In position time graphs, otherwise known as distance time graphs, acceleration/decceleration occurs when the line curves, which looks like the picture I've included.
In hours 2 to 3, the object does not accelerate, but instead stays at a constant speed of 20m per hour.
If the net force on an object is zero, then ...
a). The object's acceleration is zero.
b). The object's velocity is a constant speed in a straight line.
Note: (a) and (b) both say the same thing.
Answer:
The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s
Explanation:
Given,
Your velocity, V₁ = 2 m/
The velocity of the person, V₂ =?
The velocity of the person relative to you, V₂₁ = 3 m/s
According to the relative velocity of two
V₂₁ = V₂ -V₁
∴ V₂ = V₂₁ + V₁
On substitution
V₂ = 3 + 2
= 5 m/s
Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s
Yes, your weight can change if the force of gravity is different on a different planet.
Like on the Moon because acceleration due to gravity is 1/6 that on the Earth, your weight on the Moon would be 1/6 the value on the Earth.
But note that you mass remains the same.
Answer:
the tension in each side of the cable is 3677.57 N
Explanation:
given data
traffic light = 20 kg
cable between two poles = 30 m
sag in the cable = 0.40 m
solution
by the free body diagram
tan θ = 
.............1
θ = 1.527 °
and
tension = mg
The net force is along x - axis is express as
T2 cosθ = T1 cosθ .................2
so T2 - T1 ..............3
and
when we take it along y axis that is express as
( T1 + T2) sinθ = mg ...................4
so by equation 3 we put here
2 × T1 sin(1.527) = 20 × 9.8
T1 = 3677.57 N
