You breathe in more oxygen than you breathe out. And you breathe out more carbon dioxide than you breathe in.
What type of change is oxygen turning into carbon dioxide?
A. physical only
B. chemical only
C. physical and chemical

Answer:

amount of space occupied

Within the provided options, 'the amount of space a substance's matter occupies' best describes volume. Volume is an extensive property and the SI unit is a cubic meter (m³). It is different from mass and density, where mass refers to the amount of matter a substance contains, and density is the ratio of mass to volume.

The best description of volume among the given choices would be 'the amount of space a substance's matter occupies.' Volume, in the physical or scientific context, is an extensive property which means that its value depends on the amount of matter being considered. It is directly proportional to the amount of substance, hence, the more substance there is, the greater the volume. The standard unit of volume in the International System of Units (SI) is the cubic meter (m³), which is defined as the space occupied by a cube with sides of one meter in length.

It's important to differentiate volume from other properties like mass and density. Mass refers to the amount of matter a substance contains and density is an intensive property defined as the ratio of a substance's mass to its volume.

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Answer:

4.4 m/s

Explanation:

momentum is always conserved so we can use conversation of momentum to solve the question, also momentum is a vector quantity ( it has magnitude and direction) which is the product of the bodies mass and velocity.

conservation law of momentum relates by the formula below:

momentum before collision = momentum after collision

M1U1 + M2U2 = M1V1 + M2V2

in the case of this two, the formula becomes

M1U1 + M2U2 = V (M1 + M2) since she jumped into his arm

there masses are M1 = 75.6 kg M2 = 59 kg and their velocities are  U1 = 3.7 m/s and U2 = 5.4 m/s, their common velocity after collision = V since their motion is backward the formula becomes

-M1U1 - M2U2 = V(M1 + M2)

substitute the values into the equations

(-75.6 × 3.7 ) + (- 59 × 5.4) = V ( 75.6 + 59)

- 598.32 = 134.6 V

divide both side by 134.6

V = - 598.32 / 134.6 = -4.445 m/s  = -4.4 m/s to nearest tenth the negative means in the same backward direction

Ball Y's motion after the elastic collision with ball X should be such that it ensures the law of conservation of momentum. The direction and speed would ensure the vector sum of the momenta of the two balls equals the initial momentum of ball X.

The question is regarding an elastic collision between two identical balls on a frictionless surface, ball X moving initially with a velocity of 10 m/s, and ball Y at rest. After the collision, the velocity of ball X changes to 6 m/s at an angle of 53 degrees to its original direction. In an elastic collision, total kinetic energy and momentum are conserved.

Using the conservation of momentum: Initial momentum = Final momentum. The initial momentum was all with ball X (10*m), where m is the mass of the ball. The final momentum is a combination of the momentum of ball X and ball Y. Ball X would have a momentum of 6*m (mass*new speed), in a direction 53 degrees from its initial direction. Hence, the remaining momentum must be carried by ball Y to satisfy the law of conservation of momentum.

The exact motion of ball Y after the collision would then be determined by the pathway that would ensure the vector sum of the momenta of the two balls equals the initial momentum of ball X. This can be determined by drawing a vector diagram or by using some trigonometry.

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The distances are:

- E and F: 4 units.

- C and D: 5 units.

- G and H: 3 units.

- A and B: 6 units.

Why?

We can find the distance between each pair of points using the following formula:

[tex]d(P_1,P_2)=\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}[/tex]

So, calculating we have:

- E(-2, -1) and F(-2, -5):

[tex]d(E,F)=\sqrt{(-2-(-2))^{2}+(-5-(-1))^{2}}\\\\d(E,F)=\sqrt{0+(-5+1)^{2}}=\sqrt{0+(-4)^{2}}=\sqrt{16}=4units[/tex]

- C(-4, 1) and D(1, 1):

[tex]d(C,D)=\sqrt{(1-(-4))^{2}+(1-(1))^{2}}\\\\d(C,D)=\sqrt{(5)^{2}+0^{2}}=\sqrt{25+0}=\sqrt{25}=5units[/tex]

- G(3, -5) and H(6, -5):

[tex]d(G,H)=\sqrt{(6-(3))^{2}+(-5-(-5))^{2}}\\\\d(G,H)=\sqrt{(3)^{2}+0^{2}}=\sqrt{9+0}=\sqrt{9}=3units[/tex]

- A(5, 4) and B( 5, -2):

[tex]d(A,B)=\sqrt{(5-(5))^{2}+(-2-(4))^{2}}\\\\d(A,B)=\sqrt{0+(-6)^{2}}=\sqrt{0+36}=\sqrt{36}=6units[/tex]

Have a nice day!

