During heavy exercise, muscle cells use oxygen more quickly than normal. Which system responds by carrying extra oxygen to these muscle cells?
a. digestive b. circulatory c. nervous
d. immune

Answer:

In coastal areas, strong winds and powerful waves break off soft or grainy rocks from hardier rocks. Too much weathering occurs, it might break off parts of the cliff and be dangerous to humans, or animals.

Rain has acid in it which could eat up the cliff.

Explanation:

Answer:

1. He feels worse off.

2. His awareness of the grim situation.

Explanation:

Elie Weisel's memoir "Night" is about the persecution of the jews by the Germans during the worst genocide in world history. The events leading up to the Holocaust and the resulting after-effects through his personal experience provides one of the most prominent witness accounts of the crime.

When Weisel states that "lying down wasn't an option", he reveals how congested space was in the cabin. The "lucky ones" were able to breathe in the fresh air from the window, while the rest have to be satisfied with wherever they are. This reveals his realization of the grim situation inside the over-packed cabin, where there is hardly any space to move.

And when he said that they "never ate enough to satisfy our hunger", he presents the realization and understanding of the grim situation in the train's cabin where eating is not a luxury, but a necessity to stay alive. And for that, they know they have to "economize, to save for tomorrow" rather than just stay full once.  

Answer:

1. He feels worse off.

2. His awareness of the grim situation.

Explanation:

Given data

*The given mass of the car is m = 1475 kg

*The maximum tension is T = 2551 N

The formula for the maximum possible acceleration of the truck is given by Newton's second law as

Substitute the known values in the above expression as

Hence, the maximum possible acceleration of the truck is a_max = 1.72 m/s^2

Answer:

Height of building = 135.2 m

Granite and basalt are the two main forms of rock that make up the crust. The majority of the continental crust is made up of granite. Basalt is a type of volcanic lava rock that makes up the oceanic crust.

The outermost layer of a terrestrial planet is referred to as its "crust." All known life in the universe is contained in the thin, 40 km (25 mi) deep crust of our planet, which makes up just 1% of the planet's mass. The crust, mantle, and core are the three layers that make up the earth. Minerals and solid rocks make up the crust. The mantle, which lies below the crust, is composed primarily of solid rocks and minerals with some pliable regions of semi-solid magma. There is a heated, dense metal core in the center of the Earth.

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Answer:

b

Explanation:

Answer: less than 9.8 m/s

Explanation:

Yes, melting lead into a liquid to form pellets is a physical change.

This is because the chemical composition of lead remains the same even after it has been melted and then solidified into pellets. In other words, the molecular structure of lead does not change during the melting process, but only the physical state of the material changes from a solid to a liquid and then back to a solid. This type of change is reversible and can be undone by cooling the lead pellets to their solid state. Therefore, melting lead to form pellets is an example of a physical change rather than a chemical change.

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--The complete Question is, Is lead melting into liquid to form pellets a physical change? --

Answer:

i would say the neuron

Answer:

The moment arm or lever arm is the perpendicular distance between the line of action of the force and the center of moments.

Explanation:

Moment = Force x Distance or M = (F)(d) 

The primary effect of increasing the surface tension of the liquid is to increase the capillary force. Capillary forces arise due to the combined effects of adhesion and cohesion

When the surface tension of the liquid increases, the capillary rise will increase. It is because the increase in surface tension leads to an increase in the force that pulls the liquid upwards in a tube. is as follows;If you place a capillary tube in a beaker filled with water, the water surface inside the tube rises slightly higher than the level outside the tube.

This rise in water level is called capillary rise. The capillary rise is caused by the attraction between the molecules of the water and the molecules of the glass tube.This attraction is called capillary force or capillary action. The capillary force is due to the combined effect of adhesive and cohesive forces. The adhesive force is the attraction between the molecules of the liquid and the molecules of the solid surface, while the cohesive force is the attraction between the molecules of the liquid.

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So, the value of work that he had done is 900 J.

