Glucose is a reactant in cellular respiration.

True or False?

The mass of water that undergoes a change in temperature from 10 degrees celsius to 60 degrees celsius is 24.9 g.

Mass is the quantity of matter a body contained.

To calculate the mass of  the water, we use the formula below

Formula:

Where:

From the question,

Given:

Substitute these values into equation 1

Hence, the mass of water is 24.9 g.

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Answer:

To solve this problem, we can use the formula:

d = (v^2)/(2μg)

d = distance traveled

v = speed of the car

μ = coefficient of kinetic friction

g = acceleration due to gravity

First, let's calculate the distance traveled when the car is traveling at 23 m/s:

d = (23^2)/(2*0.7*9.81) ≈ 67.97 meters

Now, let's calculate the distance traveled when the car is going twice as fast (46 m/s):

d = (46^2)/(2*0.7*9.81) ≈ 271.88 meters

Therefore, the car would travel approximately 271.88 meters if it were going twice as fast.

The light waves reflecting off a red stop sign would be longer than the light waves reflecting off a violet-colored jacket. This is because red light has a longer wavelength than violet light.

Light waves, like all waves, are characterized by their wavelength. The wavelength of a wave determines its color, with shorter wavelengths appearing as blue and violet, and longer wavelengths appearing as red and orange.

Because the wavelength of red light is longer than the wavelength of violet light, the light waves reflecting off the red stop sign would be longer than the light waves reflecting off the violet-colored jacket.

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Answer:

The speed of the ball just before it hits the crossbar is 7.22 m/s.

Explanation:

We can use the conservation of energy to solve this problem.

At the moment the player kicks the ball, the ball has only kinetic energy, since it was at rest on the ground before being kicked. When the ball hits the crossbar, it has both kinetic energy and potential energy, since it is at a height above the ground. We can set the initial kinetic energy equal to the sum of the final kinetic and potential energy:

(1/2)mv^2 = mgh

where:

m = mass of the ball (0.500 kg)

v = initial speed of the ball (15.0 m/s)

g = acceleration due to gravity (9.80 m/s^2)

h = height of the crossbar above the ground (2.6 m)

We want to solve for the speed of the ball just before it hits the crossbar, which we can do by rearranging the equation:

v = sqrt(2gh)

v = sqrt(29.802.6) = 7.22 m/s (rounded to two decimal places

If F₁ has a greater magnitude than F₂, the box will accelerate backward because the net force is in the backward direction (1st option)

To obtain the direction in which the box will move, we shall determine the net force acting on the box. This is illustrated below:

Assumption:

Net force = Magnitude of force 1 (F₁) - Magnitude of force 2 (F₂)

Net force = F₁ - F₂

Net force = 40 - 25

Net force = 15 N backward

From the above illustration, we can see that the net force is 15 N backward.

Thus, we can conclude from the box will accelerate backward (1st option)

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The final velocity of the cylinder is 1.22 m/s when it reaches the bottom of the ramp.

To solve this problem, we need to use conservation of energy and rotational kinematics.

Calculate the gravitational potential energy (GPE) of the cylinder at the top of the ramp:

GPE = mgh = (2.9 kg)(9.81)(0.75 m) = 21.39 J

Calculate the final kinetic energy (KE) of the cylinder when it reaches the bottom of the ramp:

[tex]KE = 1/2 mv^2 + 1/2 Iω^2[/tex]

where v is the linear velocity, I is the moment of inertia, and ω is the angular velocity.

Since the cylinder rolls without slipping, we know that v = ωr, where r is the radius of the cylinder.

[tex]KE = 1/2 mv^2 + 1/4 mv^2 = 3/4 mv^2 = 3/8 mgh[/tex]

Substituting the values we have:

KE = 3/8 (2.9 kg)(9.81)(0.75 m) = 63.56 J

Finally, we can use conservation of energy to find the final velocity of the cylinder:

GPE = KE

[tex]mgh = 3/8 mgh + 1/2 mv^2 + 1/2 Iω^2[/tex]

Solving for velocity:

[tex]v = \sqrt (2gh/5) = \sqrt(29.81 m/s^20.75 m/5) = 1.22 m/s[/tex]

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the complete question is:

At the top of a ramp, a 2.9 kg solid cylinder (radius = 0.20 m, length = 0.70 m) is released from rest and allowed to roll without slipping. The ramp measures 0.75 m in height and 5.0 m in length. calculate the final velocity when it reaches the bottom of the ramp

Answer:24 volts ÷ 4 amps = 6 ohms

Explanation:

We know that the Ohm's law has given the relationship between the current, voltage and the resistance of the wire. Mathematically, it can be written as :

Where

I is the current

R is the resistance

If current flowing is 4 amps and voltage is 24 volts. The formula to find the resistance will be :

R = 6 Ohms

Hence, the correct option is (d) " 24 volts ÷ 4 amps = 6 ohms ".

