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Alina [70]
3 years ago
10

A 0.5 m3 container is filled with a fluid whose specific volume is 0.001 m3/kg. At standard gravitational acceleration, the cont

ents of this container weigh:
Physics
1 answer:
xenn [34]3 years ago
5 0

Answer: the contents of this container weighs 4905 kg.m/s²

Explanation:

Given that;

volume of a container V = 0.5 m³

we know that standard gravitational acceleration g = 9.81 m/s²

specific volume of liquid filled in the container v = 0.001 m³/kg

now we express the equation for weight of the container.

W = mg

W = (pV)g

W = Vg / ν

so we substitute

W =  (0.5 m³)(9.81 m/s ) / 0.001 m³/kg

W = 4.905 / 0.001

W = 4905 kg.m/s²

Therefore, the contents of this container weighs 4905 kg.m/s²

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The Heaviside function H is defined by H(t)={0 if t<0, 1 if t≥0 It is used in the study of electric circuits to represent the
Studentka2010 [4]

Answer:

V(t)= 240V* H(t-5)

Explanation:

The heaviside function is defined as:

H(t) =1 \quad t\geq 0\\H(t) =0 \quad t

so we see that the Heaviside function "switches on" whent=0, and remains switched on when t>0

If we want our heaviside function to switch on when t=5, we need the argument to the heaviside function to be 0 when t=5

Thus we define a function f:

f(t) = H(t-5)

The -5 term inside the heaviside function makes sure to displace the function 5 units to the right.

Now we just need to add a scale up factor of 240 V, because thats the voltage applied after the heaviside function switches on. (H(t-5) =1 when t\geq 5, so it becomes just a 1, which we can safely ignore.)

Therefore our final result is:

V(t)= 240V* H(t-5)

I have made a sketch for you, and added it as attachment.  

5 0
3 years ago
How is most of the electricity we use at home generated?
Sveta_85 [38]
Nuclear power plants, wind farms, water farms, and geothermal heating
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3 years ago
Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average
MissTica

Answer:

reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

Explanation:

given data

electricity consume = 14000 kWh

fuel consume = 900 gal

CO₂ produced of fuel = 26.4 lbm/gal

CO₂ produced of electricity  = 1.54 lbm/kWh

oil and electricity usage = 21 percent

to find out

the reduction in the amount of CO₂ emissions

solution

we calculate the amount of CO₂ produce here that is

amount of CO₂ produce = ( electricity consume×CO₂ produce electricity + fuel consume × CO₂ consume fuel )    ........................1

put here value

amount of CO₂ produce = ( 14000 × 1.54 + 900 × 26.4 )  

amount of CO₂ produce = 45320 lbm/yr

we know reduction is 21%

so

reduction in amount of CO₂ produced is

reduction in CO₂ produced = 45320 × 21%

reduction in CO₂ produced = 9517.2 lbm per year

so reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

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2 years ago
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