Answer:
1 and 2
Explanation:
It is given that, force exerted by air is negligible in any way.
Also, it is given that channel is in the shape of a segment of a circle with its center at O.
When it is within the frictionless channel at position 'Q',
A gravity exerts force on the ball in the downward direction. On the other hand, the channel pointing from Q to O also exerts a force on the ball.
However, there is no any force in the direction of motion. On the other hand, he channel pointing from O to Q does not exert a force on the ball.
The concept that must be applied to solve this problem is that of the magnetic field on Toroid described since the law of Ampere. Mathematically the magnetic field is determined as

Where
= Permeability constant
N = Number of loops
I = Current
r = Radius
As we can see the magnetic field inside a Toroid is NOT uniform, since it is highly proportional to the radius. Which means that as the radius increases the magnetic field decreases.
Therefore the correct answer is: FALSE
Answer:
Both these motions are caused by the Gravitational force of earth.
Explanation:
Both these motions are caused by the Gravitational force of earth.
Answer:
a) 
b) 
c)
d) 
e) 
Explanation:
a) Let's recall the definition of work.

Because the system is a quasi-static isobaric expansion, P is constant here, therefore:

![W=2.0\cdot 10^{5}(4.0\cdot 10^{-3}-2.0\cdot 10^{-3})= 400 [Nm^{2}]](https://tex.z-dn.net/?f=W%3D2.0%5Ccdot%2010%5E%7B5%7D%284.0%5Ccdot%2010%5E%7B-3%7D-2.0%5Ccdot%2010%5E%7B-3%7D%29%3D%20400%20%5BNm%5E%7B2%7D%5D)
b) Using the ideal gas equation we have:
(1)
and
(2)
We can note that n times R is a constant in (1) and (2), so we can equal those equations.
(3)
Let's solve T₂ for (3), let's recall that P₁ = P₂, so they canceled out


c) Using the equation of ideal gas we have:
d) We can write the internal energy as a function of Cv, and as we know the Cv is 1.5R for a monoatomic gas.



e) Using the first law of thermodynamic, we have:

Finally,

Have a nice day!
The resistors behavelike a single resistor of (21.2/4) = 5.3 ohms.
The total current through the parallel bunch of resistors is E/R = 35.8/5.3 = 6.75 Amperes.
That's the total current SUPPLIED by the battery. Current doesn't flow 'through' a battery.