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3 years ago

14

At a certain point on a heated metal plate, the greatest rate of temperature increase, 3 degrees Celsius per meter, is toward th

e northeast. If an object at this point moves directly north, at what rate is the temperature increasing?

2 answers:

Katena32 [7]3 years ago

3 0

Explanation:

Formula to determine the rate of temperature increasing will be calculated as follows.

$D_{u}F = \Delta F.\vec{u}$

= $|\Delta F|.|j| cos \theta$

= $3.|1|.cos(\frac{\pi}{4})$

= $3 \times \frac{1}{\sqrt{2}}$

= $\frac{3 \sqrt{2}}{2}$

= 2.121

Thus, we can conclude that the rate at which the temperature increasing is 2.121.

LekaFEV [45]3 years ago

3 0

Answer:

Explanation:

The rate of temperature increase towards north east is 3 °C/m

So, the component of increase in temperature in north is the component of the increase in temperature towards north east

Desired component = 3 cos 45° = 2.12 °C/m

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Answer: a. three times as large as the initial value.

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is inversely proportional to the volume of the gas at constant temperature (isothermal).

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Explanation:

It is given that,

Radius of the front sprockets, $r_f=8.4\ cm=0.084\ m$

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(a) Linear speed of the front sprockets, $v_f=r_f\times \omega$

$v_f=0.084\times 12.3$

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$v_r=0.0491\times 12.3$

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$v_r=60.393\ cm/s$

(b) Let $a_r$ is the centripetal acceleration of the chain as it passes around the rear sprocket.

$a_r=\dfrac{v_r^2}{r_r}$

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3 years ago

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Answer;

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Explanation;

-Properties of elements within a period on the periodic table change in a predictable way from one side of the table to the other. A period in the periodic table is a horizontal row. All elements in a row have the same number of electron shells.

-Elements in the same period have the same number of electron shells; moving across a period, elements gain electrons and protons and become less metallic.This arrangement reflects the periodic recurrence of similar properties as the atomic number increases.

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Answer:

40.47m/s²

Explanation:

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Acceleration = change in velocity/time

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Converting 1020km/hr to m/s we have;

(1020×1000/3600)m/s

= 1,020,000/3,600

= 283.3m/s

Acceleration = 283.3/7

Acceleration = 40.47m/s²

Deceleration is therefore 40.47m/s²

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