i want quick solution of this qs.
Attachments:
Answers
Answered by
0
Answer:
2dav²cosθ
Explanation:
given, area of cross sectional of water jet = a
density of water = d
velocity of water = v
mass of water, m = density of water × volume of water
= d × (x × a) = dax [ here x is length of water flowing through the water jet ]
a/c to question,
A water jet whose cross sectional area is a strike a wall making an angle θ with normal and rebounds elastically as shown in figure.
so, change in momentum, ∆P = mvcosθ- (-mvcosθ)
= 2mvcosθ
putting, m = dax
so, ∆P = 2daxvcosθ
force exerted on wall = change in momentum/time
= 2daxvcosθ/t9
= 2davcosθ (x/t)
= 2davcosθ (v)
= 2dav²cosθ
hope it helps :)
Attachments:
Answered by
0
Answer:
bahi pls send kar de.......
Similar questions