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:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
Answer:
Negative
Explanation:
If the box is heading right in the positive direction, the work will be negative. The spring has an opposite force to that of the box.
Hope this helped. :)
Explanation:
The given data is as follows.
Length (l) = 2.4 m
Frequency (f) = 567 Hz
Formula to calculate the speed of a transverse wave is as follows.
f = 
Putting the gicven values into the above formula as follows.
f =
567 Hz = 
v = 544.32 m/s
Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.
