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3 years ago

12

On a cool morning, Uyen’s breath can form a cloud when she breathes out. Which changes of state are most responsible for Uyen se

eing her breath in this way?

2 answers:

murzikaleks [220]3 years ago

8 0

Answer:

Condensation

Explanation

If it is cold outside the gaseous water molecules in our breath start accumulating in atomic/molecular clusters which then turns into the breath we are able to see. This process is known as condensation.

zhuklara [117]3 years ago

3 0

It's cold outside, the water vaper in your breath condenses into tiny droplets of liquid water and ice that you can see.

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1. Two charged objects have a repulsive force of 2 N. If the charge of one of the objects is doubled, then the new force will be

GrogVix [38]

Answer:

4 N

Explanation:

Electrostatic force is directly related to the charge of each object hence when the charge is doubled in only one of the objects, the force also double. If it's doubles on both objects, the force becomes four times. In this case, it's doubled on one object hence 2*2= 4 N

5 0

4 years ago

What is the kinetic energy of a 620.0 kg roller coaster moving with a velocity of 9.00 m/s? Round to three significant figures.

raketka [301]

Answer:

25110

Explanation:

5 0

3 years ago

While Barb was playing soccer, she was kicked in the anterior thigh by an opposing teammate. Due to this injury, what muscle fun

Alenkasestr [34]

Answer:

knee extension is the muscle function that will be difficult to perform.

Explanation:

Barb was kicked in the anterior thigh. Now, the thigh muscles performs a combined operation of moving the knee and leg and they reside in the following compartments.

- Anterior compartment which is composed of knee joint extension and thigh flexion.

- Lateral Compartment which is composed of the tensor fasciae latae, which is a tiny muscle that abducts and centrally will make the thigh to rotate.

-Medial compartment which involves thigh addiction which is rotating of the thigh around the hips.

- Posterior compartment which involves knee joint flexion and high extension.

Thus, from the different compartments listed above, we can see that the muscles that extend the knee and flex the thigh all lie in the anterior compartment of the upper leg.

Thus, we can conclude that knee extension is the muscle function that will be difficult to perform.

3 0

3 years ago

On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (u) 46Ti 77.600% 45.95263

allsm [11]

Answer: Average atomic mass of titanium on that planet is 46.52

Explanation:

Mass of isotope Ti-46 = 45.95263

% abundance of isotope Ti-46 = 77.600 % = $\frac{77.600}{100}=0.776$

Mass of isotope Ti- 48= 47.94795

% abundance of isotope Ti-48 = 16.100%= $\frac{16.100}{100}=0.161$

Mass of isotope Ti- 50 = 49.94479

% abundance of isotope Ti-50 = 6.300%= $\frac{6.300}{100}=0.063$

Formula used for average atomic mass of an element :

$\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})$

$Z=\sum[(45.95263 \times 0.776)+(47.94795 \times 0.161)+(49.94479\times 0.063)$

$Z=46.52$

Average atomic mass of titanium on that planet is 46.52

8 0

3 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

Computing the work done by the sprinter

$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

$x=3.8\ m$

The net force is

$F=m(8.8)=8.8m$

The work done by the sprinter is now computed as

$W=8.8m(3.8)=33.44m$

At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0

4 years ago