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.
To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.
Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.
To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.
To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.
Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.
Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.
The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192
We can further simplify this ratio:
96:192
48:96
24:48
12:24
6:12
3:6
1:2
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
<u>T₂ = 95.56°C</u>
Frequency = speed / wavelength
(6 m/s) / (12 m) = 0.5 Hz.
That's not infrared light.
Infrared light waves move about 50 million times faster than that, and they're only about 0.00000007 as long as that.
Answer
Given,
Time period of star,T = 3.37 x 10⁷ s
Radius of circular orbit,R = 1.04 x 10¹¹ m
a) Angular speed of the planet
b) tangential speed
v = 1.94 x 10⁴ m/s
c) centripetal acceleration magnitude
a = 3.62 x 10⁻³ m/s²
