Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As
For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
Answer:
52,360,000km
Explanation:
To solve this problem you use a conversion factor.
By taking into account that 1UA = 1.496*10^{8}km you obtain:
hence, 0.35UA is about 52,360,000km. This is the least distance between Mars and Earth
Answer:
B) 8 T.
Explanation:
For an ideal gas , the gas law that it follow is as follows
P₁V₁ / T₁ = P₂ V₂ /T₂
This law is followed by all the gases whether mono atomic or diatomic.
In this case following information are given
P₂ / P₁ = 3
V₂ / V₁ = 3
T₁ = T
T₂ = ?
From the gas law formula given , we have
= 3 X 3 T
= 9 T
Change in temperature = 9T - T = 8 T.