Given the formula C1V1=C2V2, where C indicates concentration and V indicates volume, which equation represents the correct way to find the concentration of the dilute solution (C2)?
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Given the formula , where indicates concentration and indicates volume, which equation represents the correct way to find the concentration of the dilute solution ()?
C2=V2C1V1C2=V1V2C1C2=C1V1V2C2=C1V1V2

A. The estimated maximum kinetic energy of the chamber with a 200 gram rock is approximately 0.00196 Joules. B. the estimated maximum acceleration of the chamber with a 200 gram rock is approximately 1.131 m/s².

C. The representation of the motion for a 200 gram rock as a function of time is x(t) = 0.1 × cos(2π × 0.30 Hz × t). D. It does not seem like a lot of force inside the rover.

To evaluate the design and determine if the device can damage the housing, go through the required steps:

a. The maximum kinetic energy of the chamber can be calculated using the formula:

K_max = (1/2) × m × v²

where K_max is the maximum kinetic energy, m is the mass of the rock, and v is the velocity of the rock.

Given:

Mass of the rock (m) = 200 grams = 0.2 kg

Velocity of the rock (v) = 14 cm/s = 0.14 m/s

Using the formula, we can calculate:

K_max = (1/2) × 0.2 kg × (0.14 m/s)²

K_max ≈ 0.00196 Joules

Therefore, the estimated maximum kinetic energy of the chamber with a 200 gram rock is approximately 0.00196 Joules.

b. The maximum acceleration of the chamber can be determined using the formula:

a_max = ω² × A

where a_max is the maximum acceleration, ω is the angular frequency, and A is the amplitude of the oscillation.

Given:

Angular frequency (ω) = 2π × frequency

Frequency = 0.30 Hz

Using the formula, we have:

ω = 2π × 0.30 Hz

Now, let's assume the amplitude of the oscillation is A = 0.1 meters.

a_max = (2π × 0.30 Hz)² × 0.1 meters

a_max ≈ 1.131 m/s²

Therefore, the estimated maximum acceleration of the chamber with a 200 gram rock is approximately 1.131 m/s².

c. The motion of the rock in simple harmonic motion can be described by the equation:

x(t) = A × cos(ω × t)

where x(t) represents the displacement of the rock at time t, A is the amplitude, ω is the angular frequency.

Using the values from earlier:

Amplitude (A) = 0.1 meters

Angular frequency (ω) = 2π × 0.30 Hz

The representation of the motion for a 200 gram rock as a function of time is:

x(t) = 0.1 × cos(2π × 0.30 Hz × t)

d. Based on the given information, the maximum kinetic energy of the chamber with a 200 gram rock is approximately 0.00196 Joules. This is a relatively small amount of energy.

The maximum acceleration of the chamber with a 200 gram rock is approximately 1.131 m/s². This acceleration is also relatively small.

Considering these values, it does not seem like a lot of force inside the rover. The energy and acceleration involved are relatively low, suggesting that the device should not cause significant damage to the housing. However, it is always essential to ensure that the design and materials used in the housing can withstand the expected forces and vibrations to ensure the rover's durability and functionality.

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Answer: soccer ball

Explanation:

The kicker has the potential energy and as the kicker kicks the ball that potential energy turns to kinetic

The takes a motor with an output of 8.0 W approximately 3.632 minutes to lift a 2.0 kg object 88 cm.

The time it takes a motor with an output of 8.0 W to lift a 2.0 kg object 88 cm can be calculated using the equation for work, which is given by:

W = F * d

where W is the work done, F is the force required to lift the object, and d is the distance the object is lifted. The force required to lift the object can be calculated using the equation for weight, which is given by:

F = m * g

where m is the mass of the object and g is the acceleration due to gravity (9.8 m/s^2).

Substituting the values for m and g into the equation for weight, we get:

F = 2.0 kg * 9.8 m/s^2 = 19.6 N

Now that we know the force required to lift the object, we can substitute this value into the equation for work to calculate the work done:

W = F * d = 19.6 N * 88 cm = 1717.28 J

Finally, we can use the equation for power, which is given by:

P = W / t

where P is the power of the motor and t is the time it takes to lift the object. We know the power of the motor (8.0 W) and the work done (1717.28 J), so we can calculate the time it takes to lift the object:

t = W / P = 1717.28 J / 8.0 W = 215.91 seconds = 3.632 minutes

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a) The isocost curve equation is G = (20000 - 100C)/500. b) The isocost curve equation is G = (30000 - 100c)/500. c) The isocost curve equation is G = (20000 - 100C)/375. d)  The isocost curve equation is G = (20000 - 120C)/500.

