Complete Question:
Check the circuit in the file attached to this solution
Answer:
Total current = 0.056 A(From left to right)
Explanation:
Let the current in loop 1 be I₁ and the current in loop 2 be I₂
Applying KVL to loop 1
30 - (I₁ - I₂)500 + I₂R + 15 = 0
45 - 500I₁ - 500I₂ + RI₂ = 0
I₁ = 30mA = 0.03 A
45 - 500(0.03) - 500I₂ + RI₂ = 0
30 -500I₂ + RI₂ = 0...............(1)
Applying kvl to loop 2
-RI₂ - 15 + 10 - 400I₁ = 0
-RI₂ = 5 + 400*0.03
RI₂ = -17 ................(2)
Put equation (2) into (1)
30 -500I₂ -17 = 0
-500I₂ = 13
I₂ = -13/500
I₂ = -0.026 A
The total current in the 500 ohms resistor = I₁ - I₂ = 0.03+0.026
Total current = 0.056 A
The current will flow from left to right
acceleration of rocket is given here as

now we know that

now integrating both sides



here since its given that rocket will accelerate for t = 10 s
so here we have


so after t = 10 s the speed of rocket will be 130 m/s upwards
Answer:
dont you have to times it
Explanation:
<u>Answer:</u>
Cannonball will be in flight before it hits the ground for 2.02 seconds
<u>Explanation:</u>
Initial height from ground = 20 meter.
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8
, we need to calculate time when s = 20 meter.
Substituting

So it will take 2.02 seconds to reach ground.