Answer:
see explanation below
Explanation:
Given that,
500°C
= 25°C
d = 0.2m
L = 10mm = 0.01m
U₀ = 2m/s
Calculate average temperature

262.5 + 273
= 535.5K
From properties of air table A-4 corresponding to
= 535.5K 
k = 43.9 × 10⁻³W/m.k
v = 47.57 × 10⁻⁶ m²/s

A)
Number for the first strips is equal to


Calculating heat transfer coefficient from the first strip


The rate of convection heat transfer from the first strip is

The rate of convection heat transfer from the fifth trip is equal to


Calculating 

The rate of convection heat transfer from the tenth strip is


Calculating

Calculating the rate of convection heat transfer from the tenth strip

The rate of convection heat transfer from 25th strip is equal to

Calculating 

Calculating 

Calculating the rate of convection heat transfer from the tenth strip

One well-known application of density is determining whether or not an object will float on water. If the object's density is less than the density of water, it will float; if its density is less than that of water, it will sink.In fact, submarines dive below the surface of the water by emptying their ballast tanks
No se ha da han dicho nada más de lo dicho y han ido de vuelta y han dicho nada más de que se pueda hacer el favor del niño
Given parameters:
Mass of object = 6.7kg
Velocity = 8m/s
Unknown parameter:
Kinetic energy = ?
Energy is defined as the ability to do work. There are two forms of energy;
Kinetic and potential energy.
Kinetic energy is the energy due to the motion of a body. Whereas, potential energy is the energy due to the position of a body usually at rest.
Kinetic energy is mathematically expressed as;
Kinetic energy = 
where m is the mass of the body
v is the velocity of the body
Since we have been given both mass and velocity, input the parameter to solve for the unknown;
Kinetic energy =
x 6.7 x 8² = 214.4J
So the kinetic energy of the body is 214.4J
Answer:
∆PE = 749.7 J
At 0.9 m high, PE = 793.8 J
At 1.75 m high, PE = 1543.5 J