A sample of 2 mol of diatomic gas is measured to have a temperature of 323
K. If the mass of the gas is 0.032 kg, what is the approximate average speed
of the molecules in the gas? (Recall that the equation for kinetic energy due
to translation in a gas is KEtranslational = 1m² = 3nRT, and R = 8.31 J/(mol-
K).)
2
2
O A. 652m/s
OB. 621 m/s
OC. 709 m/s
OD. 681 m/s

Answer:

Explanation:

The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. Therefore, we can use this theorem to calculate the final kinetic energy of the ball in each case.

We know that the work done by a constant force is given by the equation W = Fd cos(theta), where F is the magnitude of the force, d is the displacement of the ball, and theta is the angle between the force and displacement vectors.

Using the work-energy theorem, we can write:

W = ΔK = Kf - Ki

where ΔK is the change in kinetic energy, Kf is the final kinetic energy, and Ki is the initial kinetic energy.

We can rearrange this equation to solve for Kf:

Kf = Ki + W = Ki + Fd cos(theta)

a) Kf = 150 J + (10 N)(15 m)cos(90°) = 150 J

b) Kf = 300 J + (200 N)(1.5 m)cos(180°) = 0 J

c) Kf = 200 J + (25 N)(4 m)cos(0°) = 300 J

d) Kf = 450 J + (15 N)(30 m)cos(150°) = 112.5 J

Ranking from greatest to least final kinetic energy:

c) Ki=200J F=25N d=4m theta=0 degrees

a) Ki=150J F=10N d=15m theta=90 degrees

d) Ki=450J F=15N d=30m theta=150 degrees​

b) Ki=300J F=200N d=1.5m theta=180 degrees

The volume of the bubble at the surface is 34.15 cm³

•Initial volume (V₁) = 20 cm³

•Initial temperature (T₁) = 4 °C = 4 + 273 = 277 K

•Final temperature (T₂) = 200 °C = 200 + 273 = 473 K

•Final volume (V₂) =?

The final volume of the bubble can be obtained by using the Charles' law equation as shown below:

V₁ / T₁ = V₂ / T₂

20 / 277 = V₂ / 473

Cross multiply

277 × V₂ = 20 × 473

277 × V₂ = 9640

Divide both side by 277

V₂ = 9640 / 277

V₂ = 34.15 cm³

Thus, the volume of the bubble at the surface is 34.15 cm³

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Answer: Rectangular frames are easy to make but can get pulled out of shape. so if the sides are still attached , then the figure formed is parallelogram. useing the given measurement use the formula of a parallelogram.

formula : A = BASE X HEIGHT

Explanation:

The airplane's vector angular momentum relative to a wheat farmer on the ground directly below the airplane is 11.61 × 10⁹ j kg m²/s and this value does not change as the airplane continues its motion along a straight line.

Its angular momentum relative to the summit of Pikes Peak is 2.7 × 10⁶ R j kg m²/s, where R is the position vector of the wheat farmer relative to the summit of Pikes Peak.

a) The momentum of the airplane relative to the wheat farmer can be calculated as, Momentum, p = m × v

Thus, Momentum, p = 15000 × 180

Momentum, p = 2.7 × 10⁶ kg m/s

The position vector of the airplane relative to the wheat farmer is given as, 4300 j

The vector angular momentum of the airplane relative to the wheat farmer is given as,

L = r × p

L = 4300 j × 2.7 × 10⁶ kg m/s

L = 11.61 × 10⁹ j kg m²/s

Therefore, the airplane's vector angular momentum relative to a wheat farmer on the ground directly below the airplane is 11.61 × 10⁹ j kg m²/s.

(b) No, this value does not change as the airplane continues its motion along a straight line.

(c) The momentum of the airplane relative to the summit of Pikes Peak can be calculated as, Momentum, p = m × v

Thus, Momentum, p = 15000 × 180 Momentum, p = 2.7 × 10⁶ kg m/s

The position vector of the airplane relative to the summit of Pikes Peak is r' = R + 4300 j

The vector angular momentum of the airplane relative to the summit of Pikes Peak is given as,Vector angular momentum,

L' = r' × p

L' = (R + 4300 j) × 2.7 × 10⁶ kg m/s

L' = R × 2.7 × 10⁶ kg m/s + 4300 j × 2.7 × 10⁶ kg m/s

L' = (2.7 × 10⁶ R) j kg m²/s

Therefore, its angular momentum relative to the summit of Pikes Peak is 2.7 × 10⁶ R j kg m²/s, where R is the position vector of the wheat farmer relative to the summit of Pikes Peak.