To draw the isocost curve showing the different combinations of gas and coal, we need to use the cost per unit values for coal and gas, as well as the given expenditure (TC) and the changes in expenditure or prices.

Let's denote the quantity of coal as C and the quantity of gas as G. The cost per unit of coal is 100, and the cost per unit of gas is 500.

(a) Initial expenditure (TC) of 20000:

To find the combinations of gas and coal that can be purchased with an initial expenditure of 20000, we can use the following isocost equation

TC = 100C + 500G

We can rearrange the equation to solve for G in terms of C

G = (TC - 100C) / 500

Now we can plot the isocost curve with TC = 20000 using the equation above.

(b) Expenditure (TC) increases by 50%

If the expenditure increases by 50%, the new expenditure (TC_new) becomes 1.5 × TC = 1.5 × 20000 = 30000.

We can use the same isocost equation as before, but with the new expenditure value:

TC_new = 100C + 500G

Rearranging the equation to solve for G

G = (TC_new - 100C) / 500

Now we can plot the isocost curve with TC_new = 30000.

(c) Gas price reduced by 25%:

If the gas price is reduced by 25%, the new cost per unit of gas (Gas_new) becomes 0.75 × 500 = 375.

We can use the original isocost equation, but with the new cost per unit value:

TC = 100C + 375G

Rearranging the equation to solve for G

G = (TC - 100C) / 375

Now we can plot the isocost curve with the reduced gas price.

(d) Coal price rises by 20%

If the coal price rises by 20%, the new cost per unit of coal (Coal_new) becomes 1.2 × 100 = 120.

We can use the original isocost equation, but with the new cost per unit value:

TC = 120C + 500G

Rearranging the equation to solve for G:

G = (TC - 120C) / 500

Now we can plot the isocost curve with the increased coal price.

By plotting these isocost curves on a graph with G on the y-axis and C on the x-axis, we can visualize the different combinations of gas and coal that can be purchased at the given expenditures or price changes.

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Answer:

A I think hope it helps!!

The way how charged particles interact is demonstrated with the Charge Launcher Gizmo. Tiny charged particles have the ability to attract (pull together) or repel (push apart) one another, just like magnets.

The positive charge is present in the red particles.

The red particle took a slightly bent downhill route when it was launched from the upper left corner of the grid.

The blue particles are negatively charged. This particle is fixed, which means it is affixed to the grid. The blue particle took the same course downhill when it was released from the upper right corner of the grid and at the very end it curled upward.

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Answer:

Answer:

A solar eclipse results when the moon passes in between the earth and the sun hiding the sun fully or partly for some time. A lunar eclipse occurs when the earth passes in between the moon and the sun casting its shadow on the moon and thus hiding it fully or partly for some time.

Explanation:

your welcome

Answer:

La esfera no flotará pero se hundirá cuando se coloque en el agua porque su densidad es mayor que la del agua.

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Radio (r) de la esfera = 0,4 m

Masa de esfera = 300 Kg

Densidad del agua = 1000 Kg / m³

A continuación, determinaremos el volumen de la esfera. Esto se puede obtener de la siguiente manera:

Radio (r) de la esfera = 0,4 m

Pi (π) = 3,14

Volumen (V) de la esfera =?

V = 4/3 πr³

V = 4/3 × 3,14 × 0,4³

V = 12,56 / 3 × 0,064

V = 0,27 m³

A continuación, determinaremos la densidad de la esfera. Esto se puede obtener como se ilustra a continuación:

Masa de esfera = 300 Kg

Volumen de esfera = 0,27 m³

Densidad de esfera =?

Densidad = masa / volumen

Densidad de la esfera = 300 / 0,27

Densidad de la esfera = 1111,11 Kg / m³

Comparando la densidad del agua y la de la esfera.

Sustancia >>>>>>> Densidad

Agua >>>>>>>>>>> 1000 Kg / m³

Esfera >>>>>>>>>> 1111,11 Kg / m³

De la ilustración anterior, podemos ver que la densidad de la esfera es mayor que la del agua.

Por lo tanto, la esfera no flotará sino que se hundirá cuando se coloque en el agua porque su densidad es mayor que la del agua.

Answer:

Below

Explanation:

Speed at beginning = 0

speed at bottom = ?

   d = 1/2 at^2

   6 = 1/2 (9.81)(t^2 )    shows t = 1.1 s

vf = at = 9.81 ( 1.1) = 10.85 m/s

Average = ( 10.85 - 0) / 2 = 5.42  m/s  average

Answer:

6.76875 • 10 ^-10 F

Explanation:

You just follow the formula:

C = εA/d

ε = 8.85 • 10^-12

A = 3.61 • 10^-4

D = 4.72 • 10 ^-6

So:

C = (8.85 • 10^-12)(3.61 • 10^-4)/4.72 • 10 ^-6 = 6.76875 • 10 ^-10 F

Good Luck!  :)

Explanation:

\(E = \frac{BA}{t} \sin( \theta) \\ = \frac{1.23 \times 2}{1} \\ = 2.46 \: V\)

The shuttle is under a force of 1.625 105N from the fuel. The force on a body expressed as a change in momentum is,

                     F= Δ\(\frac{p}{t}\)  (1)

Where,

F is the force that the shuttle is being subjected to from the fuel.

P represents the change in the shuttle's momentum.

t stands for the overall amount of time needed.

Considering our query,

ΔP= \(=325000=3.25X 10^{5} kg-m/s\)

t = 2 seconds

Equation (1) yields the following results when the necessary values are substituted:

\(F= \frac{3.25 10^{5} }{2}\)

\(F= 1.625\) x \(10^{5}\)N

Hence, 1.625105N is the force the fuel is exerting on the shuttle. The amount of motion a body has is referred to as momentum. Given that momentum relies on both velocity and the vector of the body's motion, it really is quantified as "mass speed". As acceleration and mass are both scalar quantities, vector quantities include momentum.

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According to the given statement the correct answer is when we speak of the large-scale structure of the universe, we are referring to the distribution and arrangement of matter and energy on the largest scales, such as galaxy clusters and superclusters.

This includes the overall cosmic web of filaments and voids, as well as the patterns of cosmic microwave background radiation that provide insights into the early universe. Understanding the large-scale structure is crucial for investigating the origins and evolution of the universe.When we refer to the large-scale structure of the universe, we are talking about the distribution of matter and galaxies on a very large scale. This includes the arrangement of clusters and superclusters of galaxies, as well as the vast cosmic voids that separate them. The large-scale structure of the universe is studied in the field of cosmology, and understanding its properties can provide insight into the origin, evolution, and future of the universe. Some of the tools used to study the large-scale structure of the universe include telescopes, surveys of galaxy positions, and computer simulations.

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The chemical energy of petrol is converted to heat energy on combustion. The heat energy is converted to kinetic energy by the use of internal combustion engines in vehicles. The law of conservation of energy is maintained in each process.

The kinetic energy of an object is associated with its motion. It can be related to the mass and velocity as

K.E = 1/2 mv²

Given is a diagram of energy conversion due to combustion.

The chemical energy of petrol is converted to heat energy on combustion. The heat energy is converted to kinetic energy by the use of internal combustion engines in vehicles.

The law of conservation of energy states that the energy can neither be created nor destroyed. It can be only converted to one form to the other.

In the given process, the mass decreases but energy remains the same in all forms of energy.

Hence, the law of conservation of energy is maintained in each process

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The statement is false, that the greatest magnification on the compound light microscope can be achieved by the use of the high-power objective lens.

A microscope with numerous lenses and its own light source is referred to as a compound light microscope. In this type of microscope, the rotating nosepiece is positioned closer to the specimen and contains the objective lenses, while the binocular eyepieces house the ocular lenses.

The compound binocular microscope is now more often utilized, even though it can occasionally be found as monocular with only one ocular lens.

Zacharias Jansen's 1595 invention of the compound microscope with collapsible tubes and a 9X magnification produced the first light microscope.

Since then, microscopes have advanced significantly; the strongest compound microscopes available today have magnifications of 1,000 to 2,000X.

Hence, the statement is false that it can be achieved by using the high power objective lens.

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Answer:

is it George Lemaitre?

Explanation:

sorry if i'm not helpful

Answer:

Electromagnetic field, a property of space caused by the motion of an electric charge. A stationary charge will produce only an electric field in the surrounding space. If the charge is moving, a magnetic field is also produced. An electric field can be produced also by a changing magnetic field.

The charge will move in the direction that minimizes its electric potential energy, either towards lower potential (for positive potential energy) or towards higher potential (for negative potential energy).

When an unknown charge is released from rest at a particular location in an electric field, the direction in which the charge moves will depend on its initial electric potential energy.

The charge will move in the direction that reduces its potential energy. In other words, it will move in the direction of decreasing electric potential.

If the charge has positive electric potential energy, it will move in the direction of decreasing potential towards a region of lower potential.

Conversely, if the charge has negative electric potential energy, it will move in the direction of increasing potential towards a region of higher potential.

In summary, the charge will move in the direction that minimizes its electric potential energy, either towards lower potential towards higher potential.

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A cylindrical bit of graphite with a radius of 0.7 mm is subjected to a potential differential of 24 V along its length by the student.

