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3 years ago

10

A rock is dropped straight down from roof. It hits the ground with a velocity of 28.2 m/sec. How high was the roof, and how long

did it take to hit the ground?

1 answer:

notsponge [240]3 years ago

6 0

Answer:

18 secs

Explanation:

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Why does a person feel cooler under a rotating fan?

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The beat your doctor listens to through a sethoscope is the sound of the four values opening

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and closing .

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3 years ago

Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

antoniya [11.8K]

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

$\dfrac{dT}{dt}=c(T-T_{0})$

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Where, $dT=T_{1}-T_{2}$

Here, $T =\dfrac{T_{1}+T_{2}}{2}$

$T_{0}$ = 25°C (surrounding temperature)

dt = 1 minute

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Put the value into the formula

$\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)$

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3 years ago

Which of the following is not a petroleum product

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D. wool
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3 years ago

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

Lostsunrise [7]

Answer:

$0.06\Omega/m$

Explanation:

Firstly, when you measure the voltage across the battery, you get the emf,

E = 13.0 V

In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.

Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire ( $I=\frac{V}{R}$ ). But this is not the case.

Let the resistance of the ammeter be r

Hence, using Ohm's law we get the following 2 equations:

$\frac{13}{20z+r} =7.6$ .......(1)

$\frac{13}{40z+r} =4.5$ ......(2)

Substituting the value of r from (2) in (1), we have,

$13=152z+7.6\times\frac{13-180z}{4.5}$

which simplifying gives us, $z=0.0589\Omega/m\approx0.06\Omega/m$ (which is our required solution)

putting the value of z in either (1) or (2) gives us, r = 0.5325 $\Omega$

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3 years ago