Which component of the galaxy is shown in this image? Dust

Without effective assessment methods, teachers have no documented proof of childrens specific ____ and _____.

schepotkina [342]

Answer:

D strengths and weakneses

Explanation:

8 0

3 years ago

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea

melisa1 [442]

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

<u>Given:</u>

Linear charge density of the ring= $0.1\ \rm \mu C/m$
Radius of the ring R=0.2 m
Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let $\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

5 0

2 years ago

A lemon with mass 0.3 kg falls out of a tree from a height of 1.8 m. How much mechanical energy does the lemon have just before

Zina [86]

Mechanical energy is the energy that is possessed by an object due to its motion or due to its position. It can either be kinetics or potential. In this problem you know it starting position so you can calculate it's potential energy (PE):

<span>PE=mass∗gravity∗height=0.3kg∗9.8m/s2∗1.8m=?

</span>The answer will typically be given in joules:

1J=kg∗m2s2 Could be wrong... But I believe it is 5.3...? as a final product.

3 0

3 years ago

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When you trace the outline of your palm how do you find its area

Tcecarenko [31]

Answer:

you count the squares or messure it

Explanation:

you can raw equal squares about 1 cm wide if possible all equal and count the squares eg theres 10 squares (small hand) so that would be 10cm squared

7 0

2 years ago

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According to Newton's Second Law, the force of the club hitting the golf ball will cause it to accelerate. At the moment of impa

vazorg [7]

Answer:

Option B

Explanation:

<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>

Here in this case the force of the club hitting the golf ball will be in one direction and the force acting on club due to golf ball will be in opposite direction and magnitude of this force will be same as the magnitude of the force of the club hitting the golf ball

In this case the action will be the force of the club hitting the golf ball and reaction will be the force acting on club due to golf ball

∴ The club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club

8 0

2 years ago

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