Answer:
doppler shift's formula for source and receiver moving away from each other:
<em>λ'=λ°√(1+β/1-β)</em>
Explanation:
acceleration of spaceship=α=29.4m/s²
wavelength of sodium lamp=λ°=589nm
as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm
using doppler shift's formula:
<em>λ'=λ°√(1+β/1-β)</em>
putting the values:
700nm=589nm√(1+β/1-β)
after simplifying:
<em>β=0.17</em>
by this we can say that speed at that time is: v=0.17c
to calculate velocity at an acceleration of a=29.4m/s²
we suppose that spaceship started from rest so,
<em>v=v₀+at</em>
where v₀=0
so<em> v=at</em>
as we want to calculate t so:-
<em>t=v/a</em> v=0.17c ,c=3x10⁸ ,a=29.4m/s²
putting values:
=0.17(3x10⁸m/s)/29.4m/s²
<em>t=1.73x10⁶</em>
IMA = Ideal Mechanical Advantage
First class lever = > F1 * x2 = F2 * x1
Where F1 is the force applied to beat F2. The distance from F1 and the pivot is x1 and the distance from F2 and the pivot is x2
=> F1/F2 = x1 /x2
IMA = F1/F2 = x1/x2
Now you can see the effects of changing F1, F2, x1 and x2.
If you decrease the lengt X1 between the applied effort (F1) and the pivot, IMA decreases.
If you increase the length X1 between the applied effort (F1) and the pivot, IMA increases.
If you decrease the applied effort (F1) and increase the distance between it and the pivot (X1) the new IMA may incrase or decrase depending on the ratio of the changes.
If you decrease the applied effort (F1) and decrease the distance between it and the pivot (X1) IMA will decrease.
Answer: Increase the length between the applied effort and the pivot.
128.1-127.8= 0.3Hz
<span>129.1-128.1= 1.0Hz </span>
<span>129.1-127.8= 1.3Hz</span>
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
