Answer:
M_c = 100.8 Nm
Explanation:
Given:
F_a = 2.5 KN
Find:
Determine the moment of this force about C for the two cases shown.
Solution:
- Draw horizontal and vertical vectors at point A.
- Take moments about point C as follows:
M_c = F_a*( 42 / 150 ) *144
M_c = 2.5*( 42 / 150 ) *144
M_c = 100.8 Nm
- We see that the vertical component of force at point A passes through C.
Hence, its moment about C is zero.
Answer:
Explanation:
Given:
Charge = <em>q</em>
Electric field strength =
weight of the droplet = <em>mg</em>
The charge is suspended motionless. This is because the electric force on the charge is balanced by the weight of the droplet.
electric force on charged droplet, 
This is balanced by the weight, 
Equating the two:

Answer:
The child represented by a star on the outside path.
Explanation:
You do this one just like the other one that I just solved for you.
For this one ...
The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.
The question tells us: We have 2.0 cm³.
Each cm³ of space that the object occupies contains 2.5 gm of mass.
So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.