3 × 10⁻³ × 10⁻³ C
q = 74.3 × 10⁻⁶ C
q = 74.3 μC
Since q and q' initially attract, it implies that they initially had opposite charges.
So, q = 74.3 μC and q' = -74.3 μC
So, the charges are + 74.3 μC and - 74.3 μC
Which of the following quantities is measured by the area under the velocity time graph? (a) Magnitude of velocity (b) Magnitude of acceleration (c) Magnitude of displacement (d) Average Speed
Answer:
c.
magnitude of displacement
Andrea's near point is 20.0 cm and her far point is 2.0 m. Her contact lenses are designed so that she can see objects that are infinitely far away. What is the closest distance that she can see an object clearly when she wears her contacts?
Answer:
the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Explanation:
Given the data in the question,
near point = 20 cm
far point = 2 m = 200 cm
Now, for an object that is infinitely far away, the image is at is its far point.
so using the following expression, we can determine the focal length
1/f = 1/i + 1/o
where f is the focal length, i is the image distance and o is the object distance.
here, far point i = 2 m = 200 cm and v is ∞
so we substitute
1/f = 1/(-200 cm) + 1/∞
f = -200 cm
Also, for object at its closest point, the image appear at near point,
so
1/f = 1/i + 1/o
we make o the subject of formula
o = ( i × f ) / ( i - f )
given that near point i = 20 cm
we substitute
o = ( -20 × -200 ) / ( -20 - (-200) )
o = 4000 / 180
o = 22.2 cm
Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm
Steel railway tracks are laid at 8oC. What size of expansion gap are needed 10m long rail sections if the ambient temperature varies from -10oC to 50oC? [Linear expansivity of steel = 12 x]
Answer:
Gap left = Change in length on heating
Gap=Initial length×Coefficient of linear expansion×change in temperature
Gap=10×0.000012×15m
⟹Gap=0.0018 m
this is an example u have to put your equation in it
when a boron is added to a pure semi conducter it becomes
Answer:
it becomes a p-type conductor
Explanation:
answer from gauth math
Sunlight above the Earth's atmosphere has an intensity of 1.36 kW/m2. If this is reflected straight back from a mirror that has only a small recoil, the light's momentum is exactly reversed, giving the mirror twice the incident momentum. (a) Calculate the force per square meter of mirror (in N/m2). N/m2 (b) Very low mass mirrors can be constructed in the near weightlessness of space, and attached to a spaceship to sail it. Once done, the average mass per square meter of the spaceship is 0.170 kg. Find the acceleration (in m/s2) of the spaceship if all other forces are balanced. m/s2 (c) How fast (in m/s) is it moving 24 hours later
Answer:
a) [tex]F=9.2*10^{-6}N/m^2[/tex]
b) [tex]a=5.4*10^{-4}m/s[/tex]
c) [tex]v=46.65m/s[/tex]
Explanation:
From the question we are told that:
Intensity I= 1.36 kW/m2=>1360W/m
b)Average mass per square meter m = 0.170 kg
c) [tex]T=24hrs[/tex]
a)
Generally the equation for force per square meter is mathematically given by
[tex]F=\frac{2E}{C}[/tex]
[tex]F=\frac{2*1360}{3*10^8}[/tex]
[tex]F=9.2*10^{-6}N/m^2[/tex]
b)
Generally the equation for force is mathematically given by
F=ma
Therefore
[tex]a=\frac{F}{m}[/tex]
[tex]a=\frac{9.2*10^{-6}N/m^2}{0.0170}[/tex]
[tex]a=5.4*10^{-4}m/s[/tex]
c)
Generally the Newton's equation for Motion is mathematically given by
[tex]v=u+at[/tex]
[tex]v=0+5.4*10^{-4}m/s*(24*3600)[/tex]
[tex]v=46.65m/s[/tex]
Which of the following would most likely produce the strongest magnetic
field?
A. A single moving electron
B. A stationary electric charge
C. A current in a straight wire
D. A current in a coil
Answer:
I current in a coil,,,,,,
Answer:D?
Explanation:
Sorry if i'm wrong....
Identify the correct descriptions of alpha particles. Select one or more: Alpha particles are more massive than beta particles. An alpha particle is a helium nucleus. An alpha particle has a negative charge. An alpha particle is a form of electromagnetic radiatio
Answer:
Alpha particles are more massive than beta particles.
Explanation:
The alpha particles are also called double-positive Heilum Nuclei because they have a charge of "+2" and a mass of 4 a.m.u. The properties of the alpha particles are as follows:
1. It possesses high energy due to high velocity. It is 7.7 MeV for most energetic from Rac (i.e: Bismuth-214)
2. It has a very high ionizing power. A 7.7 MeV particle produces about 0.2 x 10⁶ ions.
3. The range of alpha particles is very small. It is about 7 x 10⁻² m and only 4 x 10⁻⁵ m in aluminum for 7.7 MeV alpha-particle.
4. Alpha particles produce fluorescence on striking certain substances, such as zinc sulphide and bariumplatinocynide.
The beta particles are fast-moving electrons, which have a negligible mass.
Hence, the correct option is:
Alpha particles are more massive than beta particles.
How many types of physics?
Answer:
Two Main Branches of Physics
it is Classical Physics and Modern Physics.
Explanation:
Further sub Physics branches are Mechanics, Electromagnetism, Thermodynamics, Optics, etc. The rapid progress in science during recent years has become possible due to discoveries and inventions in the field of physics.
hope it helped
calculate the value of 200°C in Kelvin
Answer:
473.15
Explanation:
A 1050 kg car accelerates from 11.3 m/s to 26.2 m/s . What impulse does the engine give?
Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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There are 5640 lines per centimeter in a grating that is used with light whose wavelegth is 455 nm. A flat observation screen is located 0.661 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
The minimum width of the screen is 34 cm.
For a diffraction grating, dsinθ = mλ where d = grating spacing = 1/5640 lines per cm = 1/5640 cm per line = 1/5640 × 10⁻² m per line, θ = angle between principal maximum and the center axis of the grating, m = order of maxima = 1 (since we require the position of the principal maximum) and λ = wavelength = 455 nm = 455 × 10⁻⁹ m
So, sinθ = mλ/d
Also tanθ = L/D where θ = angle between principal maximum and the center axis of the grating, L = distance between central maximum and principal maximum and D = distance between grating and screen = 0.661 m.
For small angles sinθ ≈ tanθ
So, mλ/d = L/D
making L subject of the formula, we have
L = mλD/d
L = 1 × 455 × 10⁻⁹ m × 0.661 m ÷ 1/5640 × 10⁻² m per line
L = 1 × 455 × 10⁻⁹ m × 0.661 m × 5640 × 10² line per m
L = 1696258.2 × 10⁻⁷ m
L = 0.16963 m
L ≅ 0.17 m
So, for centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is w = 2L.
So, w = 2 × 0.17 m
w = 0.34 m
w = 34 cm
So for the centers of all the principal maxima formed on either side of the central maximum fall on the screen, the minimum width of the screen is 34 cm.
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