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Answer: The rock

Explanation:

A straight piece of wire with a current I flowing through it is placed in a magnetic field

A straight piece of wire with a current I flowing through it is placed in a magnetic fielduniform and perpendicular to the magnetic field lines. Magnetic force acting on the string

A straight piece of wire with a current I flowing through it is placed in a magnetic fielduniform and perpendicular to the magnetic field lines. Magnetic force acting on the stringthere is a way

The correct option is C, the velocity between 2s and 4s is 7 meters per second.

Here we have the graph of the velocity of a T-Rex as function of time in seconds.

Here we need to find the average value between 2 seconds and 4 seconds.

At 2 seconds, the graph says that the velocity is 7m/s

And we can see an horizontal line that ends at 4s, so the veloicty at 4 seconds is 7m/s

Then the average velocity in that interval is that one (because it is constant)

Then the correct option is C.

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The compressive stress in the steel column is found to be approximately 397.6 MPa.

The formula for calculating the area of a circle can be used to determine the steel column's cross-sectional area (A),

A = π*(d/2)², diameter of the column is d,

A = π*(0.4/2)²

A = 0.1257m²

The compressive stress (σ) in the column can be calculated using the formula, σ = F/A, F is the load carried by the column is F.

σ = 50 MN/0.1257m²

σ = 397.6 MPa

Therefore, the compressive stress in the steel column is approximately 397.6 MPa.

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Answer:

B Gravitational potential energy

Explanation:

Answer:

i Believe the correct answer is "An open circuit being changed into a closed circuit"

Explanation:

The air will be when it stops moving 180J.

The generally accepted boundary where space begins, which is also the point where the atmospheric pressure is assumed to be zero, is called the Kalman line. Simply put, gases move from areas of high pressure to areas of low pressure. And the greater the pressure difference, the faster air moves from high pressure to low pressure. This draft is the wind we experience.

As the air rises the surface pressure decreases. As the air rises, it expands and cools adiabatic cooling that is, it cools by changing volume rather than by increasing or decreasing heat. As a result, condensation/sedimentation occurs. Cold air sinks. The boundary between Earth and space is called the Kalman Line, an imaginary line 100 kilometers above Earth. This line is considered the boundary between space and the Earth's atmosphere.

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To determine the image characteristics using the 3 principal rays and SALTS (Size, Attitude, Location, Type), we'll consider both lenses and mirrors separately. Here's how you can analyze each case:

Lenses:

Place an object at different distances to the left of a lens with a focal length (f).

a) Object placed beyond 2f:

In this case, the object is placed far beyond twice the focal length of the lens.

Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.

Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.

Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.

The image will be formed on the opposite side of the lens, between the focal point and twice the focal length.

SALTS:

Size: The image will be smaller than the object.

Attitude: The image will be inverted.

Location: The image will be located between the focal point and twice the focal length.

Type: The image will be real.

b) Object placed at 2f:

In this case, the object is placed at twice the focal length of the lens.

Principal ray 1: A ray parallel to the principal axis will pass through the focal point on the opposite side.

Principal ray 2: A ray passing through the optical center will continue in a straight line without any deviation.

Principal ray 3: A ray passing through the focal point on the object side will emerge parallel to the principal axis.

The image will be formed on the opposite side of the lens at twice the focal length.

SALTS:

Size: The image will be the same size as the object.

Attitude: The image will be inverted.

Location: The image will be located at twice the focal length.

Type: The image will be real.

c) Object placed between f and 2f:

In this case, the object is placed between the focal point and twice the focal length of the lens.

In this case, the object is placed far beyond twice the focal length of the mirror.

Principal ray 1: A ray parallel to the principal axis will reflect through the focal point on the same side.

Principal ray 2: A ray passing through the focal point on the object side will reflect parallel to the principal axis.

