A golf ball is hit with an initial velocity of 20m/s at an angle of 30° to the horizontal. calculate;

Physics

1 answer:

svp [43]2 years ago

3 0

Answer:

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An airplane flies with a velocity of 55.0 m/s [ 35° N of W] with respect to the air (this is

rodikova [14]

Answer:

21 m/s.

Explanation:

The computation of the wind velocity is shown below:

But before that, we need to find out the angles between the vectors

53° - 35° = 18°

Now we have to sqaure it i.e given below

v^2 = 55^2 + 40^2 - 2 · 55 · 40 · cos 18°

v^2 = 3025 + 1600 - 2 · 55 · 40 · 0.951

v^2 = 440.6

v = √440.6

v = 20.99

≈ 21 m/s

Hence, The wind velocity is 21 m/s.

6 0

3 years ago

The three components of velocity in a velocity field are given by u = Ax + By + Cz, v = Dx + Ey + Fz, and w = Gx + Hy + Jz. Dete

Alexxandr [17]

Answer:

The relationship is only between the coefficients A, E and J which is:

$A + E + J = 0$ . The remaining coefficients can be anything without any constraints.

Explanation:

Given:

The three components of velocity is a velocity field are given as:

$u = Ax + By + Cz\\\\v = Dx + Ey + Fz\\\\w = Gx + Hy + Jz$

The fluid is incompressible.

We know that, for an incompressible fluid flow, the sum of the partial derivatives of each component relative to its direction is always 0. Therefore,

$\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z}=0$

Now, let us find the partial derivative of each component.

$\frac{\partial u}{\partial x}=\frac{\partial }{\partial x}(Ax+By+Cz)\\\\\frac{\partial u}{\partial x}=A+0+0=A\\\\\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}(Dx+Ey+Fz)\\\\\frac{\partial v}{\partial y}=0+E+0=E\\\\\frac{\partial w}{\partial z}=\frac{\partial }{\partial z}(Gx+Hy+Jz)\\\\\frac{\partial w}{\partial z}=0+0+J=J$