Answer:OB=58.3m
Explanation:
So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.
now take the starting point as a origin such that cow moves in x-y co-ordinate axis.
As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.
So the final displacement is the length of cow from the origin that is length OB.
now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]
now displacement of cow= length of OB
= ![\sqrt{[37.084]^{2}+[44.984]^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5B37.084%5D%5E%7B2%7D%2B%5B44.984%5D%5E%7B2%7D%20%20%7D)
=
OB =
Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.
So 
.
Kinetic energy of system at first: 
. After: 
. The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).
Answer is A)
Answer:
true! : )
(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)
Explanation:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. <u>gravitational pull decreases with an increase in the distance between two objects.</u> Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.
When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>
On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.
What is Hard stabilization?
- Hard stabilization is the prevention of erosion through the use of artificial barriers.
- Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
- Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
- These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
- Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.
Learn more about the Hard stabilization with the help of the given link:
brainly.com/question/16022736
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