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mars1129 [50]
3 years ago
6

When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

.09x10 ^-19 C 1.53x10 ^-19 C 6.39x10 ^-19 C 4.66x10 ^-19 C I used these measurements to find the charge on the electron
Physics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C   → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C   → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C   → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first  charge , 2nd to 5 th charges can be  written as 2e,  3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of  1.6 x 10⁻¹⁹ C  exists.

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The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
3 years ago
iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms
saw5 [17]

Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

\frac{2}{a^{3} }  = \frac{N_{A} }{V_{molar} } where V_{molar}  = \frac{A}{\rho } an and A is the atomic mass of iron.

Therefore:

\frac{2}{a^{3} } = \frac{N_{A} * p }{A}

This implies that:

A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} }  }

  = \frac{4}{\sqrt{3} }r

Assuming that there is no phase change gives:

\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3}   }

  = 8.60 g/m³

3 0
3 years ago
By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles
Mademuasel [1]

Answer:

F' = (4/9)F

Explanation:

The electrostatic force between two charged objects is given by Coulomb's Law:

F = kq₁q₂/r²   -------------------- equation (1)

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of second charge

r = distance between charges

Now, when the charges and distance altered as follows:

q₁' = 2q₁

q₂' = 2q₂

r' = 3r

Then,

F' = kq₁'q₂'/r'²

F' = k(2q₁)(2q₂)/(3r)²

F' = (4/9)kq₁q₂/r²

using equation (1):

<u>F' = (4/9)F</u>

7 0
2 years ago
A wave has a wavelength of 5m and frequency of 5 Hz. What is the speed of the wave?​
Anna71 [15]
If you multiply m (the unit for wavelength) with 1s (the unit for frequency), you will get m/s, the unit for speed. Now multiply! 25 m/s is your final answer!
3 0
2 years ago
How do forest fires affect the Earth's atmosphere?
netineya [11]

It is B because the other ones are good.

5 0
2 years ago
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