Answer:
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Explanation:
Electric field due to plates
Ef = V/d
Ef = 2033 V / (2.0 * 10^-2 m )
Ef = 101650 V/m
So, we can write
Ef * q = m*g
q = m*g / E
f
The mass can be equal using the density and the volume so:
m = ρ * v
The volume can be find as:
v = 2.298 x 10 ⁻ ¹⁶ m³
q = ρ * v * g / Ef
q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Answer:
The time constant is
Explanation:
From the question we are told that
the time take to charge is
The mathematically representation for voltage potential of a capacitor at different time is
Where is the time constant
is the potential of the capacitor when it is full
So the capacitor potential will be 100% when it is full thus 100% = 1
and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so
V = 0.85
Hence
Unusual precipitation patterns
Answer:
71.85 m/s
Explanation:
Given the following :
Length of skid marks left by jaguar (s) = 290 m
Skidding Acceleration (a) = - 8.90m/s²
Final velocity of jaguar (v) = 0
Speed of Jaguar before it Began to skid =?
Hence, initial speed of jaguar could be obtained using the formula :
v² = u² + 2as
Where
v = final speed of jaguar ; u = initial speed of jaguar(before it Began to skid) ; a = acceleration of jaguar ; s = distance /length of skid marks left by jaguar
0² = u² + (2 × (-8.90) × 290)
0 = u² + (-5,162)
u² = 5162
Take the square root of both sides
u = √5162
u = 71.847 m/s
u = 71.85m/s
Answer:
(a) 2.33 A
(b) 15.075 V
Explanation:
From the question,
The total resistance (Rt) = R1+R2 = 3.85+6.47
R(t) = 10.32 ohms.
Applying ohm's law,
V = IR(t)..........equation 1
Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.
Note: Since both resistors are connected in series, the current flowing through them is the same.
Therefore,
I = V/R(t)............. Equation 2
Given: V = 24 V, R(t) = 10.32 ohms
Substitute these values into equation 2
I = 24/10.32
I = 2.33 A.
Hence the current through R1 = 2.33 A.
V2 = IR2.............. Equation 3
V2 = 2.33(6.47)
V2 = 15.075 V