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3 years ago

Raw sewage is raised vertically by 5.49m in the amountof

Physics

1 answer:

uranmaximum [27]3 years ago

5 0

Answer:

P = 1235.7646 W

Explanation:

Given data:

height of raised sewage = 5.49 m

rate of sewage =1.89*10^6 lt/day

density of sewage = 1.050 kg/m^3

power is written as

$P = \frac{work}{time}$

work = m g h

$m - mass = ( \rho V)$

h -height

mass lifted per day $= ( 1.89*10^6)(1.050)$

= 1984500 kg

time = 24 hours* 3600 seconds per hour

power $P = \frac{( 1984500)(9.8)( 5.49)}{(24)(3600)}$

P = 1235.7646 W

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Answer: 294J

Explanation:

$U_g=mg(y_f-y_i)$

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A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe

Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

= surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ / 2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀

= 6.78 X 10³ N/C

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3 years ago

Which is an example of a chemical change?

yulyashka [42]

It’s burning wax. All of the other options are physical changes .

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3 years ago

What happens to the force of gravity between two masses if one mass is decreased?

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<span>B. It stays the same</span>

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3 years ago

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A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibratio

Nikitich [7]

Answer:

If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

Explanation:

The motion of the ceiling is y = Y sinωt

y = 0.05 sin (2 π × 2) t

y = 0.05 sin 4 π t

K = 25 lb/ft × 4 sorings

K = 100 lb/ft

Amplitude of the microscope $\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]$

where;

$\epsilon = 0$

$W_n = \sqrt { \frac{k}{m}}$

= $\sqrt { \frac{100*32.2}{200}}$

= 4.0124

replacing them into the above equation and making X the subject of the formula:

$X =$ $Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}$

$X =$ $0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}$

$X =$ $5.676*10^{-3} \ mm$

Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = $5.676*10^{-3} \ mm$

8 0

3 years ago