Answer:
D
Explanation:
For this kind of problem, forces add. F = F1 + F2
F1 = 6 N
F2 = 10 N
F = 6N + 10N
F = 16N
Answer:
Please find the answer in the explanation.
Explanation:
Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?
We can make the weak one to carry less load through two different ways or means.
First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.
Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.
So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.
The x location therefore is 290*cos(53)*35 = 6108.4m
The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.
This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m
So your (x,y) coordinates equals (6108.4, 2097.5)
Answer:
I think it will back towards the earth because earth gravitional field will attract to Wards
Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;

where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

Total potential due to this charges = 4500 V + 6750 V = 11250 V