A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn
1 answer:
The magnitude of the vector C is 96.32m
<h3>How to solve for the magnitude of vector c</h3>
Ax = AcosθA
= 40 cOS 20
= 37.59
Ay = AsinθA
-40sin20
= -13.68
Bx = B cos θ B
= 75Cos50
= 48.21
By = BsinθB
= 75sin50
= 57.45
Cx = AX + Bx
= 37.59 + 48.21
= 85.8
Cy = Ay + By
= -13.65 + 57.45
= 43.77
The magnitude is solved by
|c| =
= √85.8² + 43.77²
= 96.32m
The magnitude of the vector c is 96.32m
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Answer:
5.7141 m
Explanation:
Here the potential and kinetic energy will balance each other
This is the initial velocity of the system and the final velocity is 0
t = Time taken = 0.04 seconds
F = Force = 18000 N
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
Equation of motion
From Newton's second law
Squarring both sides
The height from which the student fell is 5.7141 m
Observer A is moving inside the train
so here observer A will not be able to see the change in position of train as he is standing in the same reference frame
So here as per observer A the train will remain at rest and its not moving at all
Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body
So here observer B will see the actual motion of train which is moving in forward direction away from the platform
Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction
So the distance between them will decrease at faster rate
Now as per Newton's II law
F = ma
Now if train apply the brakes the net force on it will be opposite to its motion
So we can say
- F = ma
so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate
It is not affected by the gravity because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train
So there is no effect on train motion