Answer:
(a) Elongation of the rod==5.61×10⁻⁹m
(b) Change in diameter=1.640×10⁻⁸m
Explanation:
Given data
Diameter d=78 in=1.9812 m
Cross Area is:
Applied Load P=17 KN=17×10³N
E=29 × 106 psi=1.99947961×10¹¹Pa
Stress and Strain in x direction
Stress in x direction
σ=P/A
σ=5517.25 Pa
Strain in x direction
ε=σ/E
ε=2.76×10⁻⁸
Part (a)
Elongation of the rod=Lε
=(0.2032)(2.76×10⁻⁸)
Elongation of the rod==5.61×10⁻⁹m
Part(b) Change in diameter
Strain in y direction
ε₁= -vε
ε₁= -(0.30)(2.76×10⁻⁸)
ε₁=-8.28×10⁻⁹
Change in diameter=d×ε₁
Change in diameter=(1.9812m)×(-8.28×10⁻⁹)
Change in diameter=1.640×10⁻⁸m
The period of a simple pendulum is given by:
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
(1)
We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.
We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
And its period is the reciprocal of its frequency:
So now we can use eq.(1) to find the gravitational acceleration of the planet:
Answer:
0.782 s
Explanation:
The water flows horizontally from the hose, so its initial vertical velocity is 0.
Given:
y₀ = 3 m
y = 0 m
v₀ = 0 m/s
a = -9.8 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 0.782 s
Round as needed.
If the desk doesn't move, then it's not accelerating.
If it's not accelerating, then the net force on it is zero.
If the net force on it is zero, then any forces on it are balanced.
If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.
So the friction on the desk must be equal to your<em> 245N</em> .
