If you ran at a rate of 300 m/min for 30 minutes, how far would you have run?

At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.

The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:

vrms = √(3kT/m)

Where:

vrms is the rms speed

k is the Boltzmann constant (1.38 x 10^-23 J/K)

T is the temperature in Kelvin

m is the molar mass of the gas in kilograms

To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:

m = 16 g/mol = 0.016 kg/mol

Substituting the values into the formula, we have:

vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))

Calculating this expression yields the rms speed of the oxygen sample:

vrms ≈ 482.34 m/s

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Answer:

it would be the cord will have the most energy at the moment.

Explanation:

...

The rise in temperature of the body as determined from the specific heat capacity, mass, and heat change of the body is 12.24 °C.

The rise in temperature of the body is determined from the specific heat capacity, mass, and heat change of the body.

The formula relating the temperature rise of the body, the specific heat capacity, mass, and heat change of the body is given below:

Heat change = mass * specific heat capacity * temperature rise

Temperature rise = Heat change / mass * specific heat capacity

Temperature rise = 3.6 * 10⁶ J / 84 kg * 3,500 J/kg/°C

Temepartaure rise = 12.24 °C

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Answer:

Thats probably 11

Explanation:

a.  The force on the +3.0 mC charge is 1.35 x 10^-6 N.

b.   The electric field at (x, y) = (1.00 m, 0.50 m) is  5.49 x 10^5 N/C

c. The point where the bc is zero (other than at infinity) is located on the line that connects the two charges and at the midpoint of the line segment connecting them.

d.  The potential at the point found in part c is at point zero

An electric field is described as the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.

a.  we need to use Coulomb's law, which states that the force between two charges is given by:

F = k * (q1 * q2) / r^2

F = k * (q1 * q2) / r^2 = 8.99 x 10^9 Nm^2/C^2 * (1.0 x 10^-3 C * 3.0 x 10^-3 C) / (2.0 m)^2 = 1.35 x 10^-6 N

b. To calculate the electric field at (x, y) = (1.00 m, 0.50 m), we use the equation for the electric field due to a point charge:

E = k * q / r^2

The distance from the point charge is:

r = sqrt((1.00 m - 2.0 m)^2 + (0.50 m - 0 m)^2) = sqrt(3.25 m^2) = 1.8 m

Therefore, the electric field at (x, y) = (1.00 m, 0.50 m) is:

E = k * q / r^2 = 8.99 x 10^9 Nm^2/C^2 * (3.0 x 10^-3 C) / (1.8 m)^2 = 5.49 x 10^5 N/C

c. This point is located on the line that connects the two charges and at the midpoint of the line segment connecting them.

d.  At the point found in c, the electric field is zero, so the work done to move a test charge from infinity to that point is also zero.

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The object distance of the lens is 10 cm and the image distance of the lens is 30 cm.

The object and image distance formed by the lens is calculated by applying the following lens formula.

v + u = 40 ------- (1)

v/u = 3 ------------ (2)

v = 3u

Substitute v into equation (1);

3u + u = 40

4u = 40

u = 40/4

u = 10 cm

The image distance = 3u

= 3 x 10 cm

= 30 cm

Thus, the object distance is 10 cm and the image distance is 30 cm.

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Answer: A

Explanation: We know that f=p*n

f=50*300=15000 Hz = 15kHz.

Have a great day! <3

If the number of revolutions is 300 and the paired poles are 50 , then the frequency would be 15 kHz, therefore the correct answer is option A.

It can be defined as the number of cycles completed per second. It is represented in hertz and inversely proportional to the wavelength.

The frequency of a pendulum is the reciprocal of the time period can be given by the following relation,

F = 1 / T

As given in the problem, we have to calculate frequency if the number of revolutions is 300 and the paired poles are 50.

F = 300 × 50

  = 1500 kHz

Thus, If the number of revolutions is 300 and the paired poles are 50, then the frequency would be 15 kHz, therefore the correct answer is option A.

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The medium in which electromagnetic waves and mechanical waves travel is one of their primary distinctions. Light and other electromagnetic waves, including radio waves, can move through void space without the aid of a physical medium.

They may move through vacuum, air, or other materials and are made up of oscillating electric and magnetic fields. The propagation of mechanical waves, such as sound or water waves, on the other hand, depends on a physical medium.

To transport energy, they rely on particle interactions and displacements in the medium. Since mechanical waves need a physical medium to carry their energy, they cannot move through a vacuum.

