Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Answer: 459.14 N
Explanation:
from the question, we have
diameter = 10 m
radius (r) = 5 m
weight (Fw) = 670 N
time (t) = 8 seconds
Circular motion has centripetal force and acceleration pointing perpendicular and inwards of the path, therefore we apply the equation below
∑ F = F c = F w − Fn ..............equation 1
Fn = Fw − Fc = mg − (mv^2 / r) ...................equation 2
substituting the value of v as (2πr / T) we now have
Fn = mg − (m(2πr / T )^2) / r
Fn= mg − (4(π^2)mr / T^2) ..........equation 3
Fw (mass of the person) = mg
therefore m = Fw / g
m = 670 / 9.8 = 68.367 kg
now substituting our values into equation 3
Fn = 670 - ( (4 x (π^2) x 68.367 x 5 ) / 8^2)
Fn = 670 - 210.86
Fn = 459.14 N
because water is loosely packed but when it is cold it becomes closely packed in order to form ice and thus the force attraction between them also increase.
<h3><u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u><u>-</u></h3>
- Energy Transferre=11KJ
- Efficiency=35%
<h3>☆Usefully transferred energy:-</h3>
