Hwueuehhdhdhehdhsijsjsjdjhh
2 answers:
You might be interested in
Answer:
t=20s
Explanation:
To solve this problem we must apply the first law of thermodynamics, which indicates that the energy that enters a system is the same that must come out, resulting in the following equation
For this problem we will assume that the water is in a liquid state, since it is a domestic refrigerator
q=m.cp.(T2-T1)
q=heat
m=mass of water =600g=0.6Kg
cp=
specific heat of water=4186J/kgK
T2=temperature in state 2=20°C
T1=temperature in state 1=0°C
solving:
q=(0.6)(4186)(20-0)=50232J
A refrigerator is a device that allows heat to be removed to an enclosure (Qin), by means of the input of an electrical energy (W) and the heat output (Qout), the coefficient of performance COP, allows to know the ratio between the heat removed ( Qin) and the added electrical power (W), the equation for the COP is
To solve this exercise we must know the value of the heat removed to the water (Qin)
solving for Qin
Qin=(COP)(Win)
Qin=(5)(500W)=2500W
finally we remember that the definition of power is the ratio of work over time
w=work
p=power=500w
Answer:
A) See Annex
B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)
C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)
D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)
E) Net force (its components)
Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂
Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂
Explanation:
For calculation of d (diagonal of the square, we apply Pythagoras Theorem)
d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L
d = 4√2 units of length (we will assume meters, to work with MKS system of units)
B) Force of Q₁ exerts on charge Q₂
Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)
C) Force of Q₃ exerts on charge Q₂
Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)
D) Force of -Q₄ exerts on charge Q₂
Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)
E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)
Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and
Q₄ = -Q
Then the force is:
F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]
We should get its components
F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2
Note that this components have opposite direction than forces Fq₁₂ and
Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
In new conditions
Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and
Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]
Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively
Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx
and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny
And we get:
Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]
Fnx = (2,59/64 )* K*Q²
Fny has the same magnitude then
Fny =(2,59/64 )* K*Q²
The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces
