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yulyashka [42]
3 years ago
5

Which answer below is the formula mass of a water molecule

Physics
1 answer:
Firlakuza [10]3 years ago
5 0
Molecular formula of water molecule is H₂O.
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The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

L_T= Total signal level in db

n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
3 years ago
When low on gasoline, is it better to increase or decrease the speed of the vehicle? Explain.
sergiy2304 [10]
Yeah!! It is better to decrease the speed of the vehicle so the ejection of gases would happen in slow manner.... then you can save your gasoline
5 0
3 years ago
1. ________________electricity is the type of electricity commonly used in homes and businesses throughout the world.
amid [387]
Number 1 is letter A
3 0
3 years ago
A rubber balloon is rubbed with a PVC pipe and the rubber balloon becomes positively charged. Why is this? a) Because the rubber
pentagon [3]

Answer:

d)

Explanation:

Electrons are lost or gained when the ballon is rubbed with a PVC. As the rubber ballon lost electrons, it will have more protons, hence the positive charge. (More protons than electrons in the ballon).

4 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
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