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Answer: 1 Reaching for the Moon : the autobiography of NASA mathematician Katherine 2 Johnson by Katherine G Johnson( Book )

3 Katherine Johnson by Thea Feldman( Book )

4 Katherine Johnson by Ebony Wilkins( Book )

5Counting the stars by Lesa Cline-Ransome( Book )

6An Act to Award Congressional Gold Medals to Katherine Johnson and Dr.

Explanation:

The horizontal velocity of the airplane is,

The height of the airplane is,

The vertical initial velocity of the box is zero as the airplane is moving in the horizontal direction.

let the time to reach the victim is t.

we can write,

Substituting the values we get,

Hence the required time is 14.3 s

Answer:

The power output of the engine is 1,000 Watt

Explanation:

Mechanical Power

Power is the amount of energy transferred or converted per unit of time. In the SI, the unit of power is the watt, equal to one joule per second.

The power can be calculated as:

\(\displaystyle P=\frac {W}{t}\)

Where W is the work and t is the time.

The engine does a work of W=60,000 J during t=60 s. Calculate the power:

\(\displaystyle P=\frac {60,000}{60}\)

P = 1,000 Watt

The power output of the engine is 1,000 Watt

Answer:

0.41s

Explanation:

solve for t and y

,........

Answer:

d. convention

Explanation:

hope this helped

Ant is performing a work

what is work?

Work is the force applied on an individual with respect to displacement.

Work = Force × displacement

Unit is Nm

Elephant is pushing bt there is no displacement occurred so the work of elephant is zero.

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The ant works, but the elephant does not.

Work done = Force × Displacement.

If there are ants and houseflies,

Ants drag the house bug, so they use specific force to move the house bug from one point to another, so we can say they work.

In the case of the elephant and the tree,

When the elephant pushes the tree (applying a force), the tree does not move, i.e., there is no displacement, so there is work.

Work done = Force × Displacement

= Force × 0

= 0

Therefore,

The ant works, but the elephant does not.

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None of the added energy goes into the rotational kinetic energy of the diatomic ideal gas when 510 J of heat flows into it at constant volume.

For a diatomic ideal gas, the total internal energy can be divided into three parts: translational, rotational, and vibrational energy. When heat flows into the gas at constant volume, the internal energy of the gas increases. If the gas is initially at rest, all of the added energy goes into increasing the temperature and therefore the translational kinetic energy of the gas molecules.

Assuming that the gas is initially at rest, all of the added energy goes into the translational kinetic energy of the gas molecules. Therefore, none of the added energy goes into the rotational kinetic energy of the gas. Thus, the amount of energy going into the rotational kinetic energy of the gas is 0 J.

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Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

         Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

         Em_f = K = ½ m v²

since friction is zero, energy is conserved

          Em₀ = Em_f

          1 / 2k x² = ½ m v²

          v = \(\sqrt{ \frac{k}{m} }\)     x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

Based on the law of conservation of energy, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.

The energy in a compressed spring is elastic potential energy given by the formula:

where

The kinetic energy of a body is the energy the body the has due to it's motion.

Kinetic energy, KE, is givenby the formula below:

From the law of conservation of energy, the total energy in a closed system is conserved.

Based on this law, all the energy in the compressed spring is converted to the kinetic energy of the box just before it reaches the ground.

Therefore, the elastic potential energy of the system when the spring is fully compressed is equal to the kinetic energy of the system at the moment immediately before the box hits the ground.

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Answer and Explanation:

A well-coated structure is defined as having a coating that meets a certain standard of quality. The answer to this particular question depends on the specific criteria being used to evaluate the coating. This would typically require a coating coverage of 90% or better, if not higher.

However, in general, a well-coated structure would typically refer to a surface that has been thoroughly and evenly covered with a coating material such as paint or varnish. This ensures that the underlying material is protected from environmental factors such as moisture and UV radiation. In addition, a well-coated structure can also improve the overall appearance of the surface, making it more aesthetically pleasing. Regarding the options provided in the question, the answer would depend on the specific criteria being used to evaluate the coating. However, it is safe to say that a well-coated structure would require a high level of coating coverage, with minimal areas left uncovered or with an uneven application. This would typically require a coating coverage of 90% or better, if not higher. Ultimately, the specific answer would depend on the standards and expectations set by the evaluating body.

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A well-coated structure is defined as having a coating that meets a certain standard of quality. The answer to this particular question depends on the specific criteria being used to evaluate the coating. This would typically require a coating coverage of 90% or better, if not higher.

