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3 years ago

Please I need help with this question (see image).Show workings if necessary

Physics

1 answer:

timofeeve [1]3 years ago

5 0

<h2><u>Question</u> :</h2>

When an athlete perspires after running

Evaporation occurs and helps to cool the body.
Convection cools the body.
The body absorbs cold from the surrounding air.
Heat is conducted away from the body.

<h2><u>Answer</u> :</h2>

When an athlete perspires after running evaporation occurs and helps to cool the body.

Correct option is A.

<h2><u>Explanation</u> :</h2>

When an athlete perspires after running then sweat evaporates and we know that evaporation causes cooling as evaporation requires heat energy. The required heat energy is taken away by the molecules of sweat when they convert from liquid into gas, and this causes cooling on the surface of body. That's why it helps to cool the body.

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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

$W = U_f - U_i$

we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

7 0

3 years ago

The noise level coming from a pig pen with

sweet [91]

Answer:

The decibel of the remaining pigs is 51.5 dB.

Explanation:

Decibel (dB) is a unit of measure of the intensity of a given sound.

Number of pigs = 199, noise level = 74.3 dB.

Given that the intensity (I) of the sound from the pen is proportional to the number of pigs (N), thus:

I $\alpha$ N

I = kN

where k is the constant of proportionality.

⇒ k = $\frac{I}{N}$

= $\frac{74.3}{199}$

k = 0.3734

When 61 numbers of pigs were removed, the number of remaining pigs (N) squealing at their original level is 138.

Thus, the becibel level (I) of the remaining pigs can be determined by:

I = kN

= 0.3734 × 138

= 51.53 dB

The becibel level (I) of the remaining pigs is 51.53 dB.

6 0

4 years ago

6. Draw conclusions: Newton’s first law states that an object in motion will travel at a constant velocity unless acted upon by

gavmur [86]

You can test if it’s true by holding a pencil in mid air over a table and the table is supposed to be the unbalanced forced that stopped the pencil from moving at the constant velocity it was going by.

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3 years ago

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An electric grinder uses a grinding wheel

luda_lava [24]

(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s²

<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>

<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>

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3 years ago

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Please I need help with this question (see image).Show workings if necessary​

Please I need help with this question (see image).Show workings if necessary