Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
Answer:
A) and B) are correct.
Explanation:
Let's take a look at the attached picture. Now
The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE
The equivalent capacitance is going to be: 
And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE
And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE
Answer:
Explanation:
Work done in carrying bricks
mgh
= 207 x 9.8 x 3.65
-= 7404.4 J
Work done in compressing gas
PΔV
Pressure x change in volume
1.8 x 10⁶ ΔV = 7404.4
ΔV = 7404.4 / 1.8 x 10⁶m³
= 4113.33 x 10⁻⁶ m³
= 4113.33 cc
Answer:
Explanation:
The formula to determine the size of a capillary tube is
h = 2•T•Cos θ / r•ρ•g
Where
h = height of liquid level
T = surface tension
r = radius of capillary tube
ρ = density of liquid
θ = angle of contact = 0°
g =acceleration due to gravity=9.81m/s²
The liquid is water then,
ρ = 1000 kg / m³
Given that,
T = 0.0735 N/m
h = 0.25mm = 0.25 × 10^-3m
Then,
r = 2•T•Cos θ / h•ρ•g
r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81
r = 5.99 × 10^-3m
Then, r ≈ 6mm
The radius of the capillary tube is 6mm
So, the minimum size is
Volume = πr²h
Volume = π × 6² × 0.25
V = 2.83 mm³
The minimum size of the capillary tube is 2.83mm³
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<span>torque = rF
= 0.1(10)
=1 Nm</span>
