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Flura [38]
2 years ago
5

Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0

° with the horizontal. What is the height of the building?
Physics
1 answer:
valentina_108 [34]2 years ago
8 0

Answer:

willie tower

Explanation:

mataas kasi

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What does the scientific method help test in environmental science?
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What type of wave is infrared light
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A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance? Show
Marizza181 [45]

Answer:

Explanation:

v^2-u^2=2as\\40^2-10^2=2a*125\\1600-100=250a\\a=\frac{1500}{250}=6~m/s^2\\v=u+at\\40=10+6t\\6t=30\\ t=\frac{30}{6}=5 ~s

7 0
3 years ago
A cube of side 5.0m is in a region where the electric field is directed outward from both of two opposite cube faces, with unifo
guajiro [1.7K]

Answer:

The charge inside the cube is null.

Explanation:

If we apply the gauss theorem with a cubical gaussian surface of the size of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\frac{q_{in}}{\varepsilon_{0}}

If we consider than the direction of the electric field is \vec{E}=E_0\hat{x}, we can solve the problem differentiating  the integral for each face of the cube:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} \vec{E}\,\vec{ds_1}+\displaystyle\int_{S_2} \vec{E}\,\vec{ds_2}+\displaystyle\int_{S_3} \vec{E}\,\vec{ds_3}+\displaystyle\int_{S_4} \vec{E}\,\vec{ds_4}+\displaystyle\int_{S_5} \vec{E}\,\vec{ds_5}+\displaystyle\int_{S_6} \vec{E}\,\vec{ds_6}

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=\displaystyle\int_{S_1} E_0\hat{x}\,\hat{x}ds_1+\displaystyle\int_{S_2} E_0\hat{x}\,\hat{-x}ds_2+\displaystyle\int_{S_3} E_0\hat{x}\,\hat{y}ds_3+\displaystyle\int_{S_4} E_0\hat{x}\,\hat{-y}ds_4+\displaystyle\int_{S_5} E_0\hat{x}\,\hat{z}ds_5+\displaystyle\int_{S_6} E_0\hat{x}\,\hat{-z}ds_6

E₀ is a constant and each surface is equal to each other, so: S_1=S_2=S_i=S

Therefore:

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=E_0\displaystyle\int_{S_1} \,ds_1+E_0\displaystyle\int_{S_2} -1\,ds_2+0+0+0+0=E_0S-E_0S=0

\displaystyle\oint_{S} \vec{E}\,\vec{ds}=0=\frac{q_0}{\varepsilon_0} \longleftrightarrow q_0=0c

3 0
3 years ago
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