Answer:
3.6μF
Explanation:
The charge on the capacitor is defined by the formula
q = CV
because the charge will be conserved
q₁ = C₁V₂
q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same
q = q₁ + q₂ = C₁V₂ + C₂V₂
CV = CV₂ + C₂V₂
CV - CV₂ = C₂V₂
C ( V - V₂) = C₂V₂
C ( V/ V₂ - V₂ /V₂) = C₂
C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF
Answer: 75.02 m
Explanation:
u = 0 ( starts from rest )
v = 50 m/s
t = 3 s
( i ) a = v - u / t
= 50 - 0 /3
= 16.67
( ii ) s = ut + 1/2 at²
= 0 × 3 + 1/2 × 16.67 × 3 × 3
= <u>75.02 m</u>
Hope this helps...
Answer:
The discharge of the stream at this location is 40 cubic meters per second.
Explanation:
The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:
Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)
Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)
Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)
<u>Volume Flow Rate = 40 cubic meters per second</u>
Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>
This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.
If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be 
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
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