However, in general, a well-coated structure would typically refer to a surface that has been thoroughly and evenly covered with a coating material such as paint or varnish. This ensures that the underlying material is protected from environmental factors such as moisture and UV radiation. In addition, a well-coated structure can also improve the overall appearance of the surface, making it more aesthetically pleasing.

Regarding the options provided in the question, the answer would depend on the specific criteria being used to evaluate the coating. However, it is safe to say that a well-coated structure would require a high level of coating coverage, with minimal areas left uncovered or with an uneven application. This would typically require a coating coverage of 90% or better, if not higher. Ultimately, the specific answer would depend on the standards and expectations set by the evaluating body

Stone tools, kitchen implements, and other relevant artefacts, as well as the examinations carried out to determine their role in cannibalism.

The procedure for preparing food. I cook the majority of the meals for our household. Cuisine describes a method of preparing food. a technique used in French cooking. foods associated to cooking. The state of English cooking: The intense flavours you're looking for come from a good pan that has been heated to an extreme temperature. the relationship between cooking skills and food choices.

There are two categories of stone tools: flaked stone tools, such as knives, scrapers, and projectile points, and ground stone tools, such as manos (grinding stones-left) & hand axes (arrow heads and spear points-above). The three basic components of a point are the stem, base, and blade.

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Answer:

Explanation:

The vector in unit vector notation is:

B = (-1.33m)i + (2.81m)j

To find the direction relative to the x-axis. Do inverse tangent of opposite/adjacent

\(tan^-^1=|\frac{2.81}{1.33}| = 2.11\\180-2.11=177.89\)

This is relative to the positive x-axis

2.11 would be the correct answer if relative to the negative x axis

The magnitude of the force and the impulse applied to the car are  85714 N and  30000 N-s respectively.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

The magnitude of the force applied to the car = change in momentum/time interval

= (1200 kg × 25 m/s - 1200 kg×0)/0.35 second

= 85714 N.

The impulse applied to the car = change in momentum

= (1200 kg × 25 m/s - 1200 kg×0)

= 30000 N-s.

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The height of the object, given that the potential energy of the object doubled will be double (option D)

First, we shall list out the given parameters. Details below:

Potential energy is given as:

PE = mgh

Keeping mg constant, we have

PE₁ / h₁ = PE₂ / h₂

Inputting the given parameters, we have

P / h = 2P / h₂

Cross multiply

P × h₂ = h × 2P

Divide both sides by P

h₂ = (h × 2P) / P

h₂ = h × 2

h₂ = 2h

Thus, we can conclude that the new height of the object will also double (option D)

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The velocity in the 100-mm tube can be calculated using the principle of conservation of mass, which states that the mass flow rate of a fluid in a closed system must remain constant.

The mass flow rate of water through the larger tube is equal to the mass flow rate through the two smaller tubes combined. Therefore, we can set the equation for mass flow rate as:

ρQ = ρ1Q1 + ρ2Q2

where ρ is the density of water, Q is the volumetric flow rate (0.072 m3/s for the larger tube), ρ1 and Q1 are the density and volumetric flow rate of the 50-mm tube, and ρ2 and Q2 are the density and volumetric flow rate of the 100-mm tube.

We can rearrange this equation to solve for Q2:

Q2 = (ρQ - ρ1Q1) / ρ2

We know the values of Q, ρ, Q1, ρ1, and the velocity in the 50-mm tube (12.0 m/s), so we can use the equation for volumetric flow rate (Q = A × v) to solve for the cross-sectional area of the 50-mm tube, which is 0.001963 m2.

Using the equation for density (ρ = m/V), we can also find the density of water, which is approximately 1000 kg/m3.

Finally, plugging in all the values, we can solve for Q2 and then use the equation for velocity (v = Q/A) to find the velocity in the 100-mm tube, which is approximately 4.16 m/s.

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The velocity of the car 1 can be seen from the calculation as 20 m/s West

A coordinate system or point of view used to observe motion is known as a frame of reference. It can be used as a guide when describing an object's position, speed, and acceleration. Different frames of reference may result in various motion observations.

The relative velocity is the velocity of an object or observer as observed from a particular frame of reference.

We can see that the velocity of the car 1 is;

45 m/s - 25 m/s

= 20 m/s West

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Explanation:

I have only 4 of them ,hope they help

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
\(T_h = \frac{T}{2}\)

Now, we can determine the velocity at which the can was launched at using the following equation:
\(v_f = v_i + at\)

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
\(0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}\)

***vsinθ is the vertical component of the velocity.

Solve for 'v':
\(vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}\)

Now, recall that:
\(W = \Delta KE = \frac{1}{2}m(\Delta v)^2\)

Plug in the expression for velocity:
\(W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}\)

B.

