Answer:
- 0.86 C
Explanation:
Let the temperature of cold reservoir is T2.
T1 = 22.4 C = 295.4 K, B = 11.7
By the formula of coefficient of performance of heat pump
B = T2 / (T1 - T2)
11.7 = T2 / (295.4 - T2)
11.7 × 295.4 - 11.7 T2 = T2
T2 = 272.14 K
T2 = - 0.86 C
Answer:
–625 J
Explanation:
So, we got this formula for the work
W=mgd(Cosθ)
but remember when it's liftin somethin, its work gon be against the work of gravity, so
Cos180°= –1
W=500×1.25×(–1)
W= –625 J
The answer would be B.
(V=IR, 12-10i, 12/10=1.2)
Answer:
a) Ф = 0.016 N / C m
, b) q_{int} = 0.14 10⁻¹² C
Explanation:
a) For this problem we use Gauss's law
Ф = E .ds = /ε₀
The camp is in the x direction so it has no flow through the cylinder walls.
Ф = E A
The area of a circle is
A = π r
Ф = E π r
Ф = (x- 3.6) r
Let's calculate
Ф = (3.7 -3.6) 0.16
Ф = 0.016 N / C m
b) we clear from Gauss's law
q_{int} = Ф ε₀
Where the flow is on both sides, on the face at x = 0 the flow is zero
q_{int} = 0.016 8.85 10⁻¹²
q_{int} = 0.14 10⁻¹² C
I believe it is D a magnetic field