The current in the junction is 1350 mA or 1.350 A.
Explanation:
As per Kirchoff's first law, the algebraic sum of current meeting at any junction should be equal to the algebraic sum of current leaving the junction. As in the present case, three parallel branch circuit is given the current in 250 mA, 300 mA and 800 mA, respectively, the sum of these three current will be equal to the current in the junction.
So,
I₁+I₂+I₃ = I₄
So I₁,I₂ and I₃ are the current passed in the three parallel branches and I₄ is the current in the junction.
250 + 300 + 800 = 1350 mA
So the current in the junction is 1350 mA or 1.350 A.
Answer:
A) 37 m
Explanation:
The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:
(1)
where
v = 0 m/s is the final velocity of the car
u = 24 m/s is the initial velocity
a is the acceleration
d is the length of the skid
We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:
where
m = 1000 kg is the mass of the car
is the coefficient of friction
a is the deceleration of the car
g = 9.8 m/s^2 is the acceleration due to gravity
The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:
And we can substitute it into eq.(1) to find d:
The correct answer is A.
The cell membrane consists of a phospholipid bilayer with embedded proteins. Sometimes molecules are just too big to easily flow across the plasma membrane or dissolve in the water so that they can be filtered through the cell membrane. In these cases , the cells must put out a little energy to help get molecules in and out of the cell.
The proteins embedded in the plasma membrane form channels through which other molecules can pass. Some proteins act as carriers, that is they are 'paid" in energy to let a molecule attach to itself and then transport that molecule inside the cell. This is called active transport.