Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Here, "Wavelength is same for both waves" it is the distance between two crests or two consecutive troughs, so, it is constant for both of them, you can easily figure it out.
In short, Your Answer would be "Wavelength"
Hope this helps!
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Explanation:
1. Force=mass*acceleration
acceleration=force/mass
=100/50
=2m/s^2
2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,
So it will be= 50(9.8-2)
=50(7.8)= 390N
