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3 years ago

In which situation will the lowest resistance occur?

Physics

1 answer:

MrMuchimi3 years ago

7 0

C. Thick wire and cold temperature.

Explanation:

The resistance of a wire is given by: R = (ρL)/A

where ρ is the resistivity of the material, L is the length of the wire, A is the cross-sectional area of the wire.

From the formula, we see that the thicker the wire, the larger A, therefore the smaller the resistivity. so, a thick wire will have lower resistivity.

Moreover, the resistance of a wire increases with the temperature. In fact, high temperatures mean more motion of the atoms/electrons inside the wire, so more resistance to the flow of current through it. Therefore, colder temperature means lower resistance.

So, the correct option is thick wire and cold temperature.

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An electric motor does 900j of work for 8 hours. Calculate the power used

nataly862011 [7]

Answer:

0.031 W

Explanation:

The power used is equal to the rate of work done:

$P=\frac{W}{t}$

where

P is the power

W is the work done

t is the time taken to do the work W

In this problem, we have:

W = 900 J is the work done by the motor

t = 8 h is the time taken

We have to convert the time into SI units; keeping in mind that

1 hour = 3600 s

We have

$t=8\cdot 3600 =28,800 s$

And therefore, the power used is

$W=\frac{900}{28800}=0.031 W$

6 0

3 years ago

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

molecular mass of water, $M_w=2+16=18\ g.mol^{-1}$

∵Density of water is $0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}$

∴mass of water in 316 mL, $m_w=316\times 0.992=313.5 g$

mass of glycerin, $m_g=154\ g$

pressure of mixture, $P_x=55.32\ torr= 55.32\ mm.Hg$

temperature of mixture, $T_x=40^{\circ}C$

Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.

moles of water in the given quantity:

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

moles of glycerin in the given quantity:

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

Now the mole fraction of water:

$X_w=\frac{n_w}{n_w+n_g}$

$X_w=\frac{17.42}{17.42+1.674}$

$X_w=0.912$

Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.

$\therefore P_{sol}=X_w\times P_x$

$\therefore P_{sol}=0.912\times 55.32$

$P_{sol}=50.4\ mm.Hg$

7 0

3 years ago

Please help with these questions

lord [1]

When do you gotta turn it in?

8 0

2 years ago

A radio frequency identification application would most likely interface with a (an):_________

Irina-Kira [14]

A radio frequency identification application would most likely interface with an Operational Data Store.

The Operations Data Store (ODS) is a central database that provides the latest data snapshots from multiple transaction systems for operational reporting.

It allows organizations to combine data in its original format from various sources into a single destination to provide business reporting.

ODS contains integrated updates from operational sources and supports business intelligence (BI) tools to facilitate tactical decision making.

For example, an administrator can configure ODS to pull weekly batches of data from a billing application that is rarely updated, importing individual transaction records as they occur in the sales database(thanks to these database triggers), then combine the two into new relational tables.

As a result, querying and reporting on operational data in ODS comes with the assurance that these integrated tables contain the latest and most relevant snapshots of the business.

Learn more about Operations Data Store here : brainly.com/question/14925154

#SPJ4

4 0

2 years ago

Jason is taking a physical exam and has to do push ups for sixty seconds. He does 30 pushups in the sixty seconds. Each push up

Alexandra [31]

Answer: 2 seconds is the unit rate.

Explanation:

We know that Jason does 30 pushups in 60 seconds at a constant rate, and we know that each push up takes 2 seconds.

then, we can write this as 30 pushups/60 seconds = (1/2) pushups per second.

Here, two seconds represents the time needed to do one pushup, is the unit rate (this means that we need 2 seconds to have a unit "one pushup")

8 0

3 years ago