Answer:
Workdone is 5734.06Nm.
Explanation:
<u>Given the following data;</u>
Force applied = 125N
Angle = 35°
Distance = 56m
To find the workdone by the lawnmower, we would first of all find the horizontal component of the force applied.
Where;
- Fx represents the horizontal force.
- m is the mass of an object.
- g is the acceleration due to gravity.
- d is the angle of inclination (theta).
mg = weight = 125N
Substituting into the equation, we have;
Fx = 102.39N
Workdone is given by the formula;
Workdone = 5734.06Nm
<em>Therefore, the work done by the lawnmower is 5734.06Nm</em>.
Water set free by magma began to cool down the Earth's atmosphere, until it could stay on the surface as a liquid. ... The sun, which drives the water cycle, heats water in the oceans. Some of it evaporates as vapor into the air. Ice and snow can sublimate directly into water vapor.
0 Km , Displacement is distance from starting point. Not distance of Journey.
You haven't told us anything about the detectors being used. We don't know how the sensitivity of the detector is related to the total number of photons absorbed, and we don't even know whether you and your friend are both using the same type of detector.
All we can do, in desperation, is ASSUME that the minimum time required to just detect a star is inversely proportional to the total number of its photons that strike the detector. That is, assume . . .
(double the number of photons) ===> (detect the source in half the time) .
-- The intensity of light delivered to the prime focus of a telescope is directly proportional to the AREA of its objective lens or mirror, which in turn is proportional to the square of its radius or diameter.
So your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.
-- So we'd expect your instrument to detect the same star in
(119.5 min) / (12.96) = <em>9.22 minutes .</em>
We're simply comparing the performance of two different telescopes as they observe the same object, so the star's magnitude doesn't matter.
