Answer:
Explanation:
25 mm diameter
r₁ = 12.5 x 10⁻³ m radius.
cross sectional area = a₁
Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa
a )
velocity of blood v₁ = .6 m /s
Cross sectional area at blockade = 3/4 a₁
Velocity at blockade area = v₂
As liquid is in-compressible
a₁v₁ = a₂v₂
a₁ x .6 m /s = 3/4 a₁ v₂
v₂ = .8m/s
b )
Applying Bernauli's theorem formula
P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²
100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²
13328 +190.8 = P₂ + 339.2
P₂ = 13179.6 Pa
= 13179 / 13.6 x 10³ x 9.8 m of Hg
P₂ = .09888 m of Hg
98.88 mm of Hg
Answer:
H = 45 m
Explanation:
First we find the launch velocity of the ball by using the following formula:
v₀ = √(v₀ₓ² + v₀y²)
where,
v₀ = launching velocity = ?
v₀ₓ = Horizontal Component of Launch Velocity = 15 m/s
v₀y = Vertical Component of Launch Velocity = 30 m/s
Therefore,
v₀ = √[(15 m/s)² + (30 m/s)²]
v₀ = 33.54 m/s
Now, we find the launch angle of the ball by using the following formula:
θ = tan⁻¹ (v₀y/v₀ₓ)
θ = tan⁻¹ (30/15)
θ = tan⁻¹ (2)
θ = 63.43°
Now, the maximum height attained by the ball is given by the formula:
H = (v₀² Sin² θ)/2g
H = (33.54 m/s)² (Sin² 63.43°)/2(10 m/s²)
<u>H = 45 m</u>
Weight equals mass*gravity
W = mg
Given m = 3.1 kg, g = 9.8 m/s^2
W = (3.1)(9.8)
W = 30.38
Answer:0.69
Explanation:
Coefficient of kinetic friction=f/R=61.8/90=0.69