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Ugo [173]
3 years ago
6

Define Women Empowerment.​

Physics
1 answer:
djyliett [7]3 years ago
4 0

Answer:

Women's empowerment is all about equipping and allowing women to make life-determining decisions through the different problems in s

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If v=10v, what value of R would limit l to 10 mA? (if you have 10 volts, what resistance would be needed to limit the current to
777dan777 [17]

r=v/i =10/0.01=1000ohms B

7 0
3 years ago
A teapot with a surface area of 700 cm2 is to be plated with silver. It is attached to the negative electrode of an electrolytic
Brums [2.3K]
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An alpha particle (α), 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

5 0
3 years ago
The rest deltoid row is a back exercise true or false
bixtya [17]
False because your deltoids are in your shoulders not your back
3 0
3 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3 0
3 years ago
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