Answer:
magnitude of force on charge 2Q = 
Direction of force on charge = 61 ⁰
Explanation:
The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force
║F║ = 
= after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW
magnitude of force acting on 2Q =
The direction of the force on charge 2Q is calculated as
tan ∅ = 
= 1.8284
therefore ∅ = 
1.8284
= 61⁰
Answer: Take your pick
Explanation:
if they are all in parallel 1 /(1/100 + 1/300 + 1/50) = 30 Ω
if 50 is in parallel with 2 in series 1 / (1/(100 + 300) + 1/50) = 44.444...Ω
if 100 is in parallel with 2 in series 1 / (1/(50 + 300) + 1/100) = 77.777...Ω
if 300 is in parallel with 2 in series 1 / (1/(100 + 50) + 1/300) = 100 Ω
If 50 is in series with 2 in parallel 50 + 1/(1/100 + 1/300) = 125 Ω
If 100 is in series with 2 in parallel 100 + 1/(1/50 + 1/300) = 142.857...Ω
If 300 is in series with 2 in parallel 300 + 1/(1/50 + 1/100) = 333.333...Ω
If they are all in series 100 + 300 + 50 = 450 Ω
Answer:
The same current flows through each resistor in series. Individual resistors in series do not get the total source voltage, but divide it. The total resistance in a series circuit is equal to the sum of the individual resistances: RN(series)=R1+R2+R3+…
Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1
Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2
Now using formula of effective degree of freedom of mixture, (f), we have:
f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)
We know f=11/6,
substituting the value in above relation, we have:
Cv= R. 11/3*2
= R. 11/6
Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
= 17/6 R
Therefore, Cp/Cv = 17/11
Answer:
B. Landscape B
Explanation:
Shale is fine sediment pressed together to form rock.
Sandstone is larger (sand-grain-sized) sediment cemented together to form rock.
Shale erodes faster, as evidenced by the second attachment. That attachment shows erosion of a rock face consisting of interbedded shale and sandstone. The shale has receded significantly, leaving the sandstone layers with space between them.
