Answer:
part (a)
towards north east direction.
part (b) s = 46.60 m
Explanation:
Given,
- velocity of the river due to east =

- velocity of the boat due to the north =

part (a)
River is flowing due to east and the boat is moving in the north, therefore both the velocities are perpendicular to each other and,
Hence the resultant velocity i,e, the velocity of the boat relative to the shore is in the North east direction. velocities are the vector quantities, Hence the resultant velocity is the vector addition of these two velocities and the angle between both the velocities are 
Let 'v' be the velocity of the boat relative to the shore.

Let
be the angle of the velocity of the boat relative to the shore with the horizontal axis.
Direction of the velocity of the boat relative to the shore.
part (b)
- Width of the shore = w = 300m
total distance traveled in the north direction by the boat is equal to the product of the velocity of the boat in north direction and total time taken
Let 't' be the total time taken by the boat to cross the width of the river.
Therefore the total distance traveled in the direction of downstream by the boat is equal to the product of the total time taken and the velocity of the river
21) Acceleration from D to E: 
22) The acceleration of the bus from D to E is 
Explanation:
21)
The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, we want to measure the acceleration of the bus from point D to point E. We have:
- Initial velocity at point D: u = 0
- Final velocity at point E: v = 5 m/s
- Time elapsed from D to E: t = 21 - 16 = 5 s
Therefore, the acceleration between D and E is

22) This question is the same as 21), so the result is the same.
Learn more about acceleration:
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Answer:
B
Explanation:
Reason is that the suitcase is exerted downward and when it moves downward the equation is mgsin tita
The weights in newtowns for the given masses are
<span> masses 22.1, 33.5, 41.3, 59.2, 78
weights 216.58N 328.3N 404.74N 580.16N 764.4N
e.g, for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>
Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
= 
So the lines terminating at - 2 micro coulomb
= 
So, the number of filed lines terminates at - 2 micro Coulomb are 6.