Answer:
The current flows through the insulator is 2 mA.
Explanation:
Given that,
Resistance 
Voltage = 200 kV
We need to calculate the current
Using ohm's law


Where, I = current
V = voltage
R = resistance
Put the value into the formula



Hence, The current flows through the insulator is 2 mA.
Answer:
0.45 miles
Explanation:
since, car will be running on the edge of track, distance covered in one lap by car will be equal to circumference of track.\
we will use value of pi as 22/7
Circumference for any circular shape is given by 

Hence, in one lap
distance will be covered
in 7 laps
distance will be covered.
but is given that 7 laps should be equal to 10 miles
so 10 miles will be equal to 

Thus, diameter of track will be 0.45 miles to meet the requirement.
Basically, Newton's ideas matched up better with experiments and observations about the natural world than Aristotle's did. Newton gave a rigorous mathematical framework that made very specific predictions about our world, while Aristotle in general made more comparative laws, that even when true were less useful than the certainty Newton's laws gave us. Many of Newton's laws have been found to be at least partially incorrect now, for instance his laws of motion fall apart at speed nearing the speed of light, his laws of gravity fall apart when talking about more than two objects and in the presence of large gravitational fields that are close together, and Newton's law of cooling is just untrue in general (though can make some approximations in narrow temperature ranges).
The Ideal Gas Law
The Ideal Gas Law (PV = nRT) can be used if you have pressure, volume, and temperature.Apr
Answer:
a) ΔV = 2,118 10⁻⁸ m³ b) ΔR= 0.0143 cm
Explanation:
a) For this part we use the concept of density
ρ = m / V
As we are told that 1 carat is 0.2g we can make a rule of proportions (three) to find the weight of 2.8 carats
m = 2.8 Qt (0.2 g / 1 Qt) = 0.56 g = 0.56 10-3 kg
V = m / ρ
V = 0.56 / 3.52
V = 0.159 cm3
We use the relation of the bulk module
B = P / (Δv/V)
ΔV = V P / B
ΔV = 0.159 10⁻⁶ 58 10⁹ /4.43 10¹¹
ΔV = 2,118 10⁻⁸ m³
b) indicates that we approximate the diamond to a sphere
V = 4/3 π R³
For this part let's look for the initial radius
R₀ = ∛ ¾ V /π
R₀ = ∛ (¾ 0.159 /π)
R₀ = 0.3361 cm
Now we look for the final volume and with this the final radius
= V + ΔV
= 0.159 + 2.118 10⁻²
= 0.18018 cm3
= ∛ (¾ 0.18018 /π)
= 0.3504 cm
The radius increment is
ΔR =
- R₀
ΔR = 0.3504 - 0.3361
ΔR= 0.0143 cm