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2 years ago

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Mike understands the formula for temperature conversion, but is having difficulty imagining what various temperatures feel like.

which of these would be a good example to tell mike what 37°c feels like? a. room temperature b. a hot pan c. body temperature d. a chilly autumn evening

1 answer:

Alisiya [41]2 years ago

3 0

A good example that will help to to tell mike what 37°c feels like is body temperature.

<h3>What is room temprrature?</h3>

The average room temperature is typically around 20°C, or 68 degrees Fahrenheit.

<h3>What is body temperature?</h3>

The average body temperature is 98.6 Fahrenheit or 37 degrees Celsius.

Thus, a good example that will help to to tell mike what 37°c feels like is body temperature.

Learn more about body temperature here: brainly.com/question/5295345

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Objects with unlike charges what each other

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The answer is attract. Hope it helps! :)

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3 years ago

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nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid

loris [4]

Answer:

For Xenon fluoride, the average bond energy is 132kj/mol

For tetraflouride,the average bond energy is 150.5kj/mol.

For hexaflouride, the average bond energy is 146.5 kj/mol

Explanation:

For xenon fluoride

105/2 = 52.5

For F-F

159/2 = 79.5

Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole

For tetraflouride

284/4 = 71

For F-F

159/2 = 79.5

Average bond energy = 79.5 + 71 = 150.5kj/mol

For hexaflouride

402/6 = 67

F-F = 159/2 = 79.5

Average bond energy = 67 + 79.5 = 146.5kj/ mol

3 0

3 years ago

A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

goldfiish [28.3K]

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

$r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}$

$r_{cm} = {(37.26, 0)}{14.8}$

$r_{cm} = (2.52 cm, 0)$

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

$r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}$

here we have

$m_1 = 10.2 kg$

$m_2 = 4.6 kg$

$r_1 = (0, 0)$

$r_2 = (8.1cm, 0)$

now we have

$r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}$

$r_g_{cm} = {(37.26, 0)}{14.8}$

$r_g_{cm} = (2.52 cm, 0)$

So center of mass is same as center of gravity because value of gravity is constant here

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3 years ago

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the

Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

$x = v_0 t \frac{1}{2} at^2$

Where,

x= Displacement

$v_0$ = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

$L = \frac{1}{2} a t_1 ^2$

For the second cart

$2L \frac{1}{2} at_2^2$

When the tenth car is aligned the length will be 9 times the initial therefore:

$9L = \frac{1}{2} at_3^2$

When the tenth car has passed the length will be 10 times the initial therefore:

$10L = \frac{1}{2}at_4^2$

The difference in time taken from the second car to pass it is 5 seconds, therefore:

$t_2-t_1 = 5s$

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

$\frac{1}{2} = (\frac{t_1}{t_2})^2$

$t_1 = \frac{t_2}{\sqrt{2}}$

From the relationship when the car has passed and the time difference we will have to:

$(t_2-\frac{t_2}{\sqrt{2}}) = 5$

$t_2 (\sqrt{2}-1) = 3\sqrt{2}$

$t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Replacing the value found in the equation given for the second car equation we have to:

$\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2$

Finally we will have the time when the cars are aligned is

$18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2$

$t_3 = 36.213s$

The time when you have passed it would be:

$20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2$

$t_4 = 38.172$

The difference between the two times would be:

$t_4-t_3 = 38.172-36.213 \approx 2s$

Therefore the correct answer is C.

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3 years ago

If you add two velocity vectors going in opposite directions, will the resultant vector be smaller or larger than the vectors yo

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3 years ago

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