We can use the same process as above, where T' = 2T and Th = T.

\(v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}\)

C.

The work done in part B is 4 times greater than the work done in part A.

\(\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}\)

A. To find the work done on the can by the launching device, we need to calculate the change in kinetic energy of the can. When the can is launched, it has an initial kinetic energy of 0 (since it starts from rest).

1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]

WA = 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]

B. If the can stays in the air twice as long, the time of flight becomes 2T. We can repeat the above steps with the new value of T to find the initial velocity: Initial vertical velocity = g * (2T)

WB \(= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

C. To compare the results, we can calculate the ratio of WB to WA:

WB/WA \(= [1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]] / [1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]]\)

Notice that the mass M and the angle α0 cancel out:

WB/WA \(= [((initial velocity)^2 * cos^2(α0) + (2gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (g * T)^2)]\)

A.  At the highest point of its trajectory, the kinetic energy is again 0 (the vertical velocity component is 0). So the work done by the launching device equals the initial kinetic energy. The initial kinetic energy can be expressed as: Initial kinetic energy = \(1/2 * M * (initial velocity)^2\)

The initial velocity can be calculated using the launch angle and the time of flight T. At the highest point, the vertical component of the velocity is 0, so we only consider the horizontal component of the velocity. Horizontal component of velocity = initial velocity * cos(α0). The time T is the time taken to reach the highest point, so we can write:

T = time taken to reach the highest point = (initial vertical velocity) / g

Initial vertical velocity = g * T

Now, the initial velocity can be written as:

initial velocity = + (Initial vertical   \()√[(Horizontal component of velocity)^2\)                        

               =\(√[(initial velocity * cos(α0))^2 + (g * T)^2]\)

                =\(√[(initial velocity)^2 * cos^2(α0) + (g * T)^2]\)

Since the initial velocity is equal to the change in kinetic energy, we have: Initial kinetic energy :\(1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2] \\ WA = 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]\)

B. Initial velocity:

                     \(√[(initial velocity * cos(α0))^2 + (g * (2T))^2]\\ = √[(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

Now, the initial kinetic energy for this case is: Initial kinetic energy (new) \(= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

WB \(= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

C. This shows that the mass and angle do not affect the work done on the can; only the time of flight and the acceleration due to gravity influence it. Since we know that T is doubled in part B (2T), we can write:

WB/WA  \(= [((initial velocity)^2 * cos^2(α0) + (2g * 2T)^2)] / [((initial velocity)^2 * cos^2(α0) + (g * T)^2)]\)

WB/WA \(= [((initial velocity)^2 * cos^2(α0) + (4gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (gT)^2)]\)

Now, we can see that WB is larger than WA by a factor of \([((initial velocity)^2 * cos^2(α0) + (4gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (gT)^2)].\)

The value of this factor will depend on the specific values of the initial velocity, launch angle, and time of flight, but this ratio is greater than 1, indicating that more work is done on the can when it stays in the air for twice the time.

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The minimum coefficient of static friction with the floor is 0.3846.

To find the minimum coefficient of static friction with the floor, we need to consider the forces acting on the couch. In this case, the force of gravity is pulling the couch downward with a magnitude of mg, where m is the mass of the couch (40 kg) and g is the acceleration due to gravity (9.8 m/s²).

Since the couch does not move, the force of static friction between the couch and the floor must be equal in magnitude but opposite in direction to the horizontal pushing force of 150 N.

Therefore, we have the equation F_friction = F_push, where F_friction is the force of static friction.

The force of static friction can be calculated using the formula F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Since the couch is on a level floor and is not accelerating vertically, the normal force N is equal in magnitude but opposite in direction to the force of gravity, which is mg.

Substituting the values into the equation, we have μs * mg = 150 N.
Solving for μs, we get μs = 150 N / (mg).
Substituting the given values, we have μ_s = 150 N / (40 kg * 9.8 m/s²).
Simplifying, we find that μs = 0.3846.

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Answer:

As the charge increases, the force decreases in strength

Explanation:

Experiments with electric charges have shown that if two objects each have electric charge, then they exert an electric force on each other. The magnitude of the force is linearly proportional to the net charge on each object and inversely proportional to the square of the distance between them...

Answer: The correct answer is: As the charge increases, the force increases in strength

Explanation:

Its kind of the obvious answer and other people don't know how to answer correctly. This was years ago but if anyone needs the answer today then here it is....

Answer:

2.5 mi/s^2

Explanation:

please see paper for work!