Answer: 116.926 km/h
Explanation:
To solve this we need to analise the relation between the car and the Raindrops. The cars moves on the horizontal plane with a constant velocity.
Car's Velocity (Vc) = 38 km/h
The rain is falling perpedincular to the horizontal on the Y-axis. We dont know the velocity.
However, the rain's traces on the side windows makes an angle of 72.0° degrees. ∅ = 72°
There is a relation between this angle and the two velocities. If the car was on rest, we will see that the angle is equal to 90° because the rain is falling perpendicular. In the other end, a static object next to a moving car shows a horizontal trace, so we can use a trigonometric relation on this case.
The following equation can be use to relate the angle and the two vectors.
Tangent (∅) = Opposite (o) / adjacent (a)
Where the Opposite will be the Rain's Vector that define its velocity and the adjacent will be the Car's Velocity Vector.
Tan(72°) = Rain's Velocity / Car's Velocity
We can searching for the Rain's Velocity
Tan(72°) * Vc = Rain's Velocity
Rain's Velocity = 116.926 km/h
A factor of 50dB = a factor of 10^(50/10) = 10^5 = a factor of <em>100,000</em> .
Answer:
It involves all three methods as the handle of the pot is conducting heat via touching the pan which gets it's heat from the burner that is radiating heat to the entire pan through waves of heat from the burner. There is then convection as the heat is being transferred as the liquids temperature changes.
This should be good if I remember it correctly which I hope so much I do.
Answer:
Part a)
H = 26.8 m
Part b)
error = 7.18 %
Explanation:
Part a)
As the stone is dropped from height H then time taken by it to hit the floor is given as
now the sound will come back to the observer in the time
so we will have
so we have
solve above equation for H
Part b)
If sound reflection part is ignored then in that case
so here percentage error in height calculation is given as
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as
Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as
where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then
Resolving for F,
PART B) We need here to apply the shear strength equation, then
In such a way that the material is more resistant to tensile strength than shear force.
