Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answer:
≈ 20.35 N [newton's of tension]
Explanation:
( (2.9 × 9.8) ÷ cos(35.6°) ) × sin (35.6°) =
( (28.42) ÷ (≈0.813) ) × (≈0.582) =
(≈34.96) × (≈0.582) = 20.3449446.... ≈ 20.35
Answer:
562.5J
Explanation:
The following were obtained from the question:
F = 45N
d = 12.5m
w =?
The work done can be achieved by using
w = F x d
w = 45 x 12.5
w = 562.5J
Answer:
I'm pretty sure the answer is C . since deciduous trees are trees that loose leaves seasonally and coniferous trees are trees that don't loose leaves seaosnally and survive through the winter.
