Answer:
101.54m/h
Explanation:
Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;
Let l be the be the distance further away at which they will meet from the current points;
#The speed toward each other.

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h
Answer:
The gauge pressure of air is 110 kpa
Explanation:
Atmospheric pressure,
= 101 Kpa


where;
ρw is the density of water = 1000 kg/m³
ρo is the density of oil = 800 kg/m³
ρHg is the density of mercury = 13,600 kg/m³
g is acceleration due to gravity = 9.8 m/s²

Therefore, the gauge pressure of air is 110 kpa
Answer:
3.1 miles
Explanation:
To solve this question it is important to remember that the distance between two mile markers is approximately 1 mile
Once this is known, the question becomes very easy to solve. We make two triangle, which have the following three points
Triangle 1: Hot-Air-Balloon, Ground, Milepost 1 - With angle of depression 20
Triangle 2: Hot-Air-Balloon, Ground, Milepost 2 - With angle of depression 18
As a reminder, the angle of depression is simply the angle the balloonist's head makes with the horizontal plane to be able to see the milepost.
From this we can simply drive two formulas using the Tan function
Equation 1 - 
Equation 2 - 
Solving them simultaneously we get the value of height (h) to be 3.0852 miles or 3.1 miles
Answer:
211 mmHg
Explanation:
Absolute Pressure = Gauge Pressure + Atmospheric pressure
128 = Gauge Pressure + 100
Gauge Pressure = 28 KPa = 28 × 10³ Pa
Also Gauge Pressure = ρgh
ρ = density = 13550 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = pressure head = ?
28 × 10³ = 13550 × 9.8 × h
h = 28000/(13550 × 9.8)
h = 0.211 m = 211 mm
Answer:
Part a)
F = 0.735 N
Part b)

Explanation:
As we know that mass of the cube is given as




Now we know that cube is floating in the water
So net force due to weight of the cube must be counter balanced by buoyancy force on the liquid
so we have



Part b)
Percentage of volume submerged into the liquid is given as



now percentage of submerged liquid is given as

