Answer:
a = 4.9(1 - sinθ - 0.4cosθ)
Explanation:
Really not possible without a complete setup.
I will ASSUME that this an Atwood machine with two masses (m) connected by an ideal rope passing over an ideal pulley. One mass hangs freely and the other is on a slope of angle θ to the horizontal with coefficient of friction μ. Gravity is g
F = ma
mg - mgsinθ - μmgcosθ = (m + m)a
mg(1 - sinθ - μcosθ) = 2ma
½g(1 - sinθ - μcosθ) = a
maximum acceleration is about 2.94 m/s² when θ = 0
acceleration will be zero when θ is greater than about 46.4°
<h2>
It takes 6.78 seconds to complete 12 dribbles.</h2>
Explanation:
Frequency of dribble = 1.77 Hz
That is
Number of dribbles in 1 second = 1.77
Now we need to find how long does it take for you to complete 12 dribbles.
Time taken for 12 dribbles = 12 x Time taken for 1 dribble
Time taken for 12 dribbles = 12 x 0.565
Time taken for 12 dribbles = 6.78 seconds
It takes 6.78 seconds to complete 12 dribbles.
Answer:
the wind carries abrasive materials
Explanation:
such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first
H = 280 ft, the height of the flower pot.
g = 32 ft/s²
Neglect air resistance.
Note that 1 ft/s = 15/22 mi/h
The initial vertical velocity is zero.
Let v = the velocity with which the flower pot hits the ground.
Then
v² = 2gh
= 2*(32 ft/s²)*(280 ft)
= 17920 (ft/s)²
v = 133.866 ft/s
Also,
v = (133.866 ft/s)*(15/22 (mi/h)/(ft/s)) = 91.272 mi/h
Answer: 133.9 ft/s or 91.3 mi/h