Answer:

Explanation:

The formula relating the speed of sound, frequency, and wavelength is:

speed = frequency x wavelength

Rearranging this formula to solve for frequency:

frequency = speed / wavelength

Substituting the given values:

frequency = 342 m/s / 1.8 m

frequency = 190 Hz

Therefore, the frequency of the sound wave is 190 Hz.

Answer:

Astronomers use computer models to simulate the process of galactic evolution through mergers and collisions. These models are based on our current understanding of the physical laws that govern the behavior of matter and energy in the universe. By running simulations of galactic mergers and collisions, astronomers can test their understanding of how these physical processes work in practice and how they contribute to the formation and evolution of galaxies.

One way that astronomers might test their understanding of the physical processes of the universe is by comparing the predictions of their models to observations of real galaxies. For example, if a model predicts that a particular type of galaxy should have a certain shape, size, or distribution of stars, astronomers can compare these predictions to observations of actual galaxies to see if they match up. If there is a discrepancy between the model's predictions and the observations, this can indicate that there are some physical processes that are not well understood or included in the model.

Another way that astronomers might test their understanding is by looking for patterns or trends in the properties of galaxies that are consistent with the predictions of their models. For example, if a model predicts that galaxies that have undergone a recent merger should have a particular distribution of gas and dust, astronomers can look for evidence of this pattern in observations of real galaxies. If they find that the predicted pattern is consistently observed in a large sample of galaxies, this can provide support for the model's predictions and the physical processes that it includes.

Overall, computer models of galactic evolution through mergers and collisions provide a powerful tool for astronomers to test their understanding of the physical processes of the universe. By comparing the predictions of their models to observations of real galaxies and looking for consistent patterns and trends, astronomers can refine their understanding of how galaxies form and evolve over time.

Mechanical energy to heat energy is collision,force x distance is work done,rubbing energy friction, stewardship is using energy wisely and nuclear energy can cause heat pollution.

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The speed of the wind is 20 miles per hour.

To solve this problem, we can use the formula:

Speed = Distance/Time

Let's assume that the speed of the wind is x miles per hour.

With the wind, the plane travels at a speed of 460 miles per hour. This means that its speed relative to the ground is the sum of its airspeed and the speed of the wind:

460 = Airspeed + x

Against the wind, the plane travels at a speed of 420 miles per hour. This means that its speed relative to the ground is the difference between its airspeed and the speed of the wind:

420 = Airspeed - x

We can solve this system of equations to find the airspeed of the plane:

460 = Airspeed + x

420 = Airspeed - x

Adding the two equations gives:

880 = 2Airspeed

Dividing both sides by 2 gives:

Airspeed = 440 miles per hour

Now that we know the airspeed of the plane, we can find the speed of the wind by substituting this value into one of the equations we obtained earlier:

460 = Airspeed + x

460 = 440 + x

x = 20

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The approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.

We can use the conservation of energy, where the potential energy stored in the compressed spring is converted into the kinetic energy of the stuntman as he is launched out of the cannon.

The potential energy stored in the spring is given by:

PE = (1/2)kx²

where k is the spring constant and x is the distance the spring is compressed.

PE = (1/2)(266 N/m)(5 m)² = 3325 J

This potential energy is then converted into kinetic energy:

KE = (1/2)mv²

where m is the mass of the stuntman and v is his velocity.

3325 J = (1/2)(55 kg)v²

v² = (2*3325 J) / 55 kg

v²  = 121 m²/s²

v = √(121 m²/s²) = 11 m/s

Therefore, the approximate velocity of the stuntman, once he has left the cannon, is 11 m/s.

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The mass attached to the pulley is Kx/g.

option 1.

Assuming there is no friction, the tension in the string will be constant throughout, and the net force on the pulley-mass system will be the weight of the mass.

Let the mass of the hanging weight be m, then the weight of the mass is mg.

The spring will be extended by the same distance x, as the string is inextensible, and it will exert a force of Kx in the upward direction.

Since the pulley is in equilibrium, the net force on the pulley-mass system must be zero.

Therefore, the tension in the string must be equal to the force exerted by the spring:

T = Kx

Using the fact that the tension in the string is equal to the weight of the mass, we can write:

mg = Kx

Therefore, the mass attached to the pulley is:

m = Kx/g

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Answer:

Explanation:

To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.

Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.

Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.

A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively. Since the electric field is perpendicular to the faces, the flux through them is zero. So, Q_in = (σ1 - σ2) * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C

The magnitude of the electric field at point A is 2.31 x 10^5 N/C.

B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.

The direction of the electric field at point A is RIGHT.

C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C

The magnitude of the electric field at point B is 3.77 x 10^7 N/C.

D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.

The direction of the electric field at point B is LEFT.

E:

Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q

The net charge enclosed within this Gaussian surface is:

Q = σ1 × (2πrh)

where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:

Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C

Using Gauss's law, we can find the electric field at point C:

E × (2πrh) = Q/ε0

where ε0 is the permittivity of free space. Solving for E, we get:

E = Q / (2πε0rh)

Plugging in the values, we get:

E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C

Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.

To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.

The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:

qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)

where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.

Substituting the given values, we get:

qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC

Using Gauss's law, we have:

E * 2πrh = qenc/ε0

where ε0 is the permittivity of free space.

Solving for E, we get:

E = qenc / (2πrhε0) = 2.22 × 10^4 N/C

Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.

F:

The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.

The net flux of an electric field in a closed surface is directly proportionate to the charge contained, according to Gauss' equation.

To use Gauss's Law, we need to choose a Gaussian surface that encloses the point of interest and has symmetry such that the electric field is constant over the surface. For all points in this problem, we can choose a cylinder as our Gaussian surface with its axis perpendicular to the sheets.

Let's assume that the cylinders are tall enough such that the electric field at the top and bottom faces of the cylinder is negligible. The electric flux through the curved part of the cylinder is constant and equal to Φ_E = E*A, where A is the surface area of the curved part of the cylinder.

Using Gauss's Law, Φ_E = Q_in / ε0, where Q_in is the net charge enclosed by the Gaussian surface and ε0 is the permittivity of free space.

A: The Gaussian surface is a cylinder with radius r = 5.00 cm and height h = the distance between the sheets (20.0 cm). The net charge enclosed is Q_in = σ1 * A_top + σ2 * A_bottom, where A_top and A_bottom are the areas of the top and bottom faces of the cylinder, respectively.

Φ_E = E * A = Q_in / ε0

E = (σ1 - σ2) / (ε0 * r) = (-7.30 μC/m^2 - 5.00 μC/m^2) / (8.85 x 10^-12 C^2/Nm^2 * 0.0500 m) = -2.31 x 10^5 N/C

The magnitude of the electric field at point A is 2.31 x 10^5 N/C.

B: The electric field points from higher potential to lower potential. Since the left-hand sheet has a negative charge density and the right-hand sheet has a positive charge density, the potential decreases from left to right. Thus, the electric field at point A points from left to right.

The direction of the electric field at point A is RIGHT.

C: The Gaussian surface is a cylinder with radius r = 1.25 cm and height h = the thickness of the right-hand sheet (10.0 cm). The net charge enclosed is Q_in = σ4 * A, where A is the surface area of the curved part of the cylinder. Thus,

Φ_E = E * A = Q_in / ε0

E = σ4 / (ε0 * r) = 4.00 μC/m^2 / (8.85 x 10^-12 C^2/Nm^2 * 0.0125 m) = 3.77 x 10^7 N/C

The magnitude of the electric field at point B is 3.77 x 10^7 N/C.

D: The electric field points from higher potential to lower potential. Since the right-hand sheet has a positive charge density, the potential decreases from the right-hand sheet to the left. Thus, the electric field at point B points from right to left.

The direction of the electric field at point B is LEFT.

