Answer: c. Neither, they have the same speed.
In both the cases, there will be work done by the gravitational force. Since, the gravitational force is a conservative force, its work done does not depend on the path as it only depends on the initial and final position of the objects. Since, the initial and final positions are same for both the objects, the work done by gravitational force will be same in both the cases.
Hence, Work done by gravitational force = Change in the kinetic energy
So, the kinetic energy will be same for both the object.
Since, the masses are same: Kinetic energy = 
Hence, the speed will also be same for both the objects.
Correct option: C
Answer:
Angle and speed of wave
Explanation:
The two phenomena involved here are:
- Reflection: reflection occurs when a light wave bounces off the interface between two mediums, returning into the first medium, at a different angle
- Refraction: refraction occurs when a light wave passes through the interface between the two mediums, passing into the second medium, changing direction and speed
The coefficient of reflection tells what is the fraction of the wave that is reflected over the total; it is calculated as
where
are the index of refraction of the 1st and 2nd medium
are the angle between the ray of light and the normal to the interface in the 1st and 2nd medium
From the formula, we see that the % of wave that is reflected depends on the angle and the index of refraction. Moreover, the index of refraction is related to the speed of the wave by:
where c is the speed of light in a vacuum and v the speed of light in the medium; therefore, the correct answer is
Angle and speed of wave
Using KE = 1/2mv^2
m = mass (80kg)
v = velocity (?)
KE = kinetic energy (2500J)
2500/((1/2) x (80)) = v^2
Square root the answer to get: 7.91 m/s
-- 400 nm shifted to 430 nm . . . longer than it should be; "red shifted"; moving away from Earth
-- 610 nm shifted to 580 nm . . . shorter than at source; "blue shifted"; moving toward Earth
-- 512 nm shifted to 480 nm . . . shorter than at source; moving toward Earth
-- 670 nm shifted to 690 nm . . .longer than at source; moving away from Earth
Now I'd just like to ask one more itty bitty question, that you can think about while you're on this subject: Astronomers really do this. They measure how much the wavelength CHANGED, from the time it left the original source until the time they detect it. But HOW do they know what the wavelength WAS when it left the source ? ? ?
THIS is the part that blows my mind !
Answer:
c = 4
Explanation:
From work-energy theorem KE = workdone.
Given F = (cx - 3.00x²)i
W = ∫Fdx = ∫(cx - 3.00x²)dx = cx²/2 –3.00x³/3 + A
W = cx²/2 –x³ + A
Where A is a constant
At x = 0, KE = 20J
So W = 20J at x = 0
20 = c×0 - 0 +A
A = 20
So W = cx²/2 –x³ + 20
Also when x = 3.00m, W = KE = 11J
So
11 = c×3²/2 – 3³ + 20
11 = 4.5c – 7
4.5c = 11 + 7
4.5c = 18
c = 18/4.5 = 4
c = 4
