If you dropped a ball from any height, and measured its distance from the ground at any regular interval while it's falling, the graph of that distance versus time would be a graph that curves downward.
-- The ball is falling down. As time goes on, it gets closer and closer to the ground. Its remaining distance from the ground keeps decreasing, so the line on the graph slopes down.
-- The speed of the ball keeps increasing (it accelerates) because of the gravitational force on it. As time goes on, it covers more of the remaining distance during each interval than it did in the previous interval. The downward slope of the graph keeps increasing.
Answer:
2.5 * 10^-3
Explanation:
<u>solution:</u>
The simplest solution is obtained if we assume that this is a two-dimensional steady flow, since in that case there are no dependencies upon the z coordinate or time t. Also, we will assume that there are no additional arbitrary purely x dependent functions f (x) in the velocity component v. The continuity equation for a two-dimensional in compressible flow states:
<em>δu/δx+δv/δy=0</em>
so that:
<em>δv/δy= -δu/δx</em>
Now, since u = Uy/δ, where δ = cx^1/2, we have that:
<em>u=U*y/cx^1/2</em>
and we obtain:
<em>δv/δy=U*y/2cx^3/2</em>
The last equation can be integrated to obtain (while also using the condition of simplest solution - no z or t dependence, and no additional arbitrary functions of x):
v=∫δv/δy(dy)=U*y/4cx^1/2
=y/x*(U*y/4cx^1/2)
=u*y/4x
which is exactly what we needed to demonstrate.
Also, using u = U*y/δ in the last equation we can obtain:
v/U=u*y/4*U*x
=y^2/4*δ*x
which obviously attains its maximum value for the which is y = δ (boundary-layer edge). So, finally:
(v/U)_max=δ^2/4δx
=δ/4x
=2.5 * 10^-3
Answer:
12.17 m/s²
Explanation:
The formula of period of a simple pendulum is given as,
T = 2π√(L/g)........................ Equation 1
Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie
making g the subject of the equation,
g = 4π²L/T²................... Equation 2
Given: T = 1.8 s, l = 1.00 m
Constant: π = 3.14
Substitute into equation 2
g = (4×3.14²×1)/1.8²
g = 12.17 m/s²
Hence the acceleration due to gravity of the planet = 12.17 m/s²
B will be your answer hope this helped
Kinetic energy = (1/2)*mass*velocity^2
KE = (1/2)mv^2
KE = (1/2)(478)(15)^2
KE = 53775J