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3 years ago

Can some one help me ;-;

Physics

1 answer:

enyata [817]3 years ago

6 0

Answer: The first answer for the first problem, and the 2nd answer for the second problem

Explanation: For the first one, if it is absolute zero, the molecules would not move at all.

For the second one, the temperature of the sample will increase due to the movement.

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Some substances have the same chemical composition, but their atoms are arranged differently. A classic example is carbon. A pen

Bumek [7]

No, the density of diamond and graphite would not be the same

Explanation:

What is density?

Density is the amount of substance per unit volume. It is simply mass divided by volume.

Density is greatly influenced by the way substances are packed.

Substances that are well packed will have lower volume for the same amount of matter than another that is poorly packed.

The carbon atoms in graphite are poorly packed. They are arranged layers upon layers.
Diamond carbon atoms have a cross-linked networked pattern. They are well packed.
For the same mass of matter, graphite will take up more space than diamond.

Since:

Density = $\frac{mass}{volume}$

The one that has a lesser volume will have a higher density.

Therefore diamond will have a higher density.

learn more:

Density brainly.com/question/5055270

#learnwithBrainly

4 0

3 years ago

Read 2 more answers

Cassidy Carver's family child care center has an activities table with four chairs an easel that can accommodate a child on each

Akimi4 [234]

Answer:4

Explanation:

6 0

3 years ago

A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police

Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. $x=x_{0}+vt$

2. $x=x_{0}+v_{0}t+\frac{at^{2}}{2}$

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

$vt = v_{0}t+\frac{at^2}{2}\\\\ 16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}$

We simplify the fraction present and rearrange for our formula so that it equals 0:

$0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0$

In the very last step we factored a common factor t. There is two possible solutions to the equation at $t=0$ and:

$0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\ 0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s$

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at $t=0 s$ (when the SUV passed the police car) and $t=15.56s$ (when the police car catches up to the SUV)