Answer:
Resultant force = 8.6N
Explanation:
Using Pythogorus' theorem

Resultant force = 8.6N
Answer:
<em><u>1)A)</u></em>
<em><u>1)A)2)A)</u></em>
<h3><em><u>Hope it helps you </u></em><em><u>♡</u></em><em><u>♡</u></em></h3>
1. Ideal Mechanical Advantage (IMA): 9
Explanation:
For a wheel and axle system like the steering wheel, the IMA is given by:

where
is the radius of the wheel
is the radius of the axle
For the steering wheel of the problem, we see that
and
, so the IMA is

2. Efficiency: 88.9%
Explanation:
The efficiency of a system is defined as the ratio between the AMA (actual mechanical advantage) and the IMA:

In this problem, AMA=8 and IMA=9, so the efficiency is

Answer:
Explanation:
In case of diffraction , angular width of central maxima =2 λ/d
λ is wave length of light and d is slit width
In case of interference , angular width of each fringe
= λ /D
D is distance between two slits
No of interference fringe in central diffraction fringe
=2 λ/d x D/λ = 2 x D /d = 2 x .24/.03 = 16.
Since work is the change in kinetic energy, the efficiency of a machine can be stated as the percentage of the output work divided by the input work minus the work lost from to friction and heat. Multiply Eff by 100% to get the efficiency percentage.