A patch of ground in direct sunlight will have ______ a patch of ground shaded by clouds.
2 answers:
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Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Answer:
a) 4.98m/s²
b) 481.66N
Explanation:
a) Using the Newtons second law of motion
m is the mass of the object
g is the acceleration due to gravity
Fm is the moving force acting along the plane
Ff is the frictional force opposing the moving froce
a is the acceleration of the skier
Given
m = 60kg
g = 9.8m/s²
= 35°
Ff = 38.5N
Required
acceleration of the skier a
Substituting into the formula;
Hence the acceleration of the skier is 4.98m/s²
b) The normal force on the skier is expressed as;
N = Wcosθ
N = mgcosθ
N = 60(9.8)cos 35°
N = 588cos 35°
N = 481.66N
Hence the normal force on the skier is 481.66N