Answer:
<em>The hang time is 2.04 seconds</em>
Explanation:
<u>2-D Motion
</u>
It referred to as a situation where an object is launched in such a way it describes a curve, reaches a top height and then returns to ground level after traveling a certain distance x away from the launch point.
Let be the launching speed forming an angle with the horizontal reference. The hang time (time the object remains in the air) is given by
Since
Then
We'll use the given values
The hang time is 2.04 seconds
T1=S/V1T2=Ttot-T1=(2S/Vtot)-S/V1V2=S/T2 = S / (2S/Vtot-S/V1)Simplify in V2 = V1Vtot/(2V1-Vtot) = 48 * 90 / 6 = 720 km/hr
For the swimmer to go straight the sin component of his velocity should cancel the river velocity.
So, we have θ
This angle θ is the angle shall the swimmer point upstream from the shore.
0.32 = 0.85 sin[/tex]θ
θ =
The velocity of swimmer across the river is given by cos componet of his velocity
v = 0.85 cosθ = 0.85 cos 22 = 0.79 m/s
It’s D A solution of two neurons or more elements
Hope it helps
Answer:
7400 m/s
Explanation:
Centripetal acceleration = gravity
v² / r = GM / r²
v = √(GM / r)
Given:
G = 6.67×10⁻¹¹ m³/kg/s²
M = 5.98×10²⁴ kg
r = 9.8×10⁵ + 6.357×10⁶ = 7.337×10⁶ m
v = √(6.67×10⁻¹¹ (5.98×10²⁴) / (7.337×10⁶))
v = 7400
The orbital velocity is 7400 m/s.