7. The "3 second rule" is the time you should pause at an intersection marked with a stop sign.

Atmospheric pressure is measured to be 14.769 psia. a. What would be the equivalent reading of a water barometer (inches of H20)

Fofino [41]

Answer:

(a) water height =408.66 in.

(b) mercury height=30.04 in.

Explanation:

Given: P=14.769 psi ( 1 psi= 6894.76 $\frac{N}{m^2}$ )

we know that $P=\rho\times g\timesh$

where $\rho =Density,g=9.81\frac{m}{s^2}$

h=height.

Given that P=14.769 psi ⇒P= 101828.6 7 $\dfrac{N}{m^2}$

(a) $P=\rho_{w}\times g\times h_{w}$

$\rho_{w}=1000\frac{Kg}{m^3}$

⇒101828.67= $1000\times 9.81\times h_{w}$

$h_{w}$ =10.38 m

So water barometer will read 408.66 in. (1 m=39.37 in)

(b) $P=\rho_{hg}\times g\times h_{hg}$

$\rho_{hg}$ =13600

So 101828.67= $13600\times 9.81\times h_{hg}$

$h_{hg}$ =0.763 m

So mercury barometer will read 30.04 in.

6 0

3 years ago

Explicar el funcionamiento de un multímetro analógico.

Whitepunk [10]

Answer:

Un multímetro analógico funciona como un medidor de bobina móvil de imán permanente (PMMC) para tomar mediciones eléctricas

Explanation:

El multímetro analógico es un medidor o galvanómetro D'Arsonval que funciona según el principio de los medidores de bobina móvil de imán permanente (PMMC)

Un multímetro analógico está formado por un puntero de aguja unido a una bobina móvil colocada entre el polo norte y sur de un imán permanente dispuesto de tal manera que, cuando una corriente eléctrica fluye a través de la bobina, genera una fuerza de campo magnético que interactúa con el imán fuerza de campo de los imanes permanentes que hace que la bobina se mueva junto con el puntero de la aguja sobre un dial graduado

Para controlar el movimiento del puntero de la aguja, de modo que el par requerido para producir una cantidad de movimiento por corriente detectada por el multímetro, se colocan dos resortes a través de la bobina para proporcionar resistencia al movimiento en ambas direcciones y para permitir la calibración del multímetro analógico.

4 0

3 years ago

Water flows through a converging pipe at a mass flow rate of 25 kg/s. If the inside diameter of the pipes sections are 7.0 cm an

ser-zykov [4K]

Answer:

volumetric flow rate = $0.0251 m^3/s$

Velocity in pipe section 1 = $6.513m/s$

velocity in pipe section 2 = 12.79 m/s

Explanation:

We can obtain the volume flow rate from the mass flow rate by utilizing the fact that the fluid has the same density when measuring the mass flow rate and the volumetric flow rates.

The density of water is = 997 kg/m³

density = mass/ volume

since we are given the mass, therefore, the volume will be mass/density

25/997 = $0.0251 m^3/s$

volumetric flow rate = $0.0251 m^3/s$

Average velocity calculations:

Pipe section A:

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.07^2 = 3.85\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 3.85\times10^{-3}) = 6.513m/s$

Pipe section B:

cross-sectional area =

$\pi \times d^2\\=\pi \times 0.05^2= 1.96\times10^{-3}m^2$

mass flow rate = density X cross-sectional area X velocity

velocity = mass flow rate /(density X cross-sectional area)

$velocity = 25/(997 \times 1.96\times10^{-3}) = 12.79m/s$

7 0

2 years ago

True or False; If I was trying to find the Voltage of my computer, and I was given the Watts and Amps it uses, I would use Watt'

defon

Answer:

true

Explanation:

8 0

2 years ago

An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,

blsea [12.9K]

Answer:

the % recovery of aluminum product is 80.5%

the % purity of the aluminum product is 54.7%

Explanation:

feed rate to separator = 2500 kg/hr

in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine

of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.

After the separation, 256 kg is collected in the product stream.

of this 256 kg, 140 kg is aluminium.

% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material

% recovery of aluminium = 140kg/174kg x 100% = 80.5%

% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream

% purity of the aluminium product = 140kg/256kg

x 100% = 54.7%

8 0

2 years ago