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3 years ago

Describe where metals and nonmetals are found on the periodic table

Physics

1 answer:

Studentka2010 [4]3 years ago

4 0

Non metals are all the way on the right side

Metalloids are the stair case

Metals are on the left side

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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

2 years ago

Read 2 more answers

On the decibel scale, an increase of 10 decibels means an intensity increase of __________times?

Sonbull [250]

Answer:

Explanation:

The decibel or decibel, is a unit that relates two values of sound pressure, and electrical power. the base unit is the bel, but given the amplitude but for practicality, a submultiple, the decibel, is used. It is a scalar expression .

Zero belios is the reference value used as a base, hence a bel is equivalent to 10 decibels and means that there is a power increase of 10 times over the reference value.

3 0

2 years ago

Read 2 more answers

In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 3.40 mm . what is the differenc

Strike441 [17]

The solution would be like this for this specific problem:

Given:

diffraction grating slits = 900 slits per centimeter

interference pattern that is observed on a screen from the grating = 2.38m

maxima for two different wavelengths = 3.40mm

slit separation .. d = 1/900cm = 1.11^-3cm = 1.111^-5 m 

Whenas n = 1, maxima (grating equation) sinθ = λ/d
Grant distance of each maxima from centre = y ..
As sinθ ≈ y/D y/D = λ/d λ = yd / D 

∆λ = (λ2 - λ1) = y2.d/D - y1.d/D
∆λ = (d/D) [y2 -y1]

∆λ = 1.111^-5m x [3.40^-3m] / 2.38m .. .. ►∆λ = 1.587^-8 m

6 0

3 years ago

An athlete is working out in the weight room. He steadily holds 50 kilograms above his head for 10 seconds. Which statement is t

postnew [5]

Answer:

A. The athlete isn’t doing any work because he doesn’t move the weight.

Explanation:

We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.

So when he's holding the weight, he doesn't do any work.

3 0

2 years ago

ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly

Igoryamba

The period T of a pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$ . Using this data, we can solve the previous formula to find L:
$L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm$

4 0

2 years ago