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
Answer:
The work flow required by the compressor = 100.67Kj/kg
Explanation:
The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .
The work flow can be determined using the equation:
M1h1 + W = Mh2
U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2
Workflow = P2alpha2 - P1alpha1
Workflow = (h2 -U2) - (h1 - U1)
Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)
Workflow = ( 193.191 - 92.519)Kj/kg
Workflow = 100.672Kj/kg
Answer: 90 kgm/s
Explanation:
The momentum (linear momentum)
is given by the following equation:
Where:
is the mass of the skater
is the velocity
In this situation the skater has two values of momentum:
Initial momentum: 
Final momentum: 
Where:


So, if we want to calculate the difference in the magnitude of the skater's momentum, we have to write the following equation(assuming the mass of the skater remains constant):
Finally:
Answer:
a
The x- and y-components of the total force exerted is

b
The magnitude of the force is

The direction of the force is
Clockwise from x-axis
Explanation:
From the question we are told that
The magnitude of the first charge is 
The magnitude of the second charge is 
The position of the second charge from the first one is 
The magnitude of the third charge is 
The position of the third charge from the first one is 


The position of the third charge from the second one is



The force acting on the third charge due to the first and second charge is mathematically represented as

Substituting values



The magnitude of
is mathematically evaluated as

The direction is obtained as

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