Explanation:
Work done by winch = kinetic energy of car
∫ T ds = ½ mv²
∫ 225s ds = ½ mv²
225/2 s² = ½ mv²
225 s² = mv²
v = 15s / √m
Given s = 10 m and m = 2500 kg:
v = 15 (10) / √2500
v = 3 m/s
Answer:
The average speed of the blood in the capillaries is 0.047 cm/s.
Explanation:
Given;
radius of the aorta, r₁ = 1 cm
speed of blood, v₁ = 30 cm/s
Area of the aorta, A₁ = πr₁² = π(1)² = 3.142 cm²
Area of the capillaries, A₂ = 2000 cm²
let the average speed of the blood in the capillaries = v₂
Apply continuity equation to determine the average speed of the blood in the capillaries.
A₁v₁ = A₂v₂
v₂ = (A₁v₁) / (A₂)
v₂ = (3.142 x 30) / (2000)
v₂ = 0.047 cm/s
Therefore, the average speed of the blood in the capillaries is 0.047 cm/s.
The variable you can change in an experiment is ( B ) A dependent variable
Answer:
<em>the ball travels a distance of 8.84 m</em>
Explanation:
Range: Range is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.
R = (U²sin2∅)/g.............................. Equation 1
Where R = range, U = initial velocity, ∅ = angle of projection, g = acceleration due to gravity.
<em>Given: U = 10 m/s, ∅ = 60°</em>
<em>Constant: g = 9.8 m/s²</em>
Substituting these values into equation 1
R = [10²×sin(2×60)]/9.8
R = (100sin120)/9.8
R = 100×0.8660/9.8
R = 86.60/9.8
R = 8.84 m
<em>Therefore the ball travels a distance of 8.84 m</em>
