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4 years ago

7

How does an atom of bromine-79 become a bromide ion with a -1 charge?

2 answers:

Naily [24]4 years ago

6 0

with a -1 iron charge it becomes an electrode

wariber [46]4 years ago

4 0

with a -1 iron charge it becomes an electrode

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Which of these materials are insulators? Select the THREE (3) that apply.

Mashutka [201]

Answer: wool, rubber, & plastic

Explanation: Insulators slow transfer.

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3 years ago

An alpha particle (Î±), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec

vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2$

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

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3 years ago

Satellites can focus on specific latitudes using:

Fittoniya [83]

B- east west orbits

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3 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

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3 years ago

I’m bored please end my boredom

Xelga [282]

Explanation:

im bored too im in class rn

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3 years ago

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