Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
![(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]](https://tex.z-dn.net/?f=%280.5%29%5E%7B2%7D-%2810%29%5E%7B2%7D%20%3D%202%2Aa%2A%28100%29%5C%5C0.25-10000%3D200%2Aa%5C%5C200%2Aa%3D-9999.75%5C%5Ca%20%3D-50%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign of acceleration means that the ship slows down its velocity in order to land.
Well, you gave us the formula to calculate power from work and time,
but you didn't give us the formula for work. We have to know that.
Work = (force) x (distance)
The work to raise Sara to the top of the hill is
Work = (300 N) x (15 meters)
= 4,500 newton-meters = 4,500 joules .
Now we're ready to use the formula that you gave us. (Thank you.)
Power = (work) / (time)
= (4,500 joules) / (10 seconds)
450 joules/second = 450 watts.
The position of the sun and the moon affect how high the tide is
Answer:

Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives

For the y-direction gives

Combining both equation yields the y_component of the final velocity

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

The speed of light is: c
= 3x10^8 m/s <span>
or
c = 186,000,000 miles/sec = 1.86x10^8 mi/s
1 furlong = 0.125 mile
1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h)
= 1209600s = 1.2096x10^6 s
Therefore,
c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort)
<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>