Answers
Answer:
1.86 x 10^8 m/s
Explanation:
n = 1.61
The formula for the refractive index is given by
n = speed of light in vacuum / speed of light in material
n = c / v
v = c / n
v = (3 x 10^8) / 1.61
v = 1.86 x 10^8 m/s
Final answer:
The speed of light in a material with an index of refraction of 1.61 is calculated as approximately 1.86 * 10^8 m/s, using the equation v = c/n where c is the speed of light in vacuum and n is the index of refraction.
Explanation:
The speed of light in a given material can be calculated using the index of refraction of the material, as defined by the equation n = c/v, where n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the material.
Given that the index of refraction for the material in question is 1.61, and the speed of light in vacuum, c = 3.00 * 10^8 m/s, the speed of light v in this medium would therefore be calculated by rearranging the equation to v = c/n.
By substituting the given values into the equation, v = 3.00 * 10^8 m/s / 1.61, we find that the speed of light in the material is approximately 1.86 * 10^8 m/s.
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Related Questions
A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is 6,000 m/s. If the rocket is fired vertically upward from the surface of the Earth, determine its height after 744 kg of its total fuel load has been consumed. Since the mass of fuel consumed is small compared to the total mass of the rocket, you can consider the mass of the rocket to be constant for the time interval of interest.
A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.
Someone claps his or her hands is an example of… motion energy to sound energy sound energy to potential energy motion energy to radiant energy
Find an expression for the center of mass of a solid hemisphere, given as the distance R from the center of the flat part of the hemisphere. Express your answer in terms of R. Express the coefficients using three significant figures.
Answers
Answer:
24km/h
Explanation:
go it right on ingenuity 2020
Answers
Answer:
For 25-turn electromagnet, Number of clips = 4.1
For 50-turn electromagnet number of clips = 9.6
Explanation:
To calculate the slope of the 25-coil line and the 50-coil line to determine the average number of paper clips that a 1 V battery would pick up.
Hence;
Using the equations gotten from the graph in the previous question and 1.0 V as the value for x, we get
For 25-turn electromagnet y = 3.663x * 0.5
(rounded to one decimal place) Number of clips = 4.1
For 50-turn electromagnet y = 7.133x 2.5
(rounded to one decimal place) Number of clips = 9.6
Answers
Answer:
It will take 33 seconds to stop the car.
Explanation:
Using the first equation of kinematics we have
where
'v' is final speed of object
'u' is initial speed of object
'a' is acceleration of object
't' is time of acceleration of object
Now since it is given that since acceleration is negative and
We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get
Answers
The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M
Given data :
separation between slits ( d ) = 1.5 x 10⁻³ m
wavelength of light ( λ ) = 514 * 10⁻⁹ m
Distance from narrow slit ( D ) = 5.0 m
Determine the distance between the adjacent bright fringes
we apply the formula below
w = D * λ / d ---- ( 1 )
where : w = distance between adjacent bright fringes
Back to equation ( 1 )
w = ( 5 * 514 * 10⁻⁹ ) / 1.5 x 10⁻³
= 1.7 * 10⁻³ M
Hence we can conclude that The distance between the adjacent bright fringes is : 1.7 * 10⁻³ M
Learn more about bright fringes calculations : brainly.com/question/4449144
Answer:
m
Explanation:
d = separation between the two narrow slits = 1.5 mm = 1.5 x 10⁻³ m
λ = wavelength of the light = 514 nm = 514 x 10⁻⁹ m
D = Distance of the screen from the narrow slits = 5.0 m
w = Distance between the adjacent bright fringes on the screen
Distance between the adjacent bright fringes on the screen is given as
m
Use 1.602×10−19 C for the magnitude of the charge on an electron.
B)Find the direction of the force that this magnetic field exerts on the ball just as it enters the field.
a-from north to south
b-from south to north
Answers
Answer:
A. F=6.65*10^{-10}N
B. south - north
Explanation:
A) We use the Lorentz force
F = qv X B
|F| = qvB
to calculate the magnitude of the force we need the speed of the of the ball.
and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)
B)
b. south - north (by the rigth hand rule)
I hope this is usefull for you
regards
What will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk?
Answers
Final answer:
To find the final angular velocity when the skater pulls in his arms, we use the conservation of angular momentum.
Explanation:
To find the final angular velocity when the skater pulls in his arms, we can make use of the conservation of angular momentum. Initially, the skater's arms are outstretched, and the moment of inertia can be calculated using the parallel axis theorem. After the skater pulls in his arms, we can calculate the new moment of inertia using the same theorem. Equating the initial and final angular momentum values, we can solve for the final angular velocity.
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Final answer:
The problem involves the concept of conservation of angular momentum. The skater's spinning speed will increase when they pull their arms in. For a precise value of the final velocity, a complex calculation taking into account body mass distribution is needed.
Explanation:
This question involves the principle of conservation of angular momentum, which states that the angular momentum of an object remains constant as long as no external torques act on it. The total initial angular momentum of the skater spinning with outstretched arms is equal to his final angular momentum when he pulls his arms in.
Calculating the skater's initial and final angular momentum, you can then solve for his final velocity.
However, note that the calculation needs to take into account the skater's mass distribution. Specifically, we need to consider the percentage distributions for the arms/hands (13%), head (7%) and trunk/legs (80%), and integrate these over the skater's body.
This can result in a significantly complex calculation if done accurately, involving calculus level mathematics. However, using the qualitative knowledge that the skater's spinning speed will increase when they pull their arms in, it's reasonable to estimate, considering the mass distribution, the final velocity will be somewhere near 2 to 3 times the original rpm. But for an exact value, a detailed and complex calculation is needed.
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