Explanation:
S =ut + 1/2at^2
S = 0×6.5 + (1/2 × 9.54) × 6.5^2
S =0 + 4.77 ×42.25
S=201.5m
Answer:
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a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
from above statement we got
height = 78.4 m
since the ball is thrown, so its vertical velocity would be zero
u = 0
taking g = 9.8m/s^2
now, using the equation of motion
h = ut + gt^2/2
now putting all the values in it
we got ,
78.4 = 9.8 * t^2/ 2
by solving we got,
t = 4 sec
b) now, since along the horizontal , no force acting and accelaration is zero so
R = ut , R is RANGE
R = 5 * 4
range = 20 m
c) vertical components of the stone’s velocity just before it hits the ground = v sin θ =
horizontal components of the stone’s velocity just before it hits the ground = v cos θ
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Answer:
- The velocity component in the flow direction is much larger than that in the normal direction ( A )
- The temperature and velocity gradients normal to the flow are much greater than those along the flow direction ( b )
Explanation:
For a steady two-dimensional flow the boundary layer approximations are The velocity component in the flow direction is much larger than that in the normal direction and The temperature and velocity gradients normal to the flow are much greater than those along the flow direction
assuming Vx ⇒ V∞ ⇒ U and Vy ⇒ u from continuity equation we know that
Vy << Vx
Answer:
D.
Explanation:
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