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egoroff_w [7]
3 years ago
6

The color that an object appears to be depends on the select one:

Physics
1 answer:
ddd [48]3 years ago
3 0
B is the right answer
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It’s A because it stays in motion whenever you drop it
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Which of the following is the best definition of lifestyle activity?
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John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2
Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function cos(\omega t +\phi) is defined between -1 and 1, So it is not possible obtain a value 2\pi greater.  

In addition, if you  move the function cosine a T Value, and T is the Period,  the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have f=1+cos(\omega t +\phi), the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the f=1+cos(\omega t +\phi), and its mean that the function f=cos(\omega t +\phi), in general, is not greater than cos(\omega t +\phi).

3 0
3 years ago
What is the average de Broglie wavelength of oxygen molecules in air at a temperature of 27°C? Use the results of the kinetic th
asambeis [7]

Answer:

\lambda = 2.57 \times 10^{-11} m

Explanation:

Average velocity of oxygen molecule at given temperature is

v_{rms} = \sqrt{\frac{3RT}{M}}

now we have

M = 32 g/mol = 0.032 kg/mol

T = 27 degree C = 300 K

now we have

v_{rms} = \sqrt{\frac{3(8.31)(300)}{0.032}

v_{rms} = 483.4 m/s

now for de Broglie wavelength we know that

\lambda = \frac{h}{mv}

\lambda = \frac{6.6 \times 10^{-34}}{(5.31\times 10^{-26})(483.4)}

\lambda = 2.57 \times 10^{-11} m

7 0
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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