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guajiro [1.7K]
3 years ago
9

A typical atomic polarizability is 1 × 10-40 (C·m)/(N/C). If the q in p = qs is equal to the proton charge e, what charge separa

tion s could you produce in a typical atom by applying
Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

s = 6.25 10⁻²² m

Explanation:

Polarizability is the separation of electric charges in a structure, in the case of the atom it is the result of the separation of positive charges in the nucleus and the electrons in their orbits, macroscopically it is approximated by

                p = q s

               s = p / q

let's calculate

              s = 1 10⁻⁴⁰ / 1.6 10⁻¹⁹

              s = 0.625 10⁻²¹ m

              s = 6.25 10⁻²² m

We see that the result is much smaller than the size of the atom, therefore this simplistic model cannot be taken to an atomic scale.

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after
adoni [48]

Answer: v = 1.19 * 10^{6} m/s

Explanation: q = magnitude of electronic charge = 1.609 * 10^{-19} c

mass of an electronic charge = 9.10 * 10^{-31} kg

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = \frac{mv^{2} }{2},  potential energy = qV

hence, \frac{mv^{2} }{2} = qV

\frac{9.10 *10^{-31} * v^{2}  }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31}  * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s

7 0
3 years ago
How can I solve this?
inysia [295]
There is no equation here
3 0
3 years ago
A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

4 0
4 years ago
A box of mass m = 1.80 kg is dropped from rest onto a massless, vertical spring with spring constant k = 2.00 ✕ 102 N/m that is
Rudik [331]

Answer:

spring compressed is 0.724 m

Explanation:

given data

mass = 1.80 kg

spring constant k = 2 × 10²  N/m

initial height = 2.25 m

solution

we know from conservation of energy is

mg(h+x)  = 0.5 × k × x²       ...................1

here x is compression in spring

so put here value in equation 1 we get

1.8 × 9.8 × (2.25+x)  = 0.5 × 2× 10² × x²

solve it we get

x = 0.724344

so spring compressed is 0.724 m

3 0
3 years ago
The temperature inside my refrigerator is about 4 degrees C. If I place a balloon in my fridge that initially has a temperature
maksim [4K]

v

Convert the given temperatures from celsius to kelvin since we are dealing with gas.

To convert to kelvin, add 273.15 to the temperature in celsius.

T1 = 22 + 273.15 = 295.15 k

T2 = 4 + 273.15 = 277.15 k

V1 = 0.5 L

Let's find the final volume (V2).

To solve for V2 apply Charles Law formula below:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

5 0
1 year ago
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