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The strength of the thrust is 122 newtons.
The motion of the rocket is described by the second Newton's law, whose model is shown below:
(1)
Where:
- Thrust, in newtons
- Drag, in newtons
- Mass of the rocket, in kilograms
- Net acceleration of the rocket, in meters per square second
If we know that
,
and
, then the strength of the thrust is:
![F = D + m\cdot a](https://tex.z-dn.net/?f=F%20%3D%20D%20%2B%20m%5Ccdot%20a)
![F = 90\,N + (8\,kg)\cdot \left(4\,\frac{m}{s^{2}} \right)](https://tex.z-dn.net/?f=F%20%3D%2090%5C%2CN%20%2B%20%288%5C%2Ckg%29%5Ccdot%20%5Cleft%284%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29)
![F = 122\,N](https://tex.z-dn.net/?f=F%20%3D%20122%5C%2CN)
The strength of the thrust is 122 newtons.
To learn more on Newton's laws, we kindly invite to check this verified question: brainly.com/question/13678295
During the daytime, I have mostly line symmetry.
During the night, I often have almost spherical symmetry.
Answer:
The 40g mass will be attached at 69 cm
Explanation:
First, make a sketch of the meterstick with the masses placed on it;
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm y cm
Apply principle of moment;
sum of clockwise moment = sum of anticlockwise moment
40y = 20 (38)
40y = 760
y = 760 / 40
y = 19 cm
Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm
12cm 50 cm 69cm
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm 19 cm