Principal ray 3: A ray passing through the center of curvature will reflect back along the same path.

The image will be formed on the opposite side of the mirror, between the focal point and twice the focal length.

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Answer:

1/2mv²=0

1/2(4kg)(v²)=0

2=-v²

square root -2=v

v=1.414

Block A of mass 1 kg and block B has a mass of 3 kg, then the blocks collide and stick together so the kinetic energy of block A is one-third of the kinetic energy of the block.

Kinetic energy is a type of power that an item or particle possesses as a result of motion. When an item undergoes work—the transfer of energy—by having subject to a net force, it accelerates and consequently obtains kinetic energy. An object in motion or particle's kinetic energy, which depends on both mass and speed, is one of its features. Any combination of motions, including translation, rotation about an axis, and vibration, may be used as the type of motion.

A body's translational kinetic energy is equal to 1/2mv², or one-half of the sum of its mass, m, and the square of its velocity, v.

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The mass of Car B is -6000 kg.

To solve this problem, we can apply the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

Therefore, we can write the equation for the conservation of momentum as:

(mass of Car A * velocity of Car A) + (mass of Car B * velocity of Car B) = (mass of Car A + mass of Car B) * velocity after collision

Let's substitute the given values into the equation:

(2000 kg * 10 m/s) + (mass of Car B * 0 m/s) = (2000 kg + mass of Car B) * (-5 m/s)

Simplifying the equation:

20000 kg*m/s = -5 m/s * (2000 kg + mass of Car B)

Dividing both sides by -5 m/s:

-4000 kg = 2000 kg + mass of Car B

Subtracting 2000 kg from both sides:

mass of Car B = -4000 kg - 2000 kg

mass of Car B = -6000 kg

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

The picture is blacked out.

Answer:

If I'm correct the answer would be A) Lysosome.

The attachment is black so I can't promise you that it's the one you're looking for. But I did it on khan

Answer:

Zero

Explanation:

It is given that the charge is uniformly spread at the surface of hollow spherical shells.

There is no charge with in the hollo shell. hence, with zero charge the  electric field intensity is also zero.

Thus,

Magnitude of the electric field strength inside the hollow shell at r is zero

Answer:

62.5 m/s East

100 m/s East

Explanation:

Average velocity = displacement / time

v_avg = (2500 m East) / (40 s)

v_avg = 62.5 m/s East

Given for the mouse:

v₀ = 0 m/s

a = 2.5 m/s²

t = 40 s

Find: v

v = at + v₀

v = (2.5 m/s²) (40 s) + 0 m/s

v = 100 m/s East

Answer:

c. 7 m/s

Explanation:

3500 m / 500 s = 7 m/s

Answer:1.9 atm

Explanation: I stole it from someone O-O

So it could be wrong

Answer:

the answer is b  hydrogen atoms

Explanation:

The correct answer is D:Carbon atoms

Ethanol molecule has two carbon atoms.

A molecule can be represented in a three dimensional diagram by a ball and stick model. These models help us to conceptualize what the actual molecule looks like.

The molecule ethanol is composed of two carbon atoms, six hydrogen atoms, and one oxygen atom.

If there are two black balls in the ethanol model, they must represent carbon atoms since there are two carbon atoms.

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12. The time taken for the journey is 400 s

13. The time taken for the train is 200 s

14. The time taken is 7 h

15. The time taken for the beetle is 12 s

16. The time taken for the journey is 0.0068 h

The time taken in each case as given by the question can be obtain as follow:

12. The time taken for the journey

Time taken = Distance / Speed

Time taken = 26000 / 65

Time taken = 400 s

13. The time taken for the train

Time taken = Distance / Speed

Time taken = 3200 / 16

Time taken = 200 s

14. The time taken to travel

Time taken = Distance / Speed

Time taken = 672 / 96

Time taken = 7 h

15. The time taken for the beetle

Time taken = Distance / Speed

Time taken = 1.08 / 0.09

Time taken = 12 s

16. The time taken for the journey

Time taken = Distance / Speed

Time taken = 35 / 5147

Time taken = 0.0068 h

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Answer:

Eubacteria Kingdom :)

Explanation:

Answer: eubacteria

Explanation:

Hooke's Law can be given as follows sometimes:

The restoring force of a spring is equal to the spring constant multiplied by the displacement from its normal position:

F = -kx

Where, F = Restoring force of a spring (Newtons, N)

k = Spring constant (N/m)

x = Displacement of the spring (m)

The negative sign relates to the direction of the applied force and by convention, the minus or negative sign is present in F = -kx. The restoring force F is directly proportional to the displacement (x), according to Hooke's law. When the spring is compressed, the displacement (x) is negative. It is zero when the spring is at its original length and positive when the spring is extended.

