Answer:
The tension in string P is 25 N, while that of Q is 85 N.
Explanation:
Considering the conditions for equilibrium,
i. Total upward force = Total downward force
+ = 110 N
ii. Taking moment about P,
clockwise moment = anticlockwise moment
110 × (2.5 - 0.8) = × (3 - 0.8)
110 × 1.7 = × 2.2
187 = 2.2
=
= 85 N
From the first condition,
+ = 110 N
+ 85 N = 110 N
= 110 - 85
25 N
Therefore, the tension in string P is 25 N while that of Q is 85 N.
weight = mass*g is the formula
Answer:
Explanation:
Resistivity is given by where A is cross-sectional area, R is resistance, L is the length and is the reistivity. Substituting 0.0625 for R, 3.14 × 10-6 for A and 3.5 m for L then the resistivity is equivalent to
Answer:
1) Pm₂ = 1.9 10⁴ Pa
, b) P_feet = 5.4 10⁴ Pa
, c) Pm₄ = 4.4 10⁴ Pa
Explanation:
a) Pressure can be found using Bernoulli's equation
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rgo g y₂
The amount of blood that runs through the constant system, all the blood that reaches the brain leaves it, so we can assume that the speed of entry and exit of the total blood is the same. In this case the equation is
P₁-P₂ = rgo h (y₂-y₁)
The gauge pressure is
Pm = P₁ -P₂
Pm₂ = 1.06 10³ 9.8 1.83
Pm₂ = 19 10³ Pa
Pm₂ = 1.9 10⁴ Pa
The pressure in the heart is
Pm₁ = 246 torr (1,013 10⁵ Pa / 760 torr) = 3,279 10⁴ Pa
Therefore the gauge pressure is an order of magnitude less
Total or absolute pressure is
Pm₂ = P_heart - P_brain
P_brain = P_heart - Pm₂
P brain = 3,279 10⁴ - 1.9 10⁴
P brain = 1.4 104 Pa
b) on the feet
Pm₃ = rho g y₃
y = 2.04 m
Pm₃ = 1.06 10³ 9.8 2.04
Pm₃ = 21 10³ Pa
Pm₃ = 2.1 10⁴ Pa
Total pressure
Pm₃ = P_feet + P_heart
P_feet = Pm₃ + P_heart
P_feet = 3,279 10⁴ + 2.1 10⁴
P_feet = 5.4 10⁴ Pa
c) If you lower your head the height change is
h = 1.83 +2.04
h = 4.23 m
Pm₄ = 1.06 10³ 9.8 4.23
Pm₄ = 4.4 10⁴ Pa
Answer:
Explanation:
Refraction occurs when a ray of light passes from a medium into another medium. In this situation, the ray of light bends and changes speed, according to Snell's Law:
where
is the index of refraction of the 1st medium
is the index of refraction of the 2nd medium
is the angle of incidence (the angle between the incident ray and the normal to the boundary)
is the angle of refraction (the angle between the refracted ray and the normal to the boundary)
In this problem, we have:
is the index of refraction of air
is the index of refraction of water
is the angle of incidence in air
Solving for , we find the angle of refraction in water: