Answer:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.
Explanation:
In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.
The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.
For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:
The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.
Answer:
We know that force applied per unit area is called pressure.
Pressure = Force/ Area
When force is constant than pressure is inversely proportional to area.
1- Calculating the area of three face:
A1 = 20m x 10 m =200 Square meter
A2 = 10 mx 5 m = 50 Square meter
A3 = 20m x 5 m = 100 Square meter
Therefore A1 is maximum and A2 is minimum.
2- Calculate pressure:
P = F/ A1 = 30 / 200 = 0.15 Nm⁻² ( minimum pressure)
P = F / A2 = 30 / 50 = 0.6 Nm⁻² ( maximum pressure)
Hence greater the area less will be the pressure and vice versa.
Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So

Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g