The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000 
(40 10^-3) / π (5 10^-3)^3
σ = 305 10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305 10^6) (15 10^-3)^3 / 3(40 10^-3)
F = 17156 N.
Answer:
The amount of energy transferred to the water is 4.214 J
Explanation:
The given parameters are;
The mass of the object that drops = 5 kg
The height from which it drops = 86 mm (0.086 m)
The potential energy P.E. is given by the following formula
P.E = m·g·h
Where;
m = The mass of the object = 5 kg
g = The acceleration de to gravity = 9.8 m/s²
h = The height from which the object is dropped = 0.086 m
Therefore;
P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J
Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;
The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.
Answer:
true
Explanation:
A well designed product will increase in sells and in stock.
Answer:
7.9 kilometers per second
Explanation:
