Answer:
part (a)
towards north east direction.
part (b) s = 46.60 m
Explanation:
Given,
- velocity of the river due to east =

- velocity of the boat due to the north =

part (a)
River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,
Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 
Let 'v' be the velocity of the boat relative to the shore.

Let
be the angle of the velocity of the boat relative to the shore with the horizontal axis.
Direction of the velocity of the boat relative to the shore.
part (b)
- Width of the shore = w = 300m
total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken
Let 't' be the total time taken by the boat to cross the width of the river.
Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river
Answer:
A box sits stationary on a ramp
Explanation:
Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.
Static force of friction is calculated as follows:
F= μη
F is static force of friction.
μ is the coefficient of static friction.
η is the normal force.
Answer:
a) The shear stress is 0.012
b) The shear stress is 0.0082
c) The total friction drag is 0.329 lbf
Explanation:
Given by the problem:
Length y plate = 2 ft
Width y plate = 10 ft
p = density = 1.938 slug/ft³
v = kinematic viscosity = 1.217x10⁻⁵ft²/s
Absolute viscosity = 2.359x10⁻⁵lbfs/ft²
a) The Reynold number is equal to:

The boundary layer thickness is equal to:
ft
The shear stress is equal to:

b) If the railing edge is 2 ft, the Reynold number is:

The boundary layer is equal to:

The sear stress is equal to:

c) The drag coefficient is equal to:

The friction drag is equal to:

Answer:
450 kJ
Explanation:
Q = mCΔT
where Q is heat (energy),
m is mass,
C is specific heat capacity,
and ΔT is the temperature change.
Q = (1.2 kg) (4180 J/kg/°C) (100°C − 10°C)
Q = 451,440 J
Q ≈ 450 kJ