Answer:
The arrow is at a height of 500 feet at time t = 2.35 seconds.
Explanation:
It is given that,
An arrow is shot vertically upward at a rate of 250 ft/s, v₀ = 250 ft/s
The projectile formula is given by :
We need to find the time(s), in seconds, the arrow is at a height of 500 ft. So,
On solving the above quadratic equation, we get the value of t as, t = 2.35 seconds
So, the arrow is at a height of 500 feet at time t = 2.35 seconds. Hence, this is the required solution.
K=0.5 mu×u
K=2200J no matter the direction
Around 80 percent of the mass of the universe is made up material known as "Dark matter". It does not emit light or energy but the influence of it can be detected or observed gravitationally. Motions of stars and galaxy tell us how much mater there is, but somehow the speed of rotation of galaxy does not add up to its mass alone, there is a certain amount of matter really not accounted for. Dark matter maybe made up of non-baryonic matter, or perhaps what scientist called the WIMPS or (weakly interacting massive particles.)
Answer:
Explained
Explanation:
Photosphere: The lowest layer of the sun is called photo sphere . It is about 300 miles thick from the surface. It is the source of solar flares. It is marked by bright bubbling granules of plasma.
chromosphere emits a reddish glow as the super heated hydrogen burns off but the red rim can only be seen during total solar eclipse.
The third layer of the sun atmosphere is Corona. It can also only be seen during during a total solar eclipse. Temperature in corona can reach as high as 3.5 million degree fahrenheit. As the gases cool they become solar winds.
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
