Answer:
A. Na2S + 2KCI - 2NaCl + KZS
Explanation:
Answer:
a) 0.167 μC/m^2
b) 1.887 * 10^4 V/m
Explanation:
Hello!
First let's find the surface charge density:
a)
Since thesatellite is metallic, the accumalted charge will be uniformly distribuited on its surface. Therefore the charge density σ will be:
σ = Q/A
Where A is the area of the satellite, which is:
A=4πr^2 = πd^2 = π(1.9m)^2
Therefore:
σ = (1.9)/(π (1.9)^2) μC/m^2 = 0.167 μC/m^2
Now let's calculate the electric field
b)
Just outside the surface of the satellite the elctric field will be:
E = σ/ε0
Where ε0=8.85×10^−12 C/Vm
Therefore:
E = (0.167*10^-6 C/m^2) / (8.85*10^-12 C/Vm) = 0.01887 *10^6 V/m
E = 1.887 * 10^4 V/m
Answer:
658.16N
Explanation:
Step one:
given data
mass m= 235kg
Force F= 760N
angle= 30 degrees
Required
The horizontal component of the force
Step two:
The horizontal component of the force
Fh= 760cos∅
Fh=760cos30
Fh=760*0.8660
Fh=658.16N
Wound it be one that dissolves ?
