John carry the heaviest load.
<h3>How to find out who is carrying the heavy load?</h3>
Write down given data from questions:
Board=510cm X 510mm.
Cylinder head with dimensions=43cm X 250mm.
Cylinder lies across the board 210cm from john.
Find out: Who is carry the heaviest load?
Calculation:
We assume that mass of cylinder head = x kg
Then weight=x x 9*81
W=9.81x Newton.
Weight per unit length= Weight/Total leanth
Weight per unit length= 9.81x/43
(w/l)=0.23x N/cm
From equation contition: 
(210+21.5)




Therefore 
To learn more about mass per unit length, refer to:
brainly.com/question/24180692
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Answer:
active solar heating systems use solar energy to heat a fluid either liquid or air and then transfer the solar heat directly to the interior space or to a storage system for later use. If the solar system cannot provide adequate space heating, an auxiliary or back-up system provides the additional heat.
hope this helps : )
The answer is carbon dioxide. This primordial earths’ atmosphere was composed by gasses from degassing of the earth's interior after its formation. It is after the beginning of life that oxygen levels began to rise and levels of carbon dioxide began to reduce in the atmosphere (as a result of photosynthesis).
Answer:
1) k = 10 [N/m]
2) a-) x = 0.4 [m]
b) x = 0.075 [m]
Explanation:
To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.
F = k*x
where:
F = force [N] (units of Newtons]
k = spring constant [N/m]
x = distance = 10 [cm] = 0.1 [m]
Now, the weight is equal to the product of the mass by the gravity
W = m*g = F
where:
m = mass = 100 [g] = 0.1 [kg]
g = gravity acceleration = 10 [m/s²]
F = 0.1*10 = 1 [N]
Now clearing k
k = F/x
k = 1/0.1
k = 10 [N/m]
2)
a ) if the force is 4 [N]
clearing x
x = F/k
x = 4/10
x = 0.4 [m]
m = 75 [g] = 0.075 [kg]
W = m*g = F
F = 0.075*10 = 0.75 [N]
x = .75/10
x = 0.075 [m]
- Let, the maximum height covered by projectile be


- Projectile is thrown with a velocity = v
- Angle of projection = θ
- Velocity of projectile at a height half of the maximum height covered be

______________________________
Then –










- Now, the vertical component of velocity of projectile at the height half of
will be –


Therefore, the vertical component of velocity of projectile at this height will be–
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