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:
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Answer:
buoyant force on the block due to the water= 10 N
Explanation:
We know that
buoyant force(F_B) on a block= weight of the block in air (actual weight) - weight of block in water.
Given:
A block of metal weighs 40 N in air and 30 N in water.
F_B = 40-30= 10 N
therefore, buoyant force on the block due to the water= 10 N
<u>Answer</u>:
Effort is the unaltered force. Load is the altered force.
1750 meters.
First, determine how long it takes for the kit to hit the ground. Distance over constant acceleration is:
d = 1/2 A T^2
where
d = distance
A = acceleration
T = time
Solving for T, gives
d = 1/2 A T^2
2d = A T^2
2d/A = T^2
sqrt(2d/A) = T
Substitute the known values and calculate.
sqrt(2d/A) = T
sqrt(2* 1500m / 9.8 m/s^2) = T
sqrt(3000m / 9.8 m/s^2) = T
sqrt(306.122449 s^2) = T
17.49635531 s = T
Rounding to 4 significant figures gives 17.50 seconds. Since it will take
17.50 seconds for the kit to hit the ground, the kit needs to be dropped 17.50
seconds before the plane goes overhead. So just simply multiply by the velocity.
17.50 s * 100 m/s = 1750 m
Answer:
w = 0.943 rad / s
Explanation:
For this problem we can use the law of conservation of angular momentum
Starting point. With the mouse in the center
L₀ = I w₀
Where The moment of inertia (I) of a rod that rotates at one end is
I = 1/3 M L²
Final point. When the mouse is at the end of the rod
= I w + m L² w
As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved
L₀ = L_{f}
I w₀ = (I + m L²) w
w = I / I + m L²) w₀
We substitute the moment of inertia
w = 1/3 M L² / (1/3 M + m) L² w₀
w = 1 / 3M / (M / 3 + m) w₀
We substitute the values
w = 1/3 / (1/3 + 0.02) w₀
w = 0.943 w₀
To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s
w = 0.943 rad / s
