Answer:
Explanation:
a) I = ½mR² = ½(19)(0.15²) = 0.21375 kg•m²
b) τ = Fnet(r) = (25 - 12)(0.15) = 1.95 N•m
c) CCW
d) a = τ/I = 1.95 / 0.21375 = 9.12280701... = 9.1 rad/s²
e) CCW
The work done by the force is 47.1 J
Explanation:
The work done by a force in moving an object is given by
(1)
where
F is the magnitude of the force
d is the distance covered by the object
is the angle between the direction of the force and the motion of the object
In this problem, the force applied to the object is
F = 3.0 N
This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

And the distance covered is equal to the circumference of the circle, which is:

where r = 2.5 m is the radius.
Now we can substitute into eq.(1) to find the work done:

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Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm
Hello,
Your answer to this problem is 400/3
Hope this helps!