Answer:

G(3, -5) and H(6, -5) ------>3 units

arrowRight

E(-2, -1) and F(-2, -5)------->4 units

arrowRight

A (5, 4) and B( 5, -2)------->6 units

arrowRight

C(-4, 1) and D(1, 1)---------->5 units

arrowRight

Explanation:

Answer:

3

3.75

Explanation:

First ball

Givens

a = 5m/s^2

m = ??

F = 15N

Formula

F = m*a

Solution

15 = m * 5

15/5 = m

m = 3 kg

Second ball

Givens

a = 4m/s^2

m = ??

F = 15N

Formula

F = m*a

Solution

15 = m * 4

15/4 = m

m = 3.75 kg

Answer:

They act on different objects, so they would not appear together.

Explanation:

The second ionization energy is the energy required to remove the second electron after a valence one has been removed.

For an element, the first ionization energy is defined as the amount of energy required to remove one electron from the outermost valence shell of a neutral atom. Removing one electron increases the number of protons, making it a 1+ ion.  

The nucleus (protons) has more bonding to the electrons with negative charge and thus more energy is required if another electron needs to be removed. This higher energy required to remove second electron from a 1+ ion (after the first one has been removed) is termed as the second ionization energy. Second ionization energy leads to formation of a 2+ ion. Similarly, third ionization energy is higher than second ionization energy.

Answer:

There are more than 2 but you are probably thinking of kinetic and static friction.  Kinetic friction is the friction between objects as they are moving.  Static friction keeps something in place and resists the movement.  Static friction can be overcome by force to move the object.

Explanation:

The frequency of light changes when it is emitted from a moving source. When a galaxy is moving away from Earth, the light it emits becomes redshifted, meaning the frequency decreases. In this case, the observed frequency when the light reaches Earth is approximately 4.945 * 10^14 Hz. Option a is correct.

The frequency of light changes when it is emitted from a moving source. This change in frequency is known as the Doppler effect. When a galaxy is moving away from Earth, the light it emits becomes redshifted, meaning the frequency decreases. To calculate the new frequency, we can use the formula f' = f/(1+v/c), where f' is the observed frequency, f is the emitted frequency, v is the speed of the galaxy, and c is the speed of light.

In this case, the galaxy is moving away from Earth with a speed of 3325 km/s. The emitted frequency is 5.00 * 10^14 Hz. Plugging these values into the formula, we get f' = 5.00 * 10^14 Hz / (1 + 3325 km/s / 3 * 10^5 km/s). Simplifying this expression, we find that the observed frequency when the light reaches Earth is approximately 4.945 * 10^14 Hz.

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A) 4.945 x [tex]10^{14}[/tex] Hz is correct option. Using the Doppler effect formula, we calculate the observed frequency of the light from a galaxy receding at 3325 km/s.

To determine the frequency of light from a receding galaxy as observed on Earth, we use the Doppler effect formula for light:

[tex]f_{obs}[/tex] = [tex]f_{em}[/tex] x (1 - v/c)

Where:

Let’s substitute the values into the formula:

[tex]f_{obs}[/tex] = 5.00 x [tex]10^{14}[/tex] Hz x (1 - 3325000 m/s / 3.00 x [tex]10^{8}[/tex] m/s)

[tex]f_{obs}[/tex] = 5.00 x [tex]10^{14}[/tex] Hz x (1 - 0.01108333)

[tex]f_{obs}[/tex] ≈ 4.945 x [tex]10^{14}[/tex] Hz

Therefore, the frequency of the light observed on Earth is 4.945 x [tex]10^{14}[/tex] Hz.

Scientists measure the time between the arrival of an earthquake's __P____ and ___S____ waves to help determine the distance between the recording seismograph and the earthquake epicenter.

Explanation:

P- (compressional) and S- (shear) waves produced in earthquakes travel at different speeds. P waves are faster than S waves and hence will be detected first by a seismograph after an earthquake. The further away a seismograph is from the epicenter of an earthquake,  the longer the time difference between the two (2) waves will be.

Using several, at least 3, seismographs located at different geoghraphical locations and detecting earthquakes, geologists can extrapolate the epicenter of an earthquake using the time differences in arrivals of the two waves in each of the seismographs, using the mathematics of triangulation.

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The time between the arrival of P waves and S waves, recorded by seismographs, is used to determine the distance to an earthquake's epicenter, with the precision being affected by the accuracy of wave speed and arrival time measurements.

