A spherical balloon is being inflated. Find the rate of increase of the surface area (S = 4Ï€r2) with respect to the radius r wh
1 answer:
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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:
Learn more about work:
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Answer:
<h2>8.0995×10^-21 kgms^-1</h2>Explanation:
Mass of proton :
Speed of Proton:
Linear Momentum of a particle having mass (m) and velocity (v) :
Magnitude of momentum :
Frome equation (2), magnitude of linear momentum of the proton :