Anyone good with this type of vocabulary. I really need to get this done to study.

1 answer:

8 0

I don’t go I don’t know I’m sorry I speak Spanish

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EleoNora [17]

Answer: $a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )$

Explanation:

Given

Electric Field $\vec{E}=2\hat{i}+5.4\hat{j}$

$\vec{B}=0.4\hat{k}\ T$

velocity $\vec{v}=8\hat{i} m/s$

mass of electron $m=9.1\times 10^{-31} kg$

Force on a charge Particle moving in Magnetic Field

$F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]$

$a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}$

$a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}$

$a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2$

7 0

3 years ago

sergij07 [2.7K]

Float. This is because water has a density of 1 g/ml, so anything less than this will float.

5 0

4 years ago

seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

$v = \lambda * f (1)$

Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

$\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)$