How far away is the closest star?
2 answers:
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Answer:
V = 10.88 m/s
Explanation:
V_i =initial velocity = 0m/s
a= acceleration= gsinθ-cosθ
putting values we get
a= 9.8sin25-0.2cos25= 2.4 m/s^2
v_f= final velocity and d= displacement along the inclined plane = 10.4 m
using the equation
v_f= 7.04 m/s
let the speed just before she lands be "V"
using conservation of energy
KE + PE at the edge of cliff = KE at bottom of cliff
(0.5) m V_f^2 + mgh = (0.5) m V^2
V^2 = V_f^2 + 2gh
V^2 = 7.04^2 + 2 x 9.8 x 3.5
V = 10.88 m/s
Answer:
T = 2.83701481512 seconds
Explanation:
Hi!
The formula that you will want to use to solve this question is:
T--> period
L --> length of the pendulum
g --> acceleration due to gravity (9.8m/s^2)
since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)
so now we can plug in our given info into the formula above and solve!
T = 2*pi * sqrt(2/9.8)
T = 2.83701481512 seconds
*Note*
- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different
I hope this helped!