To solve this problem, we will apply the concepts related to the linear deformation of a body given by the relationship between the load applied over a given length, acting by the corresponding area unit and the modulus of elasticity. The mathematical representation of this is given as:
Where,
P = Axial Load
l = Gage length
A = Cross-sectional Area
E = Modulus of Elasticity
Our values are given as,
l = 3.5m
D = 0.028m
E = 200GPa
Replacing we have,
Therefore the change in length is 1.93mm
1. True
2. True
3. True
4. True
5. 10ft
Answer:
A) Elements are placed in teh same group because they have similar properties. B) Every element in a family of elements has similar atomic numbers as the others in its family. C) Alkali Metals
Explanation:
Answer:
Gravitational force will be 16 times more.
Explanation:
we know;
Gravitational force (F) = (Gm1m2)/d^2
when mass of each is doubled and distance between them is halved;
F= (G2m1×2m2)/(d/2)^2
=(4Gm1m2)/(d^2/4)
=4×4(Gm1m2)/d^2
=16(Gm1m2)/d^2
=16F
Answer:
Explanation:
For this problem we must use Newton's second law where force is gravitational attraction
F = m a
Since movement is circular, acceleration is centripetal.
a = v2 / r
Let's replace
G m M / r² = m v² / r
G M r = v²
The distance r is from the center of the planet
r = R + h
v = √ GM / (R + h)
If the friction force is not negligible
F - fr = m a
Where the friction force must have some functional relationship, for example
Fr = b v + c v² +…
Suppose we are high enough for the linear term to derive the force of friction
G m M / r - (m b v + m c v2) = m v2
G M / r - b v = v²
We see that the solution of the problem gives lower speeds and that change over time.
To compensate for this friction force, the motors should be intermittently suspended to recover speed.
