I attached a picture of the diagram associated with this question.
Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ
The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ
Answer:
2.5m/s^2
Explanation:
Step one:
given
distance = 20meters
time = 2 seconds
initial velocity u= 0m/s
let us solve for the final velocity
velocity = distance/time
velocity= 20/2
velocity= 10m/s

divide both sides by 40

Answer:
The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²
Explanation:
Data
Ta = 0.125 s
Tb = 0.08 s
Δtab = 0.5 s
distance = 1 cm
Process
1.- Calculate va
va = 1/0.125 = 8 cm/s
vb = 1/0.08 = 12.5 cm/s
2.- Calculate Δv
Δv = 12.5 - 8
Δv = 4.5 cm/s
3.- Calculate acceleration
a = Δv / Δt
a = 4.5/0.5
a = 9 cm/s²
C.
Explanation: Because mirrors are lighter, and they are easier than lenses to make perfectly smooth.
0.77 m/s2 directed 35° south of west
net force = (-17,-12)
net force = mass * acceleration
(-17,-12) = 27 * (x-acceleration,y-acceleration)
(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)
angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.
magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)