Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.
Direction of the net force (β)
β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
Answer:
Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity.
Explanation:
You're Welcome.
Answer:
40 V
Explanation:
I will assume that the resistors are
100 and 3900 and 1000 OHMS <=====(NOT W)
In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total
V = IR
I = V / R so the total current will be 200 v / 5000 ohms = .04 amps
this is the current through all of the resistors
so for the 1000 ohm resistor V = IR .04 (1000) = 40 V
Answer:
F = 75[J]
Explanation:
We know that work is defined as the product of force by distance.
In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.
where:
W = work = 50 [J]
d = distance = 2 [m]
Fweight = 50 [N]
Fupward [N]
Now replacing:
![50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]](https://tex.z-dn.net/?f=50%3D-%2850%2A2%29%2B%28F_%7Bupward%7D%2A2%29%5C%5C50%2B100%3DF_%7Bupward%7D%2A2%5C%5CF_%7Bupward%7D%3D150%2F2%5C%5CF_%7Bupward%7D%3D75%5BJ%5D)