Hi ! Here I will discuss about the effort to raise an object. Previously, when we raised an object, the energy that we put out had to be equal to the change in potential energyHi ! Here I will discuss about the effort to raise an object. Previously, when we raised an object, the energy that we put out had to be equal to the change in potential energy that happened. This is because, the higher the height of an object, the greater value of potential energy. The equation that applies is as follows :

\( \sf{\bold{W = \Delta PE}} \)

\( \boxed{\sf{\bold{W = m \times g \times \Delta h}}} \) ... (i)

\( \boxed{\sf{\bold{W = m \times g \times (h_2 - h_1)}}} \) ... (ii)

With the following condition :

However, because the value of w (weight of the object) is already mass multiplied by gravity (w = m × g). So :

\( \sf{W = m \times g \times \Delta h} \)

\( \boxed{\sf{\bold{W = w \times \Delta h)}}} \)

With the following condition :

We know that :

What was asked :

Step by step :

\( \sf{W = w \times \Delta h} \)

\( \sf{W = 600 \times 1.5} \)

\( \boxed{\sf{W = 900 \: J}} \)

So, the value of work that he had done is 900 J.

Answer:

1. chemical energy --> electrical energy --> light energy

2. Electrical Energy --> Mechanical Energy

3. Electrical Energy --> Heat Energy

4. Chemical Energy --> Thermal Energy  --> Mechanical Energy

5. Nuclear Energy --> Light energy and thermal energy

6. Mechanical Energy --> Electric Energy

7. Chemical Energy --> Electrical Energy --> Sound Energy

8. Mechanical Energy --> Sound Energy

Explanation:

The real rate of return for ten years ago, you put $5,000 in a savings account is -0.41%.

A pace of return (RoR) can be applied to any venture vehicle, from land to bonds, stocks, and artistic work. The RoR works with any resource gave the resource is bought at one particular moment and produces income sooner or later. Past rates of return, which can be compared to assets of the same type to determine which investments are most appealing, are part of how investments are evaluated. Before making an investment decision, many investors prefer to select a required rate of return.

By using the calculator of simple interest we get the value as -0.41%.

Taking into account the time value of money (TVM), which the CAGR does not take into account, is the next step in comprehending RoR over time. Limited incomes take the profit of a venture and rebate every one of the incomes in view of a markdown rate. The rebate rate addresses a base pace of return OK to the financial backer, or an expected pace of expansion. Businesses evaluate the profitability of their investments using discounted cash flows in addition to investors.

Let's say, for example, that a business uses a discount rate of 5% and is thinking about spending $10,000 on a new piece of equipment. The business uses the equipment, which results in an increase of $2,000 annually in cash inflows for five years after a $10,000 cash outflow. Each year for five years, the company applies present value table factors to the $10,000 outflow and $2,000 inflow.

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The percentage of time the operator of the single entrance booth will be free is approximately 38.462%.

To determine the percentage of time that the operator of the single entrance booth will be free, we need to calculate the service rate and the arrival rate. The service rate is given as 400 vehicles per hour, and the arrival rate is 250 vehicles per hour. The percentage of time the operator will be free can be calculated using the formula:

Free time percentage = (Service rate - Arrival rate) / Service rate * 100

Substituting the given values into the formula:

Free time percentage = (400 - 250) / 400 * 100

= 150 / 400 * 100

= 0.375 * 100

= 37.5%

Rounding off the answer to the nearest thousandths, the percentage of time the operator of the single entrance booth will be free is approximately 38.462%.

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The speed of 75 km/h expressed in:

To solve this problem the we have to convert the units with the given information.

Information about the problem:

By converting the velocity units from (km/h) to (m/s) we have:

v = 75 km/h * 1000 m/ 1km * 1 h/3600 s

v = 20.83 m/s

By converting the velocity units from (m/s) to (mi/h) we have:

v = 20.83 m/s * 1 mile/1609.34 m * 3600 s/ 1h

v = 46.59 mi/h

It is the transformation of a value expressed in one unit of measurement into an equivalent value expressed in another unit of measurement of the same nature.

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The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.

The formula for calculating the volume of a cylinder is given below.

V = πr^2 h

Get the differential of the volume as shown:

dV = V/ h dh + V / r dr

V/ h = πr^2

V/ h  = 2  πr h

Now, the differential becomes

dV =  πr^2dh +  2πrh dr

Given the following parameters

dh = 0.1 + 0.1 = 0.2 cm

dr = 0.05 cm

h = 10 cm

r = 4 cm

Substituting the values in the above equation, we get

dV = 3.14(2)^2(0.2)  + 2(3.14)(2)(10)(0.05)

dV = 2.512 + 6.28

dV = 8.792 cm

dV = 8.8 cm

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Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

m g h = 1/2 m v^2    where h is the vertical distance thru which the pendulum travels

The electric force exerted on one sphere by the other is equal to  9.27  × 10⁻⁷ N.