Scientists gauge the distance to an earthquake's epicenter by measuring the time between the arrival of an earthquake's P waves and S waves. Seismographs can record these waves with a precision of 0.100 seconds. Given that P waves travel at 7.20 km/s and S waves travel at 4.00 km/s, the time difference recorded on seismograms allows scientists to calculate the distance to an epicenter with considerable accuracy. However, the precision of this distance measurement can be limited by the exactitude of the wave speed measurements and the timing of their arrivals.

Furthermore, the tracking of seismic waves can also be significant for monitoring underground nuclear tests. If there is uncertainty in the speeds of S and P waves, or in the measurement of their arrival times, this poses a limitation in accurately determining the source of seismic energy, which could hinder detection capabilities for such underground activities.

Answer:

Speed at the bottom of the roller coaster = 49.79 m/s

Explanation:

A new ride being built at an amusement park includes a vertical drop of 126.5 meters. Starting from rest, the ride vertically drops that distance before the track curves forward.

We have to find the speed at the bottom.

Here the gravitational energy fully converts to kinetic energy, so we equate it.

Gravitational energy = [tex]m\times g\times h[/tex]

Kinetic energy = [tex]0.5 \times m\times v^{2}[/tex]

[tex]m\times g\times h[/tex] = [tex]0.5 \times m\times v^{2}[/tex]

[tex]9.8\times 126.5 = 0.5\times v^{2}[/tex]

[tex]v^{2}[/tex] = 2479.4

Velocity, v = 49.79 m/s

Answer:

3 m/s

Explanation:

average  speed = distance traveled / total time taken

                           = 6m/ 2s

                           = 3 m/s

Answer:

530 g

Explanation:

Ideal gas law:

PV = nRT

(4.0×10⁵ Pa) (0.20 m³) = n (8.31 Pa m³ /mol /K) (20+273.15 K)

n = 32.8 mol

Find the mass using the molar mass:

32.8 mol × (16 g / mol) = 525 g

Rounded to two significant figures, the mass is 530 g.

Note: I used the molar mass provided, but oxygen is usually in a diatomic state (O₂), which would make the molar mass 32 g/mol.

Answer:F= (GM1M2)/r^2

F= ((6.67•10^-11)(5.98•10^24)(70))/ (6.40•10^6)^2 = 681.6557617

Explanation:physics

The force of gravitational attraction between the Earth and a student can be calculated using Newton's law of universal gravitation. By substituting the known values into the equation, we can find the answer.

The subject of this question is the force of gravity between two bodies. The force of gravity can be calculated using the formula from Newton's law of universal gravitation: F = G * (m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two bodies, and r is the distance between the centers of the two bodies.

Using the values given in the question, we get: F = (6.67 x 10^-11 Nm^2/kg^2) * ((5.98 x 10^24 kg * 70 kg) / (6.40 x 10^6 m)^2).

Carrying out this calculation gives you the force of gravitational attraction between the student and the Earth. This Physics concept helps us understand how forces work in the universe and is further used in studies related to astronomy, engineering, and many other fields.

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In the given physics problem, the ball comes to a stop at the highest point. This is because it has transformed all of its kinetic energy into potential energy, and has no more energy to continue moving upwards, as there is no energy transfer with either the track or the air.

In this physics problem, the ball on the track is an example of a system that conserves energy. Since there is no energy transfer with either the track or the surrounding air, the total mechanical energy of the ball, which comprises potential and kinetic energy, remains constant throughout its motion. Potential energy is energy due to height, and kinetic energy is energy due to motion.

To place the dot at the highest point the ball will reach, you identify where the ball has maximum potential energy and minimum kinetic energy because at the highest point, the ball momentarily stops and hence, kinetic energy is zero.

Therefore, the ball will stop at the highest point because it has converted all its kinetic energy (energy of motion) into potential energy (energy stored due to its position) and has no more energy left to continue moving upwards. At this point, it starts converting its potential energy back into kinetic energy as it moves downwards, hence it starts moving back down the track.

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The ball will reach the highest position at the same height as position B because of the conservation of mechanical energy. At this point, all kinetic energy converts back to potential energy, causing the ball to momentarily stop before reversing direction. This principle assumes no energy losses in the system.

To determine the highest position the ball will reach on the track before stopping and reversing direction, we need to consider the conservation of energy principle. In this scenario, the total mechanical energy (sum of potential and kinetic energy) remains constant because no energy is transferred between the ball and the track or the air.

When the ball starts from rest at position B, it possesses only potential energy and no kinetic energy. As the ball moves downward, it loses potential energy but gains kinetic energy, and as it moves upward, it loses kinetic energy and gains potential energy. The ball will reach the same height on the opposite side of the track as it started, assuming no energy losses. Therefore, the highest position the ball will reach will be the same height as position B.

The ball stops at this highest position because all of its kinetic energy is converted back to potential energy at that point, making its speed momentarily zero before it reverses direction. This illustrates the conservation of mechanical energy where the initial and final heights are the same when no external forces are acting on the system.

43.48 mph

We are given;

First instance;

Second instance;

We are required to determine the average speed for the entire journey.

Time = Distance ÷ speed

First instance

Time = 50 miles ÷ 40 mph

        = 1.25 hours

Second instance;

Time = 20 miles ÷ 55 mph

        = 0.36 hours

Therefore;

Total distance = 70 miles

Total time = 1.25 hrs + 0.36 hr

               = 1.61 hrs

Thus;

Average speed = 70 miles ÷ 1.61 hrs

                          = 43.478 miles per hour

                          = 43.48 mph

To calculate the Sun's mass, we use the formula for gravitational force, substitute the given and known values, and then rearrange the equation to solve for the Sun's mass.

The question involves the calculation of the Sun's mass using the known gravitational force between Earth and the Sun. Given that, the formula used to calculate the gravitational force between two objects is: F = G * m1 * m2 / r^2, where.

Substitution and rearranging the formula results in: m2 = F * r^2 / (G * m1) which substituting the given and known values, delivers the estimated mass of the Sun.

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According to the Law of Conservation of Mass, the mass of the rust should be greater than the original mass of the iron nail, because the mass of the oxygen and water that form the rust are added to the mass of the iron nail.

The principle at play here is the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. In this scenario, an iron nail is reacting with oxygen and water from the environment to form rust, or iron(III) oxide.

If the nail were left to rust completely, the mass of the rust would be greater than the original mass of the nail. This is because the mass of the oxygen and water that combines with the iron is added to the mass of the nail to form the rust. The additional mass comes from the oxygen atoms in the air and in the water that combine with the iron atoms in the nail.

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The law of inertia states that an object:

e. Will do all of the above.

Why?

Newton's First Law or the Law of inertia, states that an object at rest will always remain at rest if a force does not act on/upon it. Also, when an object is moving describing a straight line, if a force does not act upon it, it will keep the motion forever.

According to the Law, if an object is moving at constant velocity (same speed and same direction) it will keep the motion unless an outside force acts on it.

So, the correct option will be:

e. The object will do all of the above options.

Have a nice day!

The law of inertia states that an object at rest will remain at rest unless acted on by an outside force, while an object in motion will continue moving in a straight line at a constant speed unless acted on by an outside force.

The law of inertia, also known as Newton's first law of motion, states that an object at rest will remain at rest unless acted on by an outside force. Similarly, an object in motion will continue moving in a straight line at a constant speed unless acted on by an outside force. This law applies to all objects and is a fundamental principle in physics.

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Answer:

(D) all of the above

Explanation:

All of the choices describe matter.

The student has made an incorrect conclusion that the speed of sound in water must be slower than in air because the sound is muffled.

In fact, sound waves travel faster through water than through air because water is denser (the particles are closer together). The muffled sound is due to the fact that only around 0.01% of sound is transmitted from the air into water, as water particles are harder to displace, causing most of the sound to be reflected back off the surface.

Moreover, sound under water is often heard through vibrations in the mastoid bone, behind the ear, rather than through the ear canal as on land. The use of earplugs reduces the intensity of sound and water carries low frequencies better than air, which is why aquatic animals like whales and dolphins utilize low-frequency sounds for communication.

In summary, the muffled sound is not because the speed of sound is slower in water but due to the efficient reflection at the water surface and the reduced transmission of sound from air to water.

Answer:

m, g

Explanation:

G=m.g

m mass

g gravity

Answer:

8.25 m/s

Explanation:

From the fundamental equation of motion, velocity is rate of change of displacement per unit time

[tex]Velocity=\frac {Displacement}{time}[/tex]

Given information

Displacement=2862 m

Time=347 s

Substituting the given information we obtain

[tex]Velocity=\frac {2862 m}{347 s}=8.247838617\approx 8.25 s[/tex]

Answer:

Option D)The date of future earthquakes cannot be precisely predicted.

Explanation:

Natural disasters take place without being able to be precisely predicted at times. For example, the volcanoes can more or less likely to be predicted due to the plate tectonics and profiling the history of the volcano.

However, earthquakes cannot be predicted. For example, the massive earthquake in Lisbon in Portugal hit unexpectedly. In addition, Japan was hit by an earthquake which could not be predicted. Thus, an earthquake's future is unpredictable.

D) The correct statement is that the date of future earthquakes cannot be precisely predicted. This is due to the complexity of the geological processes involved, which also explains why the other statements are false.

Among the given statements, it is true that the date of future earthquakes cannot be precisely predicted (Option D). This is due to the complex nature of geological processes. Volcanic eruptions and earthquakes do not always coincide (Option A).

Earthquakes can occur not only on dry land but also under water, at much greater depths than 100 meters (Option B). Earthquakes also do not occur at regular, predictable intervals (Option C).

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It is really important to cite units while engaging with scientific problems as they help in evaluating the right solution without any mistake.

Explanation:

While solving analytical questions to evaluate the right answers, the units play a crucial role in the equations. With the help of units in the formula, we can easily judge whether we are placing the values in the same parameters and hence, lessen the probabilities of wrong answers.

For example, while adding two measurements i.e. 4 m and 36 cm; if we don't consider the units and move on with the addition, the answer will be 40. Now, the first thing is that we are adding two measurements of different parameters. Besides this, the answers will be wrong i.e. 4.36 m is the correct answer instead of 40 and that too without mentioning the unit.

The chances of selecting the wrong answer are more when we need to choose options out of multiple choices because here we often get confused. That's why we should always make sure that we approach the scientific questions along with the units.

Answer:

Pressure = 20 MPa

Explanation:

Given:

Force acting on the shoe is, [tex]F=100\ N[/tex]

Area of shoe on which the force acts is, [tex]A= 0.05\ cm^2[/tex]

Now, first we convert the area into its standard unit of m².

We have the conversion factor as:

1 cm² = [tex]10^{-4}\ m^2[/tex]

Therefore, the area of shoe in square meters is given as:

[tex]A=0.05\times 10^{-4}\ m^2\\A=5\times 10^{-6}\ m^2[/tex]

Now, pressure on the shoe is given as:

[tex]P=\frac{Force}{Area}\\P=\frac{F}{A}[/tex]

Plug in 100 N for 'F', [tex]5\times 10^{-6}[/tex] for 'A' and solve for 'P'. This gives,

[tex]P=\frac{100\ N}{5\times 10^{-6}\ m^2}\\P=20\times 10^{6}\ N/m^2[/tex]

Now, we know that,

[tex]10^{6}\ N/m^2=1\ MPa\\\therefore 20\times 10^{6}\ N/m^2=20\times 1\ MPa=20\ MPa[/tex]

Therefore, the pressure acting on the shoe is 20 MPa.

Answer:

61 units

Explanation:

Since OA=60 units and OB= 11 units, the resultant is the hypotenuse of the two vectors

Therefore, [tex]resultant=\sqrt {(OA)^{2}+(OB)^{2}}[/tex]

Substituting the given values then

[tex]Resultant=\sqrt{60^{2}+11^{2}}=61[/tex]

Therefore, resultant between A and B is 61 units

Answer:

6.  Acceleration = 4.74 m/s^2

7.  Centripetal force = 40.5 N

Explanation:

Problem 6.

Recall that the centripetal acceleration is defined as: [tex]a_c=\frac{v^2}{r}[/tex], where V is the object's tangential velocity, and r the radius of the circular motion. Therefore, in or case, the centripetal acceleration would be:

[tex]a_c=\frac{v^2}{r}\\a_c=\frac{3.77^2}{3}\,\frac{m}{s^2} \\a_c=4.7376 \frac{m}{s^2}[/tex]

which we can round to 4.74 m/s^2 (option b in your list)

Problem 7.

Now we need to find not just the centripetal acceleration using the same formula as above, but then the centripetal force.

[tex]a_c=\frac{v^2}{r}\\a_c=\frac{4.5^2}{2.5}\,\frac{m}{s^2} \\a_c=8.1 \frac{m}{s^2}[/tex]

Now we calculate the centripetal force by multiplying this acceleration times the mass of the object following the definition of force as mass times acceleration:

Centripetal force = 5.0 kg * 8.1 m/s^2 = 40.5 N

The answers comes in Newtons (N)

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

[tex]J=\int F\Delta t[/tex]

where

F is the magnitude of the force exerted on the object

[tex]\Delta t[/tex] is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

[tex]B=0.14-0 = 0.14s\\b=8-5=3s[/tex]

and whose height is

[tex]h=900 N[/tex]

Therefore, the area (and the impulse) is

[tex]J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns[/tex]

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

[tex]J=F_{avg} \Delta t[/tex]

where

[tex]F_{avg}[/tex] is the average force exerted during the whole time [tex]\Delta t[/tex]

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

[tex]\Delta t = 0.14 s[/tex] is the time interval

Solving for the average force, we find

[tex]\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N[/tex]

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

[tex]J=\Delta p = m(v-u)[/tex]

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

[tex]v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s[/tex]

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