How many sandwiches were made in the hour?

The temperature of an air parcel and the kinetic energy of an air parcel are ___________ related. this means that as the tempera

torisob [31]

The temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.

<h3>What is temperature?</h3>

Temperature is the measure of degree of hotness or coldness of a body.

Temperature is also the measure of the average kinetic energy of a system.

When the heat is applied to body, its temperature increases as the body gains heat.

Thus, the temperature of an air parcel and the kinetic energy of an air parcel are directly related. this means that as the temperature of the air parcel increases, the kinetic energy increases.

Learn more about temperature here: brainly.com/question/25677592

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8 0

1 year ago

The temperature of a room is 77°F what would be the temperature in Celsius scale

anygoal [31]

Formulas change from F to degree C : C = ( F - 32 )/1.8

so we have (77-32)/1.8 = 25 oC

ok done. Thank to me :>

5 0

2 years ago

Compare and contrast instantaneous and average speed.

Harlamova29_29 [7]

Answer:

instantaneous velocity is a velocity covered at an instant while average velocity is the change in distance/ the change in time taken

6 0

2 years ago

A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

3 years ago

Read 2 more answers

100 POINTS! I will mark brainliest! Record your hypothesis as an “if, then” statement for the rate of dissolving the compounds:

love history [14]

Answer:

<u><em>Rate of dissolving compounds:</em></u>

If we increase the temperature of the solution, then the dissolving compound would dissolve more easily.

<u><em>Boiling Point of Compounds:</em></u>

If the inter-molecular forces of any compound is really strong, then the boiling point of the compound would be really high.

6 0

3 years ago