Answer:
The spring constant = 9.25 N/m
Explanation:
The equation of an object attached to a spring that is oscillating is
T = 2π√(m/k)
Where T = period of the oscillation, m = mass of the object, k = spring constant.
Making k the subject of the equation,
k = 4π²m/T²......................... Equation 1
Note: Period(T) is the time taken to complete one oscillation
Given: T = t/10 = 9.0/10 = 0.9 s, m = 190 g = 0.19 kg.
Constant: π = 3.14
Substitute these values into equation 1.
k = 4(3.14)²(0.19)/0.9²
k = 7.4933/0.81
k = 9.25 N/m
Thus the spring constant = 9.25 N/m
Answer:
In Step 5, you will calculate H+/OH– ratios for more extreme pH solutions. Find the concentration of H+ ions to OH– ions listed in Table B of your Student Guide for a solution at a pH = 2. Then divide the H+ concentration by the OH– concentration. Record these concentrations and ratio in Table C.
What is the concentration of H+ ions at a pH = 2?
0.01 mol/L
What is the concentration of OH– ions at a pH = 2?
0.000000000001 mol/L
What is the ratio of H+ ions to OH– ions at a pH = 2?
10,000,000,000 : 1
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I LITERALLY spent 40 MINUTES trying to figure out this question, so please, use my VERY CORRECT answers!
I hope this helps!
Answer:
a)T total = 2*Voy/(g*sin( α ))
b)α = 0º , T total≅∞ (the particle, goes away horizontally indefinitely)
α = 90º, T total=2*Voy/g
Explanation:
Voy=Vo*sinα
- Time to reach the maximal height :
Kinematics equation: Vfy=Voy-at
a=g*sinα ; g is gravity
if Vfy=0 ⇒ t=T ; time to reach the maximal height
so:
0=Voy-g*sin( α )*T
T=Voy/(g*sin( α ))
- Time required to return to the starting point:
After the object reaches its maximum height, the object descends to the starting point, the time it descends is the same as the time it rises.
So T total= 2T = 2*Voy/(g*sin( α ))
The particle goes totally horizontal, goes away indefinitely
T total= 2*Voy/(g*sin( α )) ≅∞
T total=2*Voy/g
Answer:
a) 6498.84 kW
b) 0.51
c) 0.379
Explanation:
See the attached picture below for the solution