E:Since point C is in the middle of the right-hand sheet, the electric field due to this sheet alone cancels out due to symmetry. Thus, the only electric field present is due to the left-hand sheet. The Gaussian surface is a cylinder with radius r = the radius of the sheet (10.0 cm) and height h = the thickness of the sheet (10.0 cm). The net charge enclosed is Q

The net charge enclosed within this Gaussian surface is:

Q = σ1 × (2πrh)

where h is the thickness of the left-hand sheet, r is the distance from the left-hand sheet to point C, and σ1 is the surface charge density of the left-hand sheet. Plugging in the given values, we get:

Q = (-7.30 × 10^-6 C/m^2) × (2π × 0.1 m × 0.1 m) = -4.60 × 10^-8 C

Using Gauss's law, we can find the electric field at point C:

E × (2πrh) = Q/ε0

where ε0 is the permittivity of free space. Solving for E, we get:

E = Q / (2πε0rh)

Plugging in the values, we get:

E = (-4.60 × 10^-8 C) / (2π × 8.85 × 10^-12 C^2/(N·m^2) × 0.1 m × 0.1 m) = -1.64 × 10^5 N/C

Therefore, the magnitude of the electric field at point C is 1.64 × 10^5 N/C.

To find the electric field at point C, we need to consider both sheets since point C is equidistant from both sheets. Thus, we can use Gauss's law to find the total electric field due to both sheets.

The net charge enclosed by a cylindrical Gaussian surface of radius r = 1.25 cm and height h = 20.0 cm is given by:

qenc = σ2 * (2πrh) + σ4 * (2πrh) = (σ2 + σ4) * (2πrh)

where σ2 is the charge density on the inner surface of the right-hand sheet, σ4 is the charge density on the outer surface of the left-hand sheet, and h is the distance between the two sheets.

Substituting the given values, we get:

qenc = (5.00 μC/m^2 + 4.00 μC/m^2) * (2π * 1.25 cm * 20.0 cm) = 628.32 nC

Using Gauss's law, we have:

E * 2πrh = qenc/ε0

where ε0 is the permittivity of free space.

Solving for E, we get:

E = qenc / (2πrhε0) = 2.22 × 10^4 N/C

Therefore, the magnitude of the electric field at point C is 2.22 × 10^4 N/C.

F:The direction of the electric field at point C is perpendicular to the surface of the sheet, pointing away from the positive charge density and towards the negative charge density. Since the positive charge density is on the outer surface of the left-hand sheet and the negative charge density is on the inner surface of the right-hand sheet, the direction of the electric field at point C is from left to right. Therefore, the direction of the electric field at point C is RIGHT.

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Answer:

Explanation:

Assuming that air resistance is negligible, we can use the following kinematic equations to solve for the peak height:

v_f^2 = v_i^2 + 2ad

where v_f = 0 m/s (at the peak height) and a = -9.8 m/s^2 (acceleration due to gravity)

and

d = v_i t + (1/2)at^2

where d is the displacement or the peak height we want to find, v_i is the initial velocity, t is the time it takes to reach the peak height.

First, we need to resolve the initial velocity into its vertical and horizontal components:

v_i_x = v_i cos(30°) = 121.1 m/s

v_i_y = v_i sin(30°) = 70.0 m/s

Next, we can use the vertical component of the initial velocity to find the time it takes to reach the peak height:

v_f = v_i_y + at

0 m/s = 70.0 m/s + (-9.8 m/s^2)t

t = 7.14 s

Finally, we can use the time we found and the kinematic equation for displacement to find the peak height:

d = v_i_y t + (1/2)at^2

d = (70.0 m/s)(7.14 s) + (1/2)(-9.8 m/s^2)(7.14 s)^2

d = 247.5 m

Therefore, the peak height of the football is 247.5 meters.

Answer:

Pulse transfers a single disturbance, while wave is a continuous disturbance that transfers energy.

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The voltage of the battery would be 20 volts. Option I.

According to Ohm's law, the voltage (V) across a resistor is equal to the current (I) flowing through it multiplied by its resistance (R). Mathematically,

V = I × R

In this case, the current (I) flowing through the resistor is given as 10 A and the resistance (R) of the resistor is given as 2 ohms. Substituting these values into the above formula, we get:

V = 10 A × 2 ohms = 20 volts

Therefore, the voltage of the battery is 20 volts.

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The truck's acceleration is 3.0m/s² and the momentum of the truck is  144000 kg m/s.

It is the rate at which the speed and direction of a moving object vary over time.

