Answer:
One can write F = K d^-2 where K = G M m
So dF/dd = -2 K d^-3 = -2 K / d^3 (As d increases F decreases - it is opposite to the direction of F)
Which of the following is true about resistivity of any given metal? depends on its temperature. varies nearly linearly with temperature. has units of ohm-meter. A. II and III only B. I and II only C. I and III only D. I, II and III E. III only
___________________
[tex]\huge{\underline{\sf{\blue{Answer}}}}[/tex]
___________________
[tex]\sf{C. \:I\: and \:III}[/tex]
___________________
The correct statements about resistivity of any given metal are The resistivity of metal is more than that of insulators and Metals can carry electricity more easily than insulators. Option a and c are correct answer.
Resistivity is a property that quantifies how strongly a material opposes the flow of electric current. Metals have lower resistivity compared to insulators. This means that metals allow electric current to flow more easily than insulators.
Due to their lower resistivity, metals have higher electrical conductivity and can carry electric current more easily compared to insulators. Insulators, on the other hand, have high resistivity and hinder the flow of electric current. Resistivity is a material-specific property and varies for different substances. Metals, such as copper or aluminum, have low resistivity and are often used as conductors for transmitting electricity. Insulators, such as rubber or plastic, have high resistivity and are used to prevent the flow of electricity.
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The complete question is, "Which of the following statements are true about resistivity of any given metal?
A. The resistivity of metal is more than that of insulators.
B. The insulators and metals have same resistivity.
C. Metals can carry electricity more easily than insulators.
D. The resistivity of insulators is more than that of metals.
A hydraulic vane pump has a real flow of 20 liters per minute, a pressure of 230 bar, a pump speed of 1400 rpm. Know that the input power is 10kW and the mechanical efficiency is 88%.
a) Calculate the volumetric efficiency of the pump
b) Calculate the specific volume of the pump (cm/rev). Question 3 (2,5d): Design a pneumatic transmission system to control 02 single-acting cylinders, using 02 reversing valves 3/2 acting by push button, 02 throttle valves - one-way. Describe the working principle of the system.
Explanation:
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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
If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance
Answer:
杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音
Explanation:
莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd
Answer:
B velocity
Explanation:
You want to swim from one side of a river to another side. Assume your speed is three miles per hour in the west direction, with negligible water velocity. When you reach a certain point, you will encounter water flow with a velocity of 6.2 miles per hour in the north direction. What is your resultant speed and direction
Answer:
speed = 6.71 mph and angle is 71.2 degree.
Explanation:
speed of person, u = 3 miles per hour
speed of water, v = 6 miles per hour
Resultant speed
[tex]V =\sqrt{v^2 + u^2}\\\\V = \sqrt{3^2 + 6^2}\\\\V = 6.71 mph[/tex]
The angle from the west is
tan A = 6/2 = 3
A = 71.6 degree
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms. At what depth did this reflection occur? (The average propagation speed for sound in body tissue is 1540 m/s)
Answer:
10.01 cm
Explanation:
Given that,
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.
The average propagation speed for sound in body tissue is 1540 m/s.
We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,
[tex]v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m[/tex]
or
d = 10.01 cm
So, the reflection will occur at 10.01 cm.
How much energy is stored in a spring that is compressed 0.650m if the spring constant is 725N/m?
Answer:
53.8Joule
Explanation:
hope it is helpful
please mark it as brainliest
Answer:
approximate 153.1J
Explanation:
W= 1/2k(x^2) = 1/2x725x(0.650)^2 = 153.15625 (J)
A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days?
285.3 days
Explanation:
The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write
[tex]F_c = F_G[/tex]
or
[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]
where M is the mass of the star and R is the orbital radius around the star. We know that
[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]
where C is the orbital circumference and T is orbital period. We can then write
[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]
Isolating [tex]T^2[/tex], we get
[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]
Taking the square root of the expression above, we get
[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]
which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as
[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]
[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]
27. The part of the Earth where life exists .
Mesosphere
Stratosphere
Troposphere
Biosphere
Answer:
Biosphere is the part of the earth where life exists.
If 56.5 m3 of a gas are collected at a pressure of 455 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg? *
Assuming ideal conditions, Boyle's law says that
P₁ V₁ = P₂ V₂
where P₁ and V₁ are the initial pressure and temperature, respectively, and P₂ and V₂ are the final pressure and temperature.
So you have
(455 mm Hg) (56.5 m³) = (632 mm Hg) V₂
==> V₂ = (455 mm Hg) (56.5 m³) / (632 mm Hg) ≈ 40.7 m³
An observer on Earth sees Planet X to be stationary and also sees a rocket traveling toward Planet X at 0.5c. The rocket emits a pulse of light that travels outward in all directions. According to an observer on Planet X, how fast is the light pulse traveling toward them?
a) 2c/3
b) c/2
c) 2c/3
d) 5c/6
e) c
(E) c
Explanation:
The speed of light is always equal to c regardless of the relative motion of the light source.
A nylon string on a tennis racket is under a tension of 285 N . If its diameter is 1.10 mm , by how much is it lengthened from its untensioned length of 29.0 cm ? Use ENylon=5.00×109N/m2.
Answer:
1.74×10⁻³ m
Explanation:
Applying,
ε = Stress/strain............. Equation 1
Where ε = Young's modulus
But,
Stress = F/A.............. Equation 2
Where F = Force, A = Area
Strain = e/L.............. Equation 3
e = extension, L = Length.
Substitute equation 2 and 3 into equation 1
ε = (F/A)/(e/L) = FL/eA............. Equation 4
From the question,
Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,
A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²
Substitute these values into equation 4
5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)
Solve for e
e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)
e = 82.65/4.74925×10⁴
e = 1.74×10⁻³ m
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m. What is the work done on the ball by the 34.4 N tension force in the string during one half-revolution of the ball
Answer:
the work done on the ball is 0
Explanation:
Given the data in the question;
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.
circle circumference = 2πr = 2.1 m
radius r will be; r = 2.1 m / 2π = 0.33 m
Tension force = 34.4 N
one half revolution means, displacement of the ball is;
d = 2r = 2 × 0.33 = 0.66 m
Now, Work done = force × displacement × cosθ
we know that, the angle between the tension force on string and displacement of object is always 90.
so we substitute
Work done = 34.4 N × 0.66 m × cos(90)
Work done = 34.4 N × 0.66 m × 0
Work done = 0 J
Therefore, the work done on the ball is 0
A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration of 1.8 m/s^2 toward the east. Let the x direction be eastward and the y direction be northward, and let the skateboarder be at the origin at t=0.
a. What is her x position at t=0.60s?
b. What is her y position at t=0.60s?
c. What is her x velocity component at t=0.60s?
d. What is her y velocity component at t=0.60s?
Answer:
a) The x-position of the skateboarder is 0.324 meters.
b) The y-position of the skateboarder is -2.16 meters.
c) The x-velocity of the skateboard is 1.08 meters per second.
d) The y-velocity of the skateboard is -3.6 meters per second.
Explanation:
a) The x-position of the skateboarder is determined by the following expression:
[tex]x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}[/tex] (1)
Where:
[tex]x_{o}[/tex] - Initial x-position, in meters.
[tex]v_{o,x}[/tex] - Initial x-velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a_{x}[/tex] - x-acceleration, in meters per second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:
[tex]x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}[/tex]
[tex]x(t) = 0.324\,m[/tex]
The x-position of the skateboarder is 0.324 meters.
b) The y-position of the skateboarder is determined by the following expression:
[tex]y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}[/tex] (2)
Where:
[tex]y_{o}[/tex] - Initial y-position, in meters.
[tex]v_{o,y}[/tex] - Initial y-velocity, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]a_{y}[/tex] - y-acceleration, in meters per second.
If we know that [tex]y_{o} = 0\,m[/tex], [tex]v_{o,y} = -3.6\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{y} = 0\,\frac{m}{s^{2}}[/tex], then the x-position of the skateboarder is:
[tex]y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}[/tex]
[tex]y(t) = -2.16\,m[/tex]
The y-position of the skateboarder is -2.16 meters.
c) The x-velocity of the skateboarder ([tex]v_{x}[/tex]), in meters per second, is calculated by this kinematic formula:
[tex]v_{x}(t) = v_{o,x} + a_{x}\cdot t[/tex] (3)
If we know that [tex]v_{o,x} = 0\,\frac{m}{s}[/tex], [tex]t = 0.60\,s[/tex] and [tex]a_{x} = 1.8\,\frac{m}{s^{2}}[/tex], then the x-velocity of the skateboarder is:
[tex]v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)[/tex]
[tex]v_{x}(t) = 1.08\,\frac{m}{s}[/tex]
The x-velocity of the skateboard is 1.08 meters per second.
d) As the skateboarder has a constant y-velocity, then we have the following answer:
[tex]v_{y} = -3.6\,\frac{m}{s}[/tex]
The y-velocity of the skateboard is -3.6 meters per second.
An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answer:
a) [tex]P=0.80[/tex]
b) [tex]1.25Hz[/tex]
c) [tex]A=25cm[/tex]
Explanation:
From the question we are told that:
Travel Time [tex]T=0.40s[/tex]
Distance [tex]d=50cm[/tex]
a)
Period
Time taken to complete one oscillation
Therefore
[tex]P=2*T\\\\P=2*0.40[/tex]
[tex]P=0.80[/tex]
b)
Frequency is
[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]
[tex]1.25Hz[/tex]
c)
Amplitude:the distance between the mean and extreme position
[tex]A=\frac{50}{2}[/tex]
[tex]A=25cm[/tex]
A resistor is submerged in an insulated container of water. A voltage of 12 V is applied to the resistor resulting in a current of 1.2 A. If this voltage and current are maintained for 5 minutes, how much electrical energy is dissipated by the resistor
Explanation:
Given:
[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]
[tex]V = 12 V[/tex]
[tex]I = 1.2 A[/tex]
Recall that power P is given by
[tex]P = VI[/tex]
so the amount of energy dissipated [tex]\Delta E[/tex] is given by
[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]
[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]
A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.
∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a
==> a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²
To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².
The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:
Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)
F[tex]_{horizontal}[/tex] = 150 N × cos(60°)
F[tex]_{horizontal}[/tex] = 150 N × 0.5
F[tex]_{horizontal}[/tex] = 75 N
Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)
F[tex]_{vertical}[/tex] = 150 N × sin(60°)
F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)
F[tex]_{vertical}[/tex] ≈ 129.9 N
Now, let's calculate the net force in the horizontal direction:
Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]
F[tex]_{net horizontal}[/tex] = 75 N - 15 N
F[tex]_{net horizontal}[/tex] = 60 N
Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:
F[tex]_{net horizontal}[/tex] = m × a
60 N = 20 kg × a
Now, solve for acceleration (a):
a = 60 N / 20 kg
a = 3 m/s²
So, the acceleration of the box is 3 m/s².
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1. A body moving with uniform acceleration of 10 m/s2 covers a distance of 320 m. if its initial velocity was 60 m/s. Calculate its final velocity.
Answer:
v² = u² + 2as
v² = 3600 + 6400
v² = 10000
v = 100
Explanation:
final velocity is 100 m/s
According to third equation of kinematics
[tex]\boxed{\sf v^2-u^2=2as}[/tex]
[tex]\\ \sf\longmapsto v^2=u^2+2as[/tex]
[tex]\\ \sf\longmapsto v^2=(60)^2+2(10)(320)[/tex]
[tex]\\ \sf\longmapsto v^2=3600+3400[/tex]
[tex]\\ \sf\longmapsto v^2=10000[/tex]
[tex]\\ \sf\longmapsto v=\sqrt{10000}[/tex]
[tex]\\ \sf\longmapsto v=100m/s[/tex]
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
Current is a measure of…
what are the two main types of sound like soundwave
Answer:
acoustic energy and mechanical energy
Explanation:
each type of sounds has to be tackled in their own way.
a cheetah running at a velocity of 18m/s accelerates at 1m/s² for 5sec what is the final velocity of the cheetah
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.2 m from the CD, and the second bright fringe is 0.803 m from the central maximum, what is the spacing (in m) of grooves on the CD
Answer:
[tex]d=1.29*10^{-6}m[/tex]
Explanation:
From the question we are told that:
Distance of wall from CD [tex]D=1.4[/tex]
Second bright fringe [tex]y_2= 0.803 m[/tex]
Let
Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m
Generally the equation for Interference is mathematically given by
[tex]y=frac{n*\lambda*D}{d}[/tex]
Where
[tex]d=\frac{n*\lambda*D}{y}[/tex]
[tex]d=\frac{2*431 *10^{-9}m*1.4}{0.803}[/tex]
[tex]d=1.29*10^{-6}m[/tex]
A Ball A and a Ball B collide elastically. The initial momentum of Ball A is -2.00kgm/s and the initial momentum of Ball B is -5.00kgm/s. Ball A has a mass of 4.00kg and is traveling at 2.50 m/s after the collision. What is the velocity of ball B if it has a mass of 6.50kg?
The velocity of B after the collision is obtained as -2.6 m/s.
What is the principle of conservation of momentum?Now we now that the principle of conservation of momentum states that the momentum before collision is equal to the momentum after collision.
Thus;
(-2.00kgm/s) + ( -5.00kgm/s) = ( 4.00kg * 2.50 m/s) + ( 6.50kg * v)
-7 = 10 + 6.5v
-7 - 10 = 6.5v
v = -7 - 10 /6.5
v = -2.6 m/s
Hence, the velocity of B after the collision is obtained as -2.6 m/s.
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In the lab room, you are sitting in an office chair with wheels while holding onto a force sensor, and the chair is at rest. One end of a lightweight string is attached to the force sensor, and your lab partner is holding the other end of the string. Your lab partner then moves away from you, pulling on the string. Describe how your lab partner must move for the force sensor to read a constant force. Explain
Answer:
a circular motion a constant force can be measured
Explanation:
The force is expressed by the relation
F = m a
The bold are vectros.
Therefore, when your partner moves away, he has a reading of a force, so that this force remains constant there must be an acceleration at all times, one way to achieve these is with a circular motion with constant speed, in this case the module of the velocity is constant, but the direction changes at each point and there is an acceleration at each point.
Consequently with a circular motion a constant force can be measured
The force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.
What is Force?A force can be defined as an influence that can change the motion of an object. The force is expressed by the relation
[tex]F = m a[/tex]
The force is dependent on the mass and acceleration of the object. The acceleration is a vector quantity, so the force will be a vector quantity.
Given that, in a lab room, you are sitting on a wheelchair at one end and at the other end, lab partner then moves away from you, pulling on the string that is attached to the force sensor.
When the lab partner moves away and pulled the string, there must be an acceleration during the motion. If the lab partner moves in a circular motion, then the velocity is constant but the direction changes at each point. There is an acceleration at each point that will be constant.
As the force depends on the acceleration, hence the force will be constant during the circular motion for constant acceleration.
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A block with a mass of 0.26 kg is attached to a horizontal spring. The block is pulled back from its equilibrium position until the spring exerts a force of 1.2 N on the block. When the block is released, it oscillates with a frequency of 1.4 Hz. How far was the block pulled back before being released?
Answer:
2
Explanation:
pulling force because of it force
Answer:
5.9 cm
Explanation:
f: frequency of oscillation
frequency of oscillationk: spring constant
frequency of oscillationk: spring constantm: the mass
[tex]f = \frac{1}{2\pi} \sqrt{ \frac{k}{m} } [/tex]
in this problem we know,
F= 1.4 Hz
m= 0.26 kg
By re-arranging the formula we get
[tex]k = {(2\pi \: f )}^{2} m = {(2\pi(1.4hz))}^{2} 0.26kg = 20.1 \frac{n}{m} [/tex]
The restoring force of the spring is:
F= kx
where
F= 1.2 N
k= 20.1 N/m
x: the displacement of the block
[tex]x = \frac{f}{k} = \frac{1.2 \: n}{20.1 \frac{n}{m} } = 0.059m \: = 5.9 \: cm[/tex]
Derive the explicit rule for the pattern
3, 0, -3, -6, -9,
A cylinder that is 18 cm tall is filled with water. If a hole is made in the side of the cylinder, 5.0 cm below the top level. Assume that the cylinder is large enough so that the level of the water in the cylinder does not drop significantly. How far will the stream land from the base of the cylinder?
Answer:
The distance is 22.45 cm.
Explanation:
Height of cylinder, H = 14 cm
depth of hole, h = 5 cm
The distance of landing of stream from the base of cylinder is
[tex]R = 2\sqrt{H(H-h)}\\\\R = 2\sqrt{14(14-5)}\\\\R = 22.45 cm[/tex]
Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?
Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
Light of the same wavelength passes through two diffraction gratings. One grating has 4000 lines/cm, and the other one has 6000 lines/cm. Which grating will spread the light through a larger angle in the first-order pattern
Answer:
6000 lines/cm
Explanation:
From the question we are told that:
Grating 1=4000 lines/cm
Grating 2=6000 lines/cm
Generally The Spread of fringes is Larger when the Grating are closer to each other
Therefore
Grating 2 will spread the the light through a larger angle in the first-order pattern because its the closest with 6000 lines/cm
