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kramer
3 years ago
6

Suppose a 1.2 m antenna is installed on top of a building that is 225 m tall,

Physics
1 answer:
vazorg [7]3 years ago
7 0

Answer: 226.2 m

Explanation:

If we want to find the new total height H of the building, we have to add to its initial height h_{B}=225m the height of the antenna h_{A}=1.2m, as follows:

H=h_{B}+h_{A}

H=225m+1.2m

Then:

H=226.2m This is the new total height of the building.

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How would you compare and contrast the 3 types of friction?
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Please help!!
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3 years ago
An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi
elena55 [62]

Answer:

a) \vec F_{B} = 8.766\times 10^{-14}\,T\,k, b) \vec F_{B} = -8.766\times 10^{-14}\,T\,k

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

\vec F_{B} = q\cdot \vec v \times \vec B

The charge of the electron is equal to -1.602\times 10^{-19}\,C. Then, cross product can be solved by using determinants:

\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = 8.766\times 10^{-14}\,T\,k

b) The charge of the proton is equal to 1.602\times 10^{-19}\,C. Then, cross product has the following determinant:

\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}

The magnetic force is:

\vec F_{B} = -8.766\times 10^{-14}\,T\,k

8 0
3 years ago
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