Differentiate between moments and momentum
1 answer:
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Answer:
Work done, W = 1786.17J
Explanation:
The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "
Mass of a painter, m = 75 kg
He climbs 2.75-m ladder that is leaning against a vertical wall.
The ladder makes an angle of 30 degrees with the wall.
We need to find the work done by the gravity on the painter.
The angle between the weight of the painter and the displacement is :
θ = 180 - 30
= 150°
The work done by the gravity is given by :
Hence, the required work done is 1786.17 J.
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
We must consider that it's launched from the ground () and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
And the velocity achieved in part 1:
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 () and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
Then, to get , we do:
And we substitute the values:
When you heat a certain substance with a difference of temperature 
the heat (energy) you must give to it is
where
is the specific heat of that substance (given in J/(g*Celsius))
In this case
Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a difference of 1 degree Celsius = 1 degree Kelvin
where
In this case
Observation: the specific heat of a substance is given in J/(g*Celsius) or J/(g*Kelvin) because on the temperature scale a difference of 1 degree Celsius = 1 degree Kelvin