If we pull an object vertically upwards then we need to apply a force which is equal in the magnitude of the weight of the object

now when we pull the same object upwards along an inclined plane with angle then we require a force which will balance the component of weight along the inclined
so it is given as

so as if we compare the two forces we can say that since the value of sine is always less than 1 for an angle less than 90 degree
so in the 2nd case when we pull the object along the inclined plane it will require less effort
so correct answer is
<em>A. reduce effort</em>
They are attractive
They don’t depend on charge
If the vertical component is 29.6 m/s down, and the horizontal component
is 54.8 m/s parallel to the surface, then the magnitude of the slanty vector is
√(29.6² + 54.8²) = √(876.16 + 3003.04) = √3879.2 = 62.28 m/s .
That's 139 mph ! Wow !
Answer:
0.465 kgm/s
Explanation:
Given that
Mass of the cart A, m1 = 450 g
Speed of the cart A, v1 = 0.85 m/s
Mass of the cart B, m2 = 300 g
Speed of the cart B, v2 = 1.12 m/s
Now, using the law of conservation of momentum.
It is worthy of note that our cart B is moving in opposite directions to A
m1v1 + m2v2 =
(450 * 0.85) - (300 * 1.12) =
382.5 - 336 =
46.5 gm/s
If we convert to kg, we have
46.5 / 100 = 0.465 kgm/s
Thus, the total momentum of the system is 0.465 kgm/s