Answer:
d) 289.31 m
Explanation:
Energy provided by potential energy = mgh = m x 9.8x 200 sin10.5 = 357.18m
Energy used by friction = μmgcos 10.5 x 200 = .075 x m x 9.8 x cos 10.5 x200 = 144.54 m .
Energy used by friction on plain surface = μmg x d.( dis distance covered on plain ) =.075x m x 9.8 xd = .735 m d
To equate
357.18 m -144.54 m = .735 m d
d = 289.31 m .
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:
95 J
Explanation:
You can calculate efficiency by dividing useful output by total input, then multiplying it to 100.
So the foumula goes like:
Efficiency= (Useful output/Total input)x100
In this question,
Efficiency= 95%
Useful output= x
Total input= 200
Therefore;
95=(x/200)x100
0.95=x/100
x=0.95x100
x=95 Joules
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