Pencils, lubricants, crucibles, casting facings, polishes, arc lights, batteries, electric motor brushes, and nuclear reactor cores all include graphite. China, India, Brazil, N.korea, and Canada all mine large amounts of it. Graphite was inadvertently created for the first time by Edward G.

It is special in that it possesses traits common to both metals and non-metals, including flexibility without being elastic, high thermal and electrical conductivity, high refractoriness, and chemical inertness. Due to its low X-ray and neutron absorption, graphite is a particularly useful material.

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Therefore, the nucleus experiences an acceleration of 1.93 × 10¹¹ m/s² in the positive x-direction, and its displacement after 1.70 s is 1.64 × 10¹¹m in the positive x-direction.

To solve this problem, we'll use the following formulas:

(a) Acceleration (a):

The electric force (F(e)) experienced by the helium nucleus can be calculated using the formula:

F(e) = q × E

where q is the charge of the nucleus and E is the magnitude of the electric field.

The force ((F)e) acting on the nucleus is related to its acceleration (a) through Newton's second law:

F(e) = m × a

where m is the mass of the nucleus.

Setting these two equations equal to each other, we can solve for the acceleration (a):

q × E = m × a

a = (q × E) / m

(b) Displacement (d):

To find the displacement, we can use the kinematic equation:

d = (1/2) × a × t²

where t is the time interval.

Given:

m = 6.64 × 10²⁷ kg

q = 3.20 × 10¹⁹ C

E = 4.00 ×10⁻⁷ N/C

t = 1.70 s

(a) Acceleration (a):

a = (q × E) / m

= (3.20 × 10¹⁹ C ×4.00 × 10⁻⁷ N/C) / (6.64 × 10⁻²⁷ kg)

= 1.93 ×10¹¹ m/s² (in the positive x-direction)

(b) Displacement (d):

d = (1/2) × a × t²

= (1/2) × (1.93 × 10¹¹ m/s²) ×(1.70 s)²

= 1.64 × 10¹¹ m (in the positive x-direction)

Therefore, the nucleus experiences an acceleration of 1.93 × 10¹¹ m/s² in the positive x-direction, and its displacement after 1.70 s is 1.64 × 10¹¹m in the positive x-direction.

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Answer: Figure it out yourself-

Explanation:

The time of motion of the satellite is 1.65 hours. The speed of the satellite from the centre of the Earth is 26,945.35 km/h.

This is the motion of an object in which the object travels in a straight line and its velocity remains constant along that line as it covers equal distances in equal intervals of time, regardless of time duration.

The time of motion of the satellite in hours is calculated as follows;

t = ( 99 min / 1 ) x ( 1 hour / 60 min )

t = 1.65 hours

The speed of the satellite from the centre of the Earth in km/h is calculated as follows;

v = ( 2πr ) / ( t )

where;

r is the distance of the satellite from the centre of the Earth

The position of the satellite above the surface of the Earth = 705 km

The radius of Earth = 6,371 km

The total distance of the satellite from the centre of the Earth = 705 km + 6,371 km = 7,076 km

v = ( 2π x 7076 ) / ( 1.65 )

v = 26,945.35 km/h

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The complete question is attached with the answer below.

Answer:

Look at the mirror from the positive axis and  suppose the incident ray is at angle at 40 from the y-axis

Incident ray at 50 deg

reflected ray at 130 deg

total angle of deflection = 130 - 50 = 80 deg

Now rotate the mirror by 20 deg

incident ray at 30 deg

reflected ray at 150 deg    (angle of incidence must equal angle of reflection)

total angle of deflection = 150 - 30 = 120 deg

the difference between 2 reflected rays = 120 - 80 = 40 deg

So if you increase (decrease) the incident angle the reflected ray will  be increased by twice the angle of rotation of the mirror - or the perpendicular to the mirror increases by X degrees the angles of incidence and reflection will each increase by X deg

When light enters air from water, the angle of refraction will be greater than the angle of incidence and is denoted as option B.

This is defined as the angle between a refracted ray and the normal to a refracting surface. Since light entered air from water, the water has the incident ray while the air has the refracted part.

According to Snell's law: n1sinθ1 = n2sinθ2

Refractive index of water = 1.333

Refractive index of air = 1.000293

Since air is denser than water, the angle of refraction will be greater than incidence because most refraction occurs for the transmission of light across a water-air boundary

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The average force exerted by the club on the ball is 838,400 N. Force can be characterized by its magnitude, direction, and point of application.

It can be a push or pull, and it can cause an object to start moving, stop moving, or change its direction of motion.

Force is indeed a physical factor that alters or has the potential to alter an object's state at rest or motion as well as its shape. Newton is the SI unit of force.

Finally, the average force exerted by the club on the ball is:

F = I / t = (1676.8 N·s) / (0.0020 s) = 838,400 N

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