Weight of the carriage
Normal force
Frictional force
Acceleration
Explanation:
We have to look into the FBD of the carriage.
Horizontal forces and Vertical forces separately.
To calculate Weight we know that both the mass of the baby and the carriage will be added.
- So Weight(W)
To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with , force of acting vertically downward.Both are downward and Normal is upward so Normal force
- Normal force (N)
- Frictional force (f)
To calculate acceleration we will use Newtons second law.
That is Force is product of mass and acceleration.
We can see in the diagram that and component of forces.
So Fnet = Fy(Horizontal) - f(friction)
- Acceleration (a) =
So we have the weight of the carriage, normal force,frictional force and acceleration.
Newtons third law (inertia) is to blame
Answer:
120 m
Explanation:
Given:
wavelength 'λ' = 2.4m
pulse width 'τ'= 100T ('T' is the time of one oscillation)
The below inequality express the range of distances to an object that radar can detect
τc/2 < x < Tc/2 ---->eq(1)
Where, τc/2 is the shortest distance
First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'
f = c/λ (c= speed of light i.e 3 x m/s)
f= 3 x / 2.4
f=1.25 x hz.
As, T= 1/f
time of one oscillation T= 1/1.25 x
T= 8 x s
It was given that pulse width 'τ'= 100T
τ= 100 x 8 x => 800 x s
From eq(1), we can conclude that the shortest distance to an object that this radar can detect:
= τc/2 => (800 x x 3 x )/2
=120m
Answer:
Explanation:
We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector
D₁ = - 255 cos 49 i + 255 sin49 j
= - 167.29 i + 192.45 j
Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is
D = 125 i
So
D₁ + D₂ = D
- 167.29 i + 192.45 j + D₂ = 125 i
D₂ = 125 i + 167.29 i - 192.45 j
= 292.29 i - 192.45 j
Angle of D₂ with x axes θ
tan θ = -192.45 / 292.29
= - 0.658
θ = 33.33 south of east
Magnitude of D₂
D₂² = ( 192.45)² + ( 292.29)²
D₂ = 350 km approx
Tan