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:
As carbon dioxide concentrations increase, so too does the rate of photosynthesis until a certain point where the graph levels off. At lower carbon dioxide concentrations carbon dioxide is the limiting factor because an increase in carbon dioxide causes an increase in photosynthesis.
Explanation:
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This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute