Which formula describes Boyle's law?

Find the deBroglie wavelength of an electron with 4.0 eV.

pantera1 [17]

Explanation:

It is given that,

Voltage, $V=4 eV=4\times 1.6\times 10^{-19}\ V= 6.4\times 10^{-19}\ V$

De broglie wavelength in terms of voltage is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

m and e are the mass and charge on electron. So,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{6.4\times 10^{-19}} }\ A$

$\lambda=1.53\times 10^{10}\ A$

$\lambda=1.53\ m$

So, the De broglie wavelength of an electron is 1.53 meters. Hence, this is the required solution.

3 0

3 years ago

How does charging by conduction occur?

lozanna [386]

Hello~

The answer to your question is C.

A charged object touches a neutral object.

Explanation:

Even though it would have to be a (negative) charge, that point is that it is charge, so weather it is negative or positive, they are both charged what so ever.

~Hope this Helps!~

4 0

3 years ago

he block is released, and it slides 2.0 m (from the point at which it is released) across a horizontal surface before friction s

alex41 [277]

Answer:

0.245

Explanation:

When the block is released, the initial elastic potential energy stored in the spring is entirely converted into kinetic energy of the block.

Therefore, we can calculate the initial speed of the block:

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

where the term on the left is the potential energy and where the term on the right is the kinetic energy, and where

k = 4500 N/m is the spring constant

x = 8.0 cm = 0.08 m is the compression of the spring

m = 3.0 kg is the mass of the block

v is the initial velocity

Solving for v,

$v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(4500)(0.08)^2}{3.0}}=3.1 m/s$

Then, after the block is released, all its kinetic energy is converted into thermal energy as the block slows down, due to friction. Therefore, the work done by friction is equal to the initial kinetic energy of the block.

The force of friction is

$F=\mu mg$

where

$\mu$ is the coefficient of friction

$g=9.8 m/s^2$ is the acceleration of gravity

So the work done by it is (in magnitude)

$W=Fd=\mu mg d$

where

d = 2.0 m is the distance covered

Therefore,

$\frac{1}{2}mv^2 = \mu mg d$

And solving for $\mu$ ,

$\mu = \frac{v^2}{2gd}=\frac{3.1^2}{2(9.8)(2.0)}=0.245$

8 0

3 years ago

Please help me with my physics I do not understand! Please provide work thank you.

Studentka2010 [4]

Fe=K Q1/Q2/d2
Q1 is the first charge
Q2 is the second charge
d is the distance
K= 9x10^9 NM^2/C2
Now let’s plug the numbers
Fe=9x10^9NM^2/C2 (2x10^-4C)(8x10^-4C) / (0.3m^2) you notice we took away the negative charges when we plugged the charges
Ok now we notice that we have C2 which is C to the power 2 we can write it as C^2 and we have two CSU’s beside each one of the charges we can get rid of them all by curtailment
And we can curtailment the M^2and the other M^2
Now we left with only 9x10^9N (2x10^-4)(8x10^-4)/ 0.3
Let’s multiply the (9)(2)(8)=144
And add the exponents (9)+(-4)+(-4)=1
So now we got 144x10N divide by the distance which is 0.3
144x10N / 0.3 = 4800N
Hope it helps u understand :)

3 0

3 years ago

Which sediments typically make up the C horizon?

Nastasia [14]

B) sand, silt, and clay!

5 0

3 years ago

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