We can use the following equation to calculate the acceleration of the truck:

a = (v - u) / t

where

a = acceleration

v = final velocity = 18.0 m/s

u = initial velocity = 0 m/s (the truck starts from rest)

t = time taken = 6.00 s

Substituting the values, we get:

a = (18.0 m/s - 0 m/s) / 6.00 s

a = 3.00 m/s²

Therefore, the acceleration of the truck is 3.00 m/s².

We can use the following equation to calculate the momentum of the truck:

p = m * v

where

p = momentum

m = mass of the truck = 8000.0 kg

v = final velocity = 18.0 m/s

Substituting the values, we get:

p = 8000.0 kg * 18.0 m/s

p = 144000 kg m/s

Therefore, the momentum of the truck after it has reached its final velocity is 144000 kg m/s.

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Why the ocean near Christchurch is a different temperature than we'd expect for its latitude (distance from the equator)? Water moving from the equator is warmer than would be expected based on latitude, and so is warmer than the air it passes.

Changes to prevailing winds affect ocean currents. Changes to ocean currents affect how much energy is brought to (or taken away from) a location. In El Niño years, the prevailing winds that normally drive a warm current from the Equator past New Zealand are disrupted and may stop or even reverse.

Real, inverted, and same size are the features of the image. when A concave mirror with a curvature of 8 is displayed on a ruler with a range of 0 to 14 cm.

The mirror formula may be used to calculate the image distance for an item located 4 cm from a 1.5 cm focal length mirror.

1/f = 1/u+1/v

f is the focal length

u is the object distance

v is the image distance

Keep in mind that the concave mirror's image distance and focal length are both positive.

Given:

u = 4cm

f = 1.5cm

1/v = 1/1.5-1/4

1/v = 0.67-0.25

1/v = 0.42

v = 1/0.42

v = 2.38cm

The picture is Genuine and INVERTED since the image distance value is positive.

We shall find its magnification and see if it is magnified or lessened. It is amplified if the magnification is larger than 1, and it is decreased if it is less.

Magnification = v/u

Magnification = 2.38/4

Magnification = 0.595 or. 0.6

The picture is reduced in size since the magnification is less than one (SMALLER).

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Answer:The current in a lightbulb with a voltage of 35.0 V and a resistance of 175 ohm is 0.2 A.

Find the current in a lightbulb?

Given:

The voltage in a lightbulb is given by the equation V=IR

V is the voltage, I is current, and R is the resistance.

The voltage of the lightbulb is given as 35.0 V.

The resistance of the lightbulb is given as 175 Ohm.

As the equation is given,

V= IR

where I is current, R is resistance and V is the voltage.

Now, I = V/R

As the value of Voltage and resistance of the lightbulb is given, we will put in the above equation, we get;

I = 35.0/ 175 A

I = 0.2 A.

Hence, the current of the lightbulb is 0.2 A.

Therefore, Option C is the correct answer.

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Explanation:

The statement that correctly defines an evolutionary stage of a star like the sun is that after leaving the main sequence, its core is stable due to electron degeneracy. That is option C.

The stages of the life cycle of a star include the following:

The evolutionary stage is also called the main sequence stage of the life cycle of the star.

In this stage, the core temperature reaches the point for the fusion to occur whereby the protons of hydrogen are converted into atoms of helium. This leads to the stability of the core of the newly formed start due to electron degeneracy.

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Her average speed was about 4.6 km/hr, and her average velocity was about 3.3 km/hr.

The entire distance travelled divided by the total time taken is the definition of average speed. In this case, the total distance travelled was 7.0 km, and the total time taken was 1.5 hours. Hence, the average speed can be determined as follows:

Average Speed = [tex]\frac{7.0 km }{ 1.5 \ hours }= 4.6 km/hr[/tex]

The displacement divided by the whole time travelled is the average velocity. In this case, the displacement was 3.0 km (from Mary's home to Sheila's home), and the total time taken was 1.5 hours.The average velocity can therefore be determined as follows:

Average Velocity = [tex]\frac{3.0 km }{1.5 \ hours} = 3.3 km/hr[/tex]

Therefore,Her average velocity was roughly 3.3 km/hr, and her average speed was roughly 4.6 km/hr.

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Answer:

15 hours

Explanation:

formula: f(a) = a(0.5)^(T/t)

fill in known values: 13=208(0.5)^(60/t)

use natural log to isolate t:    ln(13/208)=ln(0.5)(60/t)

solve for t: t=15

It takes 0.1311 seconds to hear the echo of the stone.