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2 years ago

According to the graph, during which time interval are the particles in the air slowing down?

Physics

2 answers:

shusha [124]2 years ago

8 0

Answer b i just checked

labwork [276]2 years ago

5 0

Answer:

6 pm to 12 am (option B)

Explanation:

When particle movement slows down, the temperature gets colder (movement of particles produces heat)

So, by looking at when the temperature is decreasing, we can tell when the particles in the air are slowing down.

This downward / negative slope is during the 6 pm - 12 am time frame

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Which of the following statements about the

Komok [63]

Answer: the answer is A

Explanation: just did it

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3 years ago

I need help with this question how to solve it for Brass and Cooper

Ksenya-84 [330]

Take into account that density and relative density are given by:

$\begin{gathered} \text{density}=\text{ mass/volume} \\ \text{relative density = density/density of water} \end{gathered}$

Take into account that the volume associated to each of the given sustances in the table is determined by the Level Difference (because it is the change in the volume of the water of the recipient in which the substance is immersed).

The density of water in kg/m^3 is 1000 kg/m^3.

Due to the density must be given in kg/m^3, it is necessary to express the volumes of the table in m^3 and mass in kg, then, consider the following conversion factor:

1 m^3 = 1000000 ml

1 kg = 1000 g

Then, you obtain the following results:

Brass:

$\begin{gathered} 53.2g\cdot\frac{1kg}{1000g}=0.0532kg \\ 6ml\cdot\frac{1m^3}{1000000ml}=0.000006m^3 \\ \text{density}=\frac{0.0532kg}{0.000006m^3}\approx8866.67\frac{kg}{m^3} \\ \text{relative density=}\frac{(\frac{8866.66kg}{m^3})}{(1000\frac{kg}{m^3})}\approx8.87 \end{gathered}$

Cooper:

$\begin{gathered} 57.4g=0.0574kg \\ 6ml=0.000006m^3 \\ \text{density}=\frac{0.0574kg}{0.000006m^3}\approx9566.67\frac{kg}{m^3} \\ \text{relative density=}\frac{\frac{9566.67kg}{m^3}}{1000kg}=9.57 \end{gathered}$

3 0

1 year ago

A storm cloud has an electric charge of (2.1400x10^1) C near the top of the cloud and (2.99x10^1) C near the bottom of the cloud

blagie [28]

Answer:

hey

Explanation:

4 0

3 years ago

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0

otez555 [7]

The given question is incomplete. The complete question is as follows.

A box of oranges which weighs 83 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and the floor.

Explanation:

The given data is as follows.

$F_{1}$ = 20 N, $F_{2}$ = 25 N, a = -0.9 $m/s^{2}$