Answer:
<em>The velocity of the ball as it hit the ground = 19.799 m/s</em>
Explanation:
Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.
v² = u² + 2gs.......................... Equation 1
Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball
<em>Given: s = 20 m, u = 0 m/s</em>
<em>Constant: g = 9.8 m/s²</em>
<em>Substituting these values into equation 1,</em>
<em>v² = 0 + 2×9.8×20</em>
<em>v² = 392</em>
<em>v = √392</em>
<em>v = 19.799 m/s.</em>
<em>Therefore the velocity of the ball as it hit the ground = 19.799 m/s</em>
Answer:
μ = tan θ
Explanation:
For this exercise let's use the translational equilibrium condition.
Let's set a datum with the x axis parallel to the plane and the y axis perpendicular to the plane.
Let's break down the weight of the block
sin θ = Wₓ / W
cos θ = W_y / W
Wₓ = W sin θ
W_y = W cos θ
The acrobat is vertically so his weight decomposition is
sin θ = = wₐₓ / wₐ
cos θ = wₐ_y / wₐ
wₐₓ = wₐ sin θ
wₐ_y = wₐ cos θ
let's write the equilibrium equations
Y axis
N- W_y - wₐ_y = 0
N = W cos θ + wₐ cos θ
X axis
Wₓ + wₐ_x - fr = 0
fr = W sin θ + wₐ sin θ
the friction force has the formula
fr = μ N
fr = μ (W cos θ + wₐ cos θ)
we substitute
μ (Mg cos θ + mg cos θ) = Mgsin θ + mg sin θ
μ =
μ = tan θ
this is the minimum value of the coefficient of static friction for which the system is in equilibrium.
Answer:
-2R
Explanation:
The focal length can be calculated using this formula
1/f = (n1/n2-1)(1/r1-1/r2)
1/f = (1/1.33-1)(1/r + 1/r)
= -0.133/1.33(2/R)
F = -2R
The focal length is this
Answer:
E 6.0sin132 = 4.46 km East
N 6.0cos132 = - 4.01 km = 4.01 km South
Explanation:
Speed = distance / time
speed = (121-88) / 30
speed = 1.1 meters? per minute