According to Coulomb’s law, the force of attraction between two charged bodies is the product of their charges and is inversely proportional to the square of the distance by which they are separated. This force acts along the line joining the two-point charges.

The magnitude of the electric force is given by:

\(F = k\frac{q_1q_2}{r^2}\)

where k is constant proportionality and has a value of 8.99 × 10⁹ N.m²/C².

Given the charge on one sphere, q₁ = + 6 ×10⁻⁹ C

The charge on the other sphere, q₂ = -11 × 10⁻⁹C

The distance between these two charges, r = 0.80 m

The magnitude of electric force between the spheres will be:

\(F = 8.99 \times 10 ^9(\frac{6\times 10^{-9}\times 11\times 10^{-9}}{(0.80)^2} )\)

F = 927.09  × 10⁻⁹ N

F = 9.27  × 10⁻⁷ N

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Answer:

\(0.833\ \text{m/s}^2\)

Explanation:

m = Mass of block = 6 kg

F = Force on block = 5 N

\(\mu\) = Coefficient of friction = 0

a = Acceleration of the block

From Newton's second law we have

\(F=ma\\\Rightarrow a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{5}{6}\\\Rightarrow a=0.833\ \text{m/s}^2\)

The acceleration of the block is \(0.833\ \text{m/s}^2\)

The difference between the momentum before and after the bounce was 1.407 kg/s.

initial energy = mg(3) J = 29.4J

After bouncing ,energy = mg(2) =19.6J

Ball's change in energy = 29.4J - 19.6J = 9.8J

This energy is dissipated in the form of heat energy or sound energy.

work done during the bounce = 9.8J

change in momentum = mvf-mvi

29.4 = 1/2m\(vi^{2}\)

vf =7.668m/s

19.6J =1/2m\(vf^{2}\)

vf=6.261m/s

Change in momentum = m(vf-vi) = 1kg x (-1.407m/s) = -1.407

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The maximum wind velocity for a 30° crosswind with a maximum crosswind component of 15 knots is 30 knots.

To determine the maximum wind velocity for a 30° crosswind when the maximum crosswind component for the airplane is 15 knots, follow these steps:

1. Identify the maximum crosswind component: 15 knots.
2. Determine the crosswind angle: 30°.
3. Use the formula: Wind velocity = Maximum crosswind component / sin(crosswind angle).

In this case:

Wind velocity = 15 knots / sin(30°)

The sine of 30° is 0.5, so:

Wind velocity = 15 knots / 0.5

Wind velocity = 30 knots

So, the maximum wind velocity for a 30° crosswind with a maximum crosswind component of 15 knots is 30 knots.

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Answer: 480 N is required to maintain the arrangement .

Explanation:

Given, 320 N  force is required to hold 2 protons in a particular arrangement .

Then the force required to keep 1 proton in a particular arrangement = 320/2

= 160 N

Then if  a proton is replaced by an alpha particle , it is given that a alpha contains 2 protons , then we conclude that after replcament , the arrangement has 1 alpha and 1 proton = 3 protons

Then , the force exerted by 3 protons = 3×160 = 480 N

The force exerted by the arrangement , if one of the protons is replaced with alpha is 480 N

, The density of the metal with a simple cubic structure is approximately \(8.93 g/cm^3.\)

To determine the density of the metal with a simple cubic structure, we can use the following formula:

Density = (Atomic weight)/(Volume of the unit cell x Avogadro's number)

For a simple cubic structure, the volume of the unit cell can be calculated as:

The volume of unit cell = \(a^3\)

where a is the length of the edge of the cube.

In a simple cubic structure, the atoms touch along the edge of the cube. So, the edge length can be calculated as:

a = 2 x Atomic radius

Substituting the given values, we get:

a = 2 x 0.126 nm = 0.252 nm

The volume of the unit cell is:

Volume of unit cell = \(a^3\)= \((0.252 nm)^3\) = 0.016 \(nm^3\)

Now, we can substitute the values into the density formula:

Density = (70.4 g/mol)/(0.016 \(nm^3\) x 6.022 x \(10^23\)/mol)

Density = 8.93 \(g/cm^3\)

Therefore, the density of the metal with a simple cubic structure is approximately\(8.93 g/cm^3.\)

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Translated Question: If a metal had the simple cubic structure, how is it shown in the figure. Yes its atomic weight is 70. 4 g/mol and the atomic radius is 0.126 nm. determine the density

Answer:

As altitude increases and atmospheric pressure decreases, the boiling point of water decreases.

Answer:

This demonstration is often done following a discussion of the ideal gas equation of state, PV=nRT.

We begin by weighing a balloon, then blowing it up and weighing it again. In the photo shown on right, the mass indication increased from 3.4 to 3.5 grams. At this point, it is important to note that the scale measures force, even though it reports a conclusion about mass based on the force measurement.

One assumption made in reaching the conclusion is that the buoyant force on the object being weighed is negligible. In the case of the balloon, this is incorrect. The buoyant force on this balloon is equal to the weight of the air displaced.

Since the volume of air inside the balloon is essentially the same as the volume of air displaced, we should expect that the buoyant force would support the weight of the air inside the balloon: The reported mass should not go up at all, because the force required of the scale should not change.

The increase in reported mass of .1 gram is attributed to the higher density of the air inside the balloon: The tension in the balloon compresses the air inside, as attested by the pressure required to blow the balloon up. Evidently, for this experiment, the pressure inside is greater than atmospheric by about 2%.

In the picture at right, the balloon is being pressed into a pan of liquid nitrogen. (The pan is the styrofoam lid of a small lunch box.) The balloon floats lightly on the liquid nitrogen unless pressed down. Pressing down places more surface area in contact with the cold nitrogen and speeds the demonstration. It is interesting to note the buoyant force by this liquified constituent of air.

The balloon shrinks dramatically, as indicated below. When left in contact with the liquid nitrogen long enough (perhaps 5 minutes) the oxygen inside the balloon liquifies, and then the nitrogen liquefies also. Close observation of the photo at the upper left corner of the pan shows some liquid nitrogen bubbles may forming above the dark spot in the center of the pan. One can also make out a faint line at the upper left corner of the pan which is the liquid nitrogen surface. The balloon still floats, riding rather high on that surface. Evidently, some of the balloon contents remain in the gas phase, making the mass of the balloon less than the mass of the displaced liquid nitrogen.

Next, we take the shrunken balloon and place it back on the scale, as above. In this instance, the reported mass is 8.7 grams, an increase of 5.2 grams.

A look at the figure on the right shows a faint line near the bottom of the cold balloon. Above that line, the balloon contains gas; below the liquid. That line represents the top surface of the liquid air inside the balloon. With this evidence, the easy thing to say would be, "Of course, liquids are heavier than gases," but that would be incorrect. We assert that the amount of air inside the balloon has not changed and that the mass of that air is not dependent on temperature.

If these assertions are true, then the force of gravity on the balloon has not changed. The scale reading is determined by the force which it must exert on the balloon in order to keep it stationary. Evidently, the required force is larger when the balloon is shrunken. The reason is that the buoyant force (upward) has decreased to practically zero, leaving the scale alone to balance the downward force by gravity.

From the data, we can say that the change in the buoyant force is equal to the weight associated with the apparent change in mass. The weight of 5.2 grams is about .052 newtons. The buoyant force is less now because the balloon displaces less air. If we could measure the change in volume of the balloon as DV, then the buoyant force would be (r g DV) upwards, where r is the density of air that was displaced by the balloon, and g is the gravitational field strength, 9.8 Newton/kg.

Note that the .052 newton force is not the weight of the air inside the balloon. Rather, it is the weight of the air that was displaced by the balloon. If we ignore the compression of air inside the balloon, the two numbers are the same. However, the two samples are completely different.

We can estimate the volume of the balloon by assuming that the hand in the photograph is about .1meters across. For purposes of estimation, we say that the volume shrank to almost zero when the balloon was cold so that the change in volume was nearly equal to the original volume. Plugging in numbers gives fair agreement with the book value of 1kg/cubic meter for the density of air.

The value for the density of air is secondary to two main features of this demonstration:

Large changes in temperature produce the large changes in volume that are indicated by the ideal gas equation.

The mass of air in a volume equal to the volume of a balloon can be determined provided that the buoyant force is understood.

Answer:

Na+ca+CO=NACo^3

Explanation:

NA Burns violently with explosions that may spatter the material. Used for making gasoline additives, electric power cable, sodium lamps, other chemicals.

Answer:

0.5 mL/min

Explanation:

There are 1000 mL in a L, and 60 minutes in an hour.

0.03 L/hr × (1000 mL/L) × (1 hr / 60 min) = 0.5 mL/min