Two units for measuring magnetic flux density are:

A sewage lagoon that has a surface area of 10 ha and a depth of 1 m is receiving 8,640 m^3 /d of sewage containing 100 mg/L of b

Marysya12 [62]

Answer: Coefficient= 0.35 per day

Explanation:

To find the bio degradation reaction rate coefficient, we have

k= $\frac{(Cin)(Qin)-(Cout)(Qout)}{(Clagoon)V}$

Here, the C lagoon= 20 mg/L

Q in= Q out= 8640 m³/d

C in= 100 mg/L

C out= 20 mg/L

V= 10 ha* 1* 10

V= 10⁵ m³

So, k= $\frac{8640*100-8640*20}{20*10^5}$

k= 0.35 per day

6 0

3 years ago

The Stefan-Boltzmann law can be employed to estimate the rate of radiation of energy H from a surface, as in

Mazyrski [523]

Explanation:

A.

H = Aeσ^4

Using the stefan Boltzmann law

When we differentiate

dH/dT = 4AeσT³

dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³

= 8.4085

Exact error = 8.4085x20

= 168.17

H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴

= 1366.376watts

B.

Verifying values

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴

= 1542.468

H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴

= 1205.8104

Error = 1542.468-1205.8104/2

= 168.329

ΔT = 40

H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴

= 1735.05

H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴

= 1735.05-1059.83/2

= 675.22/2

= 337.61

5 0

3 years ago

An alternating current E(t) =120 sin(12t) has been running through a simple circuit for a long time. The circuit has an inductan

german

Answer:

Explanation:

we have given E(t)=120 sin(12t)

R=5 ohm

L=0.2 H

ω=12 ( from expression of E)

$X_L=0.2\times 12=2.4$ ohm

$X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm$

$Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}$

$Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}$

=5.021 ohm

so amplitude of current = $\frac{v}{z}=\frac{120}{5.021}=23.89$

4 0

3 years ago

3. (a) (5 points) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queue

Bezzdna [24]

Answer:

(N-1) × (L/2R) = (N-1)/2

Explanation:

let L is length of packet

R is rate

N is number of packets

then

first packet arrived with 0 delay

Second packet arrived at = L/R

Third packet arrived at = 2L/R

Nth packet arrived at = (n-1)L/R

Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R

Now

L / R = (1000) / (10^6 ) s = 1 ms

L/2R = 0.5 ms

average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2

the average queuing delay of a packet = 0 ( put N=1)

4 0

3 years ago

Determine the maximum mass of the crate so

Tpy6a [65]

Answer:

293 kg

Explanation:

Let's say the tension in each cable is Tb, Tc, and Td.

First, find the length of cable AD:

r = √(2² + 2² + 1²)

r = 3

Using similar triangles:

Tdx = 2/3 Td

Tdy = 2/3 Td

Tdz = 1/3 Td

Sum of the forces in the x direction:

∑F = ma

Tb − 2/3 Td = 0

Td = 3/2 Tb

Sum of the forces in the y direction:

∑F = ma

2/3 Td − Tc = 0

Td = 3/2 Tc

Sum of the forces in the z direction:

∑F = ma

1/3 Td − mg = 0

Td = 3mg

From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.

8600 N = 3mg

m = 8600 N / (3 × 9.8 m/s²)

m ≈ 292.5 kg

Rounded to three significant figures, the maximum mass of the crate is 293 kg.

7 0

3 years ago