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3 years ago

A: Transformers work with either AC or DC electricity. True/False

Physics

1 answer:

lapo4ka [179]3 years ago

5 0

The answers are as follows:

A. False – Transformers are designed to work on AC

B. True – This is to compensate the losses as the length of wire increases

C. Transformer – the transformer works in the principle of electromagnetic induction

D. Superconducting Magnet – magnet made from coils of a superconducting wire

E. Relay – This electrical switch is operated by a magnet

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An electron in the first energy level of the electron cloud has an electron in the third energy level

lukranit [14]

Answer:

a lower energy than

Explanation:

3 0

4 years ago

Read 2 more answers

Agri-Chem's contract with Enerco specified a maximum of 90,000 cu.ft. X 103 per day for its complexes. However, curtailments are

zvonat [6]

Answer:

Caustic soda would be least affected by a gas curtailment

Explanation:

Let,

X1= ammonia

X2= ammonium phosphate

X3= ammonium nitrate

X4 = Urea

X5= hydro-fluoric acid

X6= Chlorine

X7= Caustic soda

X8= Vinyl chloride monomer.

Agri chem’s current natural gas usage

= (1200 × 8 + 540 × 10 + 490 × 12 + …)

= 85,680,000 cu. ft. per day

When, the curtailment is 20%, availability is

= 0.8 X 85,680

= 68,554,000 cu. ft. per day

Therefore, the gas constant

= 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8 ≤68,544

When, the curtailment is 40% availability is

= 0.6 X 85,680

= 51,408,000 cu. ft. per day

Constant = 8X1 + 10X2 + 12X3 + 12X4 + 7X5 + 18X6 + 20X7 + 14X8≤ 51,408

7 0

3 years ago

Please help Math Phys

Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land. Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0

3 years ago

Un ciclista recorre 10 km en 2 h.Calcula su velocidad media en km/h.¿cuantos metros recorre cada segundo

lys-0071 [83]

Answer:

La velocidad media es 5 $\frac{km}{h}$ , que equivale a 1.389 $\frac{m}{s}$

Explanation:

La velocidad es una magnitud física que expresa la relación entre el espacio recorrido por un objeto y el tiempo empleado para ello.

La velocidad media relaciona el cambio de la posición con el tiempo empleado en efectuar dicho cambio. Por lo que se calcula como la distancia recorrida por un objeto dividido por el tiempo transcurrido:

$velocidad=\frac{distancia}{tiempo}$

En este caso:

distancia= 10 km= 10,000 m (siendo 1 km= 1,000 m)
tiempo= 2 h= 7,200 s (siendo 1 h= 3,600 s)

Entonces, reemplazando en la definición de velocidad media:

$velocidad=\frac{10 km}{2 h}= \frac{10,000 m}{7,200 s}$

Resolviendo se obtiene:

$velocidad=5 \frac{km}{h}= 1.389\frac{m}{s}$

La velocidad media es 5 $\frac{km}{h}$ , que equivale a 1.389 $\frac{m}{s}$

4 0

3 years ago

For a particular casting setup, the top of the sprue has a diameter of 0.030 m, and its length is 0.200 m. The volume flow rate

faust18 [17]

Answer with Explanation:

We are given that

Diameter=0.030 m

Length of sprue= $h_1$ =0.200 m

Metal volume flow rate,Q=0.03 $m^3/min$

Q= $\frac{0.03}{60}=5\times 10^{-4}m^3/s$ because 1 minute=60 seconds

Let 1 for the top and 2 for the bottom

$d_=0.030 m$

$h_2=0$

$A_1=\frac{\pi d^2}{4}=\frac{3.14\times (0.030)^2}{4}$

$A_1=7.065\times 10^{-4} m^2$

$v_1=\frac{Q}{A_1}=\frac{5\times 10^{-4}}{7.065\times 10^{-4}}$

$v_1=0.708 m/s$

Pressure at the top and bottom of the sprue is atmospheric

$h_1+\frac{v^2_1}{2g}=h_2+\frac{v^2_2}{2g}$

Substitute the values

$0.2+\frac{(0.708)^2}{2\cdot 9.8}=0+\frac[v^2_2}{2\cdot 9.8}$

$v^2_2=2\cdot 9.8\cdot \frac{0.2\cdot 9.8\cdot 2+0.501264}{2\cdot 9.8}=4.421264$

$v_2=\sqrt{4.421264}=2.1 m/s$

$Q=A_2v_2$

$5\times 10^{-4}=A_2\times 2.1$

$A_2=\frac{5\times 10^{-4}}{2.1}=2.381\times 10^{-4} m^2$

Reynolds number= $\frac{v_2D\rho}{\eta}$

$\eta=0.004 N.s/m^2$

$\rho=2700 kg/m^3$

Substitute the values then we get

Reynolds number= $\frac{2.1\times 0.03\times 2700}{0.004}$

Reynolds number=42525

The Reynolds number is greater than 4000 .Therefore, the flow is turbulent.

8 0

3 years ago