F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer:
Here ya gooooooooooooooo
Explanation:
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Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
Answer:
8 m/s²
Explanation:
Given,
Force ( F ) = 4 N
Mass ( m ) = 0.5 kg
To find : -
Acceleration ( a ) = ?
Formula : -
F = ma
a = F / m
= 4 / 0.5
= 40 / 5
a = 8 m/s²
It's acceleration is 8 m/s².
Answer:
W = 19.845 J
Explanation:
Work is defined as W = Fdcos
, where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:
W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m
W = 19.845 J
