Answer:
u = 104.68 m/s
Explanation:
given,
horizontal distance = 150 m
elevation of 12.4 m
angle = 8.6°
horizontal motion = x = u cos θ. t .............(1)
vertical motion =
................(2)
from equation(1) and (2)
..........{3}




u = 104.68 m/s
The initial speed of the ball is u = 104.68 m/s
Answer:
w = 1.976 rpm
Explanation:
For simulate the gravity we will use the centripetal aceleration
, so:

where w is the angular aceleration and r the radius.
We know by the question that:
r = 60.5m
= 2.6m/s2
So, Replacing the data, and solving for w, we get:

W = 0.207 rad/s
Finally we change the angular velocity from rad/s to rpm as:
W = 0.207 rad/s = 0.207*60/(2
)= 1.976 rpm
Let R be radius of Earth with the amount of 6378 km h = height of satellite above Earth m = mass of satellite v = tangential velocity of satellite
Since gravitational force varies contrariwise with the square of the distance of separation, the value of g at altitude h will be 9.8*{[R/(R+h)]^2} = g'
So now gravity acceleration is g' and gravity is balanced by centripetal force mv^2/(R+h):
m*v^2/(R+h) = m*g' v = sqrt[g'*(R + h)]
Satellite A: h = 542 km so R+h = 6738 km = 6.920 e6 m g' = 9.8*(6378/6920)^2 = 8.32 m/sec^2 so v = sqrt(8.32*6.920e6) = 7587.79 m/s = 7.59 km/sec
Satellite B: h = 838 km so R+h = 7216 km = 7.216 e6 m g' = 9.8*(6378/7216)^2 = 8.66 m/sec^2 so v = sqrt(8.32*7.216e6) = 7748.36 m/s = 7.79 km/sec
If it is a headwind it means it's travelling against the motion of the plane. This means it's velocity is simply v=720-16=704 km/h due east.
Mercury has a high boiling point of 357 degrees C.
Mercury has a freezing point of −39 degrees C.