Answer:
0.31
Explanation:
horizontal force, F = 750 N
mass of crate, m = 250 kg
g = 9.8 m/s^2
The friction force becomes applied force = 750 N
According to the laws of friction,
Friction force = μ x Normal reaction of the surface
here, μ be the coefficient of friction
750 = μ x m g
750 = μ x 250 x 9.8
μ = 0.31
Thus, the coefficient of static friction is 0.31.
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V =
(1)
Total cost,
(2)
From eqn (1):

Using the above value in eqn (1):


Differentiating the above eqn w.r.t 'a':

For the required solution equating the above eqn to zero:


a = 4
Also

The path in order to minimize the cost must be a rectangle.
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
Answer:
The frictional torque is 
Explanation:
From the question we are told that
The mass attached to one end the string is 
The mass attached to the other end of the string is 
The radius of the disk is 
At equilibrium the tension on the string due to the first mass is mathematically represented as

substituting values


At equilibrium the tension on the string due to the mass is mathematically represented as



The frictional torque that must be exerted is mathematically represented as

substituting values


The conservation of the mass of fluid through two sections (be they A1 and A2) of a conduit (pipe) or current tube establishes that the mass that enters is equal to the mass that exits. Mathematically the input flow must be the same as the output flow,

The definition of flow is given by

Where
V = Velocity
A = Area
The units of the flow of flow are cubic meters per second, that is to say that if there is a continuity, the volume of input must be the same as that of output, what changes if the sections are modified are the proportions of speed.
In this way

