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3 years ago

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What's your normal body temperature? it may not be 98.6°f, the oft-quoted average that was determined in the nineteenth century.

a more recent study has reported an average temperature of 98.2°f. what is the difference between these averages, expressed in celsius degrees?

3 answers:

harkovskaia [24]3 years ago

6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

Igoryamba3 years ago

6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC

givi [52]3 years ago

6 0

(98.6 - 98.2)ºF = .4ºF

.4ºF* 5ºC/9ºF = 0.222 ºC OR

98.6*5ºC/9ºF = 54.80 98.25ºC/9ºF = 54.60 54.80 - 54.60 = .2ºC

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Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

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2 years ago

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Tema [17]

Answer:

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Explanation:

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

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Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

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3 years ago

An airplanes lands with a velocity of +75 m/s and comes to a stop in 20 seconds. What is the acceleration of the airplane while

erma4kov [3.2K]

Answer:

The acceleration of the car, a = -3.75 m/s²

Explanation:

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Using the I equations of motion

v = u + at

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3 years ago