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3 years ago

12

Can someone give me 8-9

1 answer:

uysha [10]3 years ago

7 0

8). A rubber band is storing more elastic energy in it when it's stretched.
It's the work you had to do to stretch it.

9). Radio, TV, microwave, infrared (heat), visible light, ultraviolet light, X-rays,
and gamma rays are all part of the electromagnetic spectrum, and all of them
carry energy from one place to another.

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Does anyone know how to solve this??

Scrat [10]

I uploaded the answer to a file hosting. Here's link:

tinyurl.com/wtjfavyw

7 0

3 years ago

Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface

salantis [7]

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

R = 2r

= $2 \times 0.1 m$

= 0.2 m

where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

U = $\frac{k \times Q \times Q}{R}$

= $\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}$

= 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, $\frac{U}{2}$ .

Hence, K.E = $\frac{U}{2}$

= $\frac{0.081}{2}$

= 0.0405 J

Now, we will calculate the speed of balls as follows.

V = $\sqrt{\sqrt{\frac{2 \times K.E}{m}}$

= $\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}$

= 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

6 0

3 years ago

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

3 years ago

If the temperature of the metal oxide and water solution is increased,this would most likely..

kiruha [24]

Chemical Reaction between metal oxide and water solution

7 0

3 years ago

How will the force of friction affect a wood block being pushed over a table?

olasank [31]

Answer:

D

friction acts in the opposite direction of motion but does not affect the motion of the object

5 0

2 years ago