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Anna11 [10]
3 years ago
8

Four aqueous solutions and their concentrations are shown in the above illustration. which of the solutions is most likely to be

the strongest conductor of electricity?
Physics
1 answer:
Elza [17]3 years ago
3 0
The solution that would most likely be a strongest conductor of electricity is the solution that is most saturated or concentrated. This is because the atoms that are found within the aqueous solutions have become positively charged resulting to the attraction of negatively charged ions that are found in electricity. On the other hand, the least conductive from the aqueous solutions would be the most unsaturated one because of less conductive ions present.
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The gravitational force experienced by Earth due to the Moon is ________ the gravitational force experienced by the Moon due to
Vsevolod [243]

The gravitational force experienced by Earth due to the Moon is <u>equal to </u>the gravitational force experienced by the Moon due to Earth.

<u>Explanation</u>:

The force that attracts any two objects/bodies with mass towards each other is defined as gravitational force. Generally the gravitational force is attractive, as it always pulls the masses together and never pushes them apart.

The gravitational force can be calculated effectively using the following formula: F=GMmr^2  

where “G” is the gravitational constant.

Though gravity has the ability to pull the masses together, it is the weakest force in the nature.

The mass of the Earth and moon varies, but still the gravitational force felt by the Earth and Moon are alike.

5 0
3 years ago
Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass
salantis [7]
Use Scoratic it works with any time of subject
5 0
3 years ago
An Olympic diver springs off of a high dive that is 3 m above the surface of the water. When she lands in the water she is trave
klemol [59]

Answer:u=4.51 m/s\ at\ angle\ of\ \theta =59.34^{\circ}

Explanation:

Given

height of building h=3 m

Landing velocity of diver v=8.90 m/s at an angle of 75^{\circ}

Let u be the initial velocity of diver at an angle of \theta with horizontal

Since there is no acceleration in horizontal direction therefore horizontal component of velocity will remain same

u\cos \theta =8.9\cos (75)        ---- -----1

Considering Vertical motion

v^2-u^2=2as

here v=8.9\sin (75)

u=u\sin \theta

s=3 m

a=9.8 m/s^2

(8.9\sin (75))^2-(u\sin \theta )^2=2\times 9.8\times 3

u\sin \theta =\sqrt{(8.9\sin 75)^2-(2\cdot 9.8\cdot 3)}

u\sin \theta =3.886         ----------------2

Divide 2 and 1 we get

\tan \theta =\frac{3.886}{8.9\cos (75)}

\tan \theta =1.687

\theta =59.34^{\circ}

Thus u\cos (59.34)=8.9\cos (75)

u=4.51 m/s

7 0
3 years ago
Solve the below problems being sure to provide the correct significant figures.
morpeh [17]


Sum of the first ten digits is:
7+3+4+5+4+1+7+8+0 = 39

39, when divided by 9 give you the remainder of 3

9 x4, is 36
36 +3 equals 39
(39 = 9 x 4 + 3)

So X= 0
5 0
4 years ago
Write the dimensional formula of<br>presor and gravitation constant<br>​
deff fn [24]
Or, G = [M1 L1 T-2] × [L]2 × [M]-2 = [M-1 L3 T-2]. Therefore, the gravitational constant is dimensionally represented as M-1 L3 T-2.
8 0
3 years ago
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