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pickupchik [31]
3 years ago
12

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:
torisob [31]3 years ago
8 0

Answer:

Explanation:

We put the charges in the ascending order as follows

1.53 P

3.26 P

4.66 P

5.09 P

6.39 P

where P is equal to 10⁻¹⁹

we round off given charges as follows

1.53 P → 1.6 P

3.26 P → 3.2 P

4.66 P → 4.8 P

5.09 P → 4.8 P

6.39 P → 6.4 P

We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .

Hence this charge may exist on single electron.

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If the image
Zina [86]

Answer:

The lens is a concave lens

Explanation:

In concave lens, the image formed is erect. diminished and virtual. The more the distance of the object from the mirror, the more diminished the image formed will be. Concave lens are used in binoculars and telescopes and are installed before or in the eye piece to enable a clear focus.

4 0
3 years ago
With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c
Alinara [238K]

Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

            Where: m is the object's mass

                         g is the  gravitational acceleration in the surface earth

                         g = 9.8 m/s2

The the ultralight car's weight is:

w_{uc} = (540)(9.8)

w_{uc} = 5292 N

And the Honda Accord's weight is:

w_{HA} = (1450)(9.8)

w_{HA} = 14210 N

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

4 0
4 years ago
98. In Fig. 24-71, a metal sphere
yarga [219]

Answer:

(a) The potential difference between the spheres is 750 kVA

(b) The charge on the smaller sphere is 6.\overline 6 μC

(c) The charge on the smaller sphere, Q₁ = 13.\overline 3 μC

Explanation:

(a) The given parameters are;

The charge on the inner sphere, q = 5.00 μC

The radius of the inner sphere, r = 3.00 cm = 0.03 m

The charge on the larger sphere, Q = 15.0 μμC

The radius of the larger sphere, R = 6.00 cm = 0.06 m

The potential difference between two concentric spheres is given according to the following equation;

V_r - V_R = k \times q \times \left ( \dfrac{1}{r} - \dfrac{1}{R} \right)

Where;

R = The radius of the larger sphere = 0.06 m

r = The radius of the inner sphere = 0.03 m

q = The charge of the inner sphere = 5.00 × 10⁻⁶ C

Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C

k = 9 × 10⁹ N·m²/C²

Therefore, by plugging in the value of the variables, we have;

V_r - V_R = 9 \times 10^9  \times 5.00 \times 10^{-6} \times \left ( \dfrac{1}{0.03} - \dfrac{1}{0.06} \right) = 750,000

The potential difference between the spheres, V_r - V_R = 750,000 N·m/C = 750 kVA

(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;

Q_f = Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C

Q_f = 20 × 10⁻⁶ C

From which we have;

Q₁/Q₂ = R/r

Where;

Q₁ = The new charge on the on the larger sphere

Q₂ = The new charge on the on the smaller sphere

Q_f = 20 × 10⁻⁶ C = Q₁ + Q₂

∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂

∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2

20 μC - Q₂ = 2·Q₂

20 μC = 3·Q₂

Q₂ = 20 μC/3

The charge on the smaller sphere, Q₂ = 20 μC/3 = 6.\overline 6 μC

(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3

The charge on the smaller sphere, Q₁ = 40 μC/3 = 13.\overline 3 μC.

5 0
3 years ago
Calculate the density of Saturn. Show your work. How does it compare with the density of water? Explain how this can be.
stepladder [879]

Answer:

The density of Saturn is 686.81 kg/m³.

Explanation:

Mass of Saturn,  m=5.68\times 10^{26}\ kg  

Volume of Saturn, V=8.27\times 10^{23}\ m^3

Density of Saturn is given by its mass divided by its volume i.e.

d=\dfrac{m}{V}

d=\dfrac{5.68\times 10^{26}}{8.27\times 10^{23}}

d=686.81\ kg/m^3

So, the density of Saturn is 686.81 kg/m³.

The density of water is 1000 kg/m³. It is clear that the density of Saturn is less than water.

4 0
3 years ago
What units should be used when measuring the mass of a ladybug?
Aleks [24]
You could use grams hope this helps
3 0
3 years ago
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