Answer: 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).
Explanation:
1) Since the mass of <span>155 g is suspended at 210 degrees, you need to find the components of its weight on the orthogonal coordinate system (0° and 90°).
</span>
<span>2) You do that using the trignometric ratios sine and cosine.
</span>
<span>Weight is mass × g.
</span>
<span>Weight of the object = 155g × g
</span>
<span>Angle, α = 210°
</span>
<span>Horizontal component (0°)
</span>
<span>cosα = horizontal / hypotenuse ⇒ horizontal = hypotenuse × cosα
</span>
⇒ horizontal = 155g × g × cos(210°) = - 134.23g × g
Vertical component
sinα = vertical / hypotenuse ⇒ vertical = hypotenuse × sinα
⇒ vertical = 155g × g × sin(210°) = -77.5g × g
3) Conclusion:
Therefore, the masses that must be suspended to balance the forces of the 155g mass are 134.23g at 0° (horizontal) and 77.5g at 90° (vertical).
Answer:
Only if the acceleration is uniform
Answer:
A case is an experimental unit. ... Another synonym is experimental unit. A variable is a characteristic that is measured and can take on different values. In other words, something that varies between cases.
Explanation:
Answer:
If the wind is offshore (blowing away from the dock), one should carefully approach the dock at a 20 to 30 degree angle. A bow line is then passed ashore and secured. In boats having an outboard, or inboard/outboard engine, the engine is turned towards the dock and put in reverse. This invariably will bring the stern into the dock.
The distance covered by the object is 42.4 m
Explanation:
The motion of the object is a uniformly accelerated motion (at constant acceleration), therefore we can use the following suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the object in this problem, we have:
u = 0 (it starts from rest)
v = 24 m/s (final velocity)

Solving for s, we find the distance travelled by the object:

Learn more about accelerated motion:
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