We know, dₓ α n² d₁
For n = 15 (15th Orbit)
It would be: d₁₅ = 15² * 1.06
d₁₅ = 238.5 A
In short, Your Answer would be: 238.5 A
Hope this helps!
Answer:
a) 0.323m/s b) 54.278 m/s
Explanation:
The shell is moving upward with a speed of 17m/s
1.7s after it was launch
Using equation of motion on a straight line,
Vfinal = Vintial + acceleration due to gravity *time
Where V represent the speed(velocity)
Vinital = V(at 1.7s) - 9.81* 1.7
Vinital = 17 - 16.677
a) Vinital = 0.323m/s
b) using the same equation,
V(after 5.5s) = Vinital + 9.81*5.5s
Vafter5.5s = 0.323 + 53.955
V after 5.5s = 54.278m/s
Answer:
a)
1.73 m/s
b)
6.43 x 10⁻⁴ m
Explanation:
m = mass of the flea = 1.8 x 10⁻⁴ kg
v₀ = initial speed of the flea = 0 m/s
v = final speed of the flea
W = work done by the force on the flea = 2.7 x 10⁻⁴ J
Using work-change in kinetic energy, Work done is given as
W = (0.5) m (v² - v₀²)
Inserting the values
2.7 x 10⁻⁴ = (0.5) (1.8 x 10⁻⁴) (v² - 0²)
v = 1.73 m/s
b)
d = distance moved by the flea while pushing off
F = Upward force applied on the flea by ground = 0.42 N
Work done is also given as
W = F d
2.7 x 10⁻⁴ = (0.42) d
d = 6.43 x 10⁻⁴ m
They signed the <em>Halibut</em> Treaty of 1937.
Answer:
x = 10.53 m
Explanation:
Let's analyze this problem a bit, the time that the cowboy must take to fall must be the time that the horse takes to arrive
Let's start by looking for the cowboy's time, which starts from rest and the point where the chair is is y = 0
y =y₀ + v₀ t - ½ g t²
0 = y₀ - ½ g t²
t =
we calculate
t = √(2 3.22 / 9.8)
t = 0.81 s
the horse goes at a constant speed
x = t
x = 13 0.81
x = 10.53 m
this is the distance where the horse should be when in cowboy it is left Cartesian