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3 years ago

13

On a hot summer day, you turn the thermostat in your house way down. Assume that your house is well sealed and that no air enter

s or leaves the house. When the air temperature in your house falls a short while later, what statement is correct regarding the air pressure?

1 answer:

Debora [2.8K]3 years ago

4 0

Answer: The temperature in the house will equally drop

Explanation: Temperature -pressure relationship, since the relationship between temperature and pressure is directly proportional when the air temperature in the house fall short the air pressure also fall short, seeing also that the house is a controlled environment, since it is well sealed that no air enters and leaves

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Answer:

The answer is D. density.

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How is NASA a catalyst to make the crew work better?

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Answer:

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3 years ago

What is the number of electrons that move past a point in a wire carrying 500 A of current in 4.0 minutes

mr Goodwill [35]

The current is defined as the amount of charge Q that passes through a given point of a wire in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Since I=500 A and the time interval is
$\Delta t=4.0 min=240 s$
the charge is
$Q=I \Delta t=(500 A)(240 s)=1.2 \cdot 10^5 C$

One electron has a charge of $q=1.6 \cdot 10^{-19}C$ , therefore the number of electrons that pass a point in the wire during 4 minutes is
$N= \frac{Q}{q}= \frac{1.2 \cdot 10^5 C}{1.6 \cdot 10^{-19}C}=7.5 \cdot 10^{23}$ electrons

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3 years ago

A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform

madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is $\epsilon_{max}= 26.8 V$

The emf induced at t = 1.00 s is $\epsilon = 24.1V$

The maximum rate of change of magnetic flux is $\frac{d \o}{dt}|_{max} =26.8V$

Explanation:

From the question we are told that

The number of turns is N = 44 turns

The length of the coil is $l = 15.0 cm = \frac{15}{100} = 0.15m$

The width of the coil is $w = 8.50 cm =\frac{8.50}{100} =0.085 m$

The magnetic field is $B = 745 \ mT$

The angular speed is $w = 64.0 rad/s$

Generally the induced emf is mathematically represented as

$\epsilon = \epsilon_{max} sin (wt)$

Where $\epsilon_{max}$ is the maximum induced emf and this is mathematically represented as

$\epsilon_{max} = N\ B\ A\ w$

Where $\o$ is the magnetic flux

N is the number of turns

A is the area of the coil which is mathematically evaluated as

$A = l *w$

Substituting values

$A = 0.15 * 0.085$

$= 0.01275m^2$

substituting values into the equation for maximum induced emf

$\epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0$

$\epsilon_{max}= 26.8 V$

given that the time t = 1.0sec

substituting values into the equation for induced emf $\epsilon = \epsilon_{max} sin (wt)$

$\epsilon = 26.8 sin (64 * 1)$

$\epsilon = 24.1V$

The maximum induced emf can also be represented mathematically as

$\epsilon_{max} = \frac{d \o}{dt}|_{max}$

Where $\o$ is the magnetic flux and $\frac{d \o}{dt}|_{max}$ is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

$\frac{d \o}{dt}|_{max} =26.8V$

8 0

3 years ago

calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2

igor_vitrenko [27]

First, balance the reaction:

_ KClO₃ ==> _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃ ==> 2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃ ==> 2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0

3 years ago