Answer:
A: No because it is nor changing speed or direction
B: Yes because it changes direction even though the speed is constant
Please Give Brainliest
Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
Answer:
2. You must be able to precisely measure variations in the star's brightness with time.
5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).
6. You must repeatedly obtain spectra of the star that the planet orbits.
Explanation:
The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:
1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.
2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.
3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.
4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
There are lots of variables that directly and indirectly contribute to the presence of gas on a surface
if the size of a planet is relatively small it will in turn be that of a smaller area which results in the less area to be covered for gas which basically means higher presence
I can go in depth more but I don't think that would be necessary all you need to know is this ...based on the size and gas will in turn be parallel to it's conformity
