'Ampere' is the unit of current. That's the rate at which
electrons travel in the circuit ... the number of electrons
every second. If you wanted the actual amount or number
of electrons, you'd need to know the length of time too.
It doesn't matter whether we're talking about a parallel or
series circuit.
Based on Newton's second law of motion, the net force applied to an object is equal to the product of the mass of the object and the acceleration it experiences. That is,
F = ma
If we are to assume that the net force is constant and that the mass is increased, the acceleration should therefore decrease in order to make constant the value at the right-hand side of the equation.
Answer: 2. Solution A attains a higher temperature.
Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.
In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.
Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.
<em>We have a formula for such condition,</em>
.....................................(1)
where:
= temperature difference
- c= specific heat of the body
<u>Proving mathematically:</u>
<em>According to the given conditions</em>
- we have equal masses of two solutions A & B, i.e.

- equal heat is supplied to both the solutions, i.e.

- specific heat of solution A,

- specific heat of solution B,

&
are the change in temperatures of the respective solutions.
Now, putting the above values


Which proves that solution A attains a higher temperature than solution B.
Answer:
Q=185.84C
Explanation:
We have to take into account the integral

In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4

but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!