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1 year ago

A reasonable estimate for the height of the walls in an ordinary american home is?

Physics

1 answer:

Olegator [25]1 year ago

8 0

A reasonable estimate for the height of the walls in an ordinary American home is 2.5m.

<h3>What is wall estimation?</h3>

Building gauge is of two sorts one is known as Rough Cost Estimate. This gauge won't depict the specific expense of the structure, however it can provide you with the rough expense worth of the structure that will help adequately in overseeing cash for the structure. Harsh Cost Estimate is directed in various ways for various sort of structures. The rough complete wall length is found in running meters in this framework and this all out length accumulated by the rate per running meter of the wall gives an actually solid cost. Assessment hypothesis is a part of insights that arrangements with assessing the upsides of boundaries in light of estimated experimental information that has an irregular part. The boundaries depict a fundamental actual setting so that their worth influences the circulation of the deliberate information.

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A vector is 14.4 m long and

MaRussiya [10]

Answer:

Explanation:

The x-component is found in the magnitude of the vector times the cosine of the angle.

$A_x=14.4cos133$ and, to 3 sig dig,

$A_x=-9.82m$

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3 years ago

The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?

kipiarov [429]

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e I = $\frac{Q}{t}$ ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

Q = I × t

= 1.5 × 2

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The amount of electrons that flow in the given time is 3.0 C.

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2 years ago

How does the scientific process generally begin?

Murljashka [212]

There are several approaches. The most favourable one (in my opinion) is this one:
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3. Creating a hypothesis (NOT a thesis!)
4. Experimenting (to prove the hypothesis)
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5 0

3 years ago

Read 2 more answers

A piston-cylinder device contains Helium gas initially at 150 kPa, 20 o C, and 0.5m 3 . The helium is now compressed in a polytr

Molodets [167]

Answer:

Explanation:

Given

$P_1=150 kPa$

$T_1=20^{\circ}C$

$V_1=0.5 m^3$

$T_2=140^{\circ}C$

$P_2=400 kPa$

R for Helium $R=2.076$

$c_v=3.115 kJ/kg-K$

mass of gas $m=\frac{P_1V_1}{RT_1}$

$m=\frac{150\times 0.5}{2.076\times 293}$

$m=0.123 kg$

Similarly $V_2$ can be found

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=0.264 m^3$

Work done $W=\int_{V_1}^{V_2}PdV$

$W=\frac{P_2V_2-P_1V_1}{n-1}$

$W=\frac{mR(T_2_T_1)}{n-1}$

Since it is a polytropic Process

therefore $PV^n=c$

$P_1V_1^n=P_2V_2^n$

$(\frac{V_1}{V_2})^n=\frac{P_2}{P_1}$

$(\frac{0.5}{0.264})^n=\frac{400}{150}$

$n=\frac{\ln 2.66}{\ln 1.893}$

$n=1.533$

$W=\frac{0.123\times 2.076(140-20)}{1.533-1}$

$W=57.48 kJ$

From Energy balance

$E_{in}-E_{out}=\Delta E_{system}$

Neglecting kinetic and Potential Energy change

$Q_{in}+W_{in}=change\ in\ Internal\ Energy$

Change in Internal Energy $\Delta U=u_2-u_1$

$\Delta U=mc_v(T_2-T_1)$

$\Delta U=0.123\times 3.115(140-20)$

$\Delta U=45.977 kJ$

$Q_{in}+57.48=45.977$

$Q_{in}=-11.50 kJ$

i.e. Heat is being removed

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3 years ago

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2 years ago