Practically, Hooke's Law is applicable only within a limited frame of reference, and through experimenting, this statement proves to be true. Because materials cannot be compressed beyond a certain size or expanded beyond a certain size without some permanent deformation or change of their original state.

The law only applies under some conditions such as a limited amount of force or deformation. Factually, many materials will noticeably deviate from Hooke's law even before those elastic limits are reached.

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Answer:

J

Explanation:

The daughter moves with greater acceleration backwards because of her weight.

Both mother and daughter move backward, but the daughter moves with greater acceleration. Therefore, option (J) is correct.

This law states that the acceleration of a body depends on two factors. The first one is the mass of the body and the net force acting on the body. The acceleration is inversely proportional to the mass and is directly proportional to the net force acting on the body and

Newton’s second law can be written in an expression as follows:

F = ma

or,   a = F/m

According to Newton's third law of motion, there is an equal and opposite reaction for every action. So when the two skaters push each other. They both will move in a backward direction.

But the mass of the daughter is 50 Kg while the mass of the mother is 100 Kg. As the mass of the daughter is less. Therefore daughter moves in a backward direction with greater acceleration.

Therefore, both move backward but the daughter moves with greater acceleration in comparison to the mother.

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Answer:

They are trying to jump-start a car using a kite and lightning connected to the battery. This will not work because there is no positive or negative charged side to the circuit.

Explanation:

The magnitude of the force P provided that the stress in the part AB is two times that of BC part is 0.8 kN.

The magnitude of the force P provided that the stress in the part AB is two times that of BC part is calculated as follows;

Take moment about the joint to determine the magnitude of the force along part BC.

120 kN x 750 mm  =  F x 1000 mm

F = ( 120 kN x 750 mm ) / ( 1000 mm )

F = 90 kN

Stress is given as force divided by area. The following equation can be used to determine the magnitude of force P.

Stress in AB = 2 times stress in BC

P/A₁  = 2F/A₂

where;

A₁ =  πd²/4  =  π(50 x 10⁻³)²/4

A₁ = 1.96 x 10⁻⁵ m²

A₂ = πd²/4  =  π(75 x 10⁻³)²/4

A₂ = 4.42 x 10⁻³ m²

P/A₁ = 2F/A₂

P = (2F x A₁) / (A₂)

P = (2 x 90 kN x 1.96 x 10⁻⁵ m² ) / ( 4.42 x 10⁻³ m² )

P = (2 x 90,000 N x 1.96 x 10⁻⁵ m² ) / ( 4.42 x 10⁻³ m² )

P = 798.2 N

P = 0.798 kN

P ≈ 0.8 kN

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Answer:

We can use Hooke's Law to find the spring constant k:

F = kx

where F is the force applied to the spring, x is the displacement of the spring from its equilibrium length, and k is the spring constant.

We can find the force applied to the spring by using the weight of the 4.0 kg fish:

F = mg = (4.0 kg)(9.81 m/s^2) = 39.24 N

The displacement of the spring is the difference between its length with the fish and its equilibrium length:

x = 12.0 cm - 10.0 cm = 2.0 cm = 0.02 m

Now we can solve for k:

k = F/x = 39.24 N / 0.02 m = 1962 N/m

To find the length of the spring with an 8.0 kg fish suspended from it, we can use the same formula with the new weight:

F = mg = (8.0 kg)(9.81 m/s^2) = 78.48 N

We can solve for x, which is the new displacement of the spring:

x = F/k = 78.48 N / 1962 N/m = 0.04 m

Therefore, the length of the spring will be:

10.0 cm + 4.0 cm = 14.0 cm

Explanation: