Answer:
(iv), (v), (vi) would be incorrect.
Explanation:
(iv) Force isn't transferred from one colliding object to another, but momentum can be.
(v) An object doesn't stop immediately a force stops acting on it. Think of a thrown ball.
(vi) For an object not to move, it means that the net force on the object is zero, and not necessarily that there are no forces acting on the object. For example, an object could be pushed on one side, and be pushed on the other side with an equal force in the opposite direction. The forces would cancel each other and the net force would be zero.
The rest should be correct.
Answer:
Gravitational force affects weight, weight changes with change in gravity
Hope it helped u,
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Answer:Expression given below
Explanation:
Given mass of spring
Compression in the spring
Let the spring constant be K
Using Energy conservation
potential energy stored in spring =Kinetic energy of Block


now conserving momentum


where
is the final velocity
You need to set their position functions equal to one another and so for the time t when that is true. That is when the tiger and the deer are in the same place meaning the tiger catches the dear
Xdear= 2t+15 deer position function.
(I integrated the velocity function )
To get the Tigers position function you must integrate the acceleration twice. This becomes
Xtiger=t^2
Now t^2=2t+15
Time t is when the tiger catches the deer
t^2-2t-15=0
(t-5)(t+3)=0 factored
t=5s is the answer you use (t=-3 is a meaningless solution)
-- Bob covered a distance of (32m + 45m) = 77 meters.
-- His displacement is the straight-line distance and direction
from his starting point to his ending point.
The straight-line distance is
D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters
The direction is the angle whose tangent is (32/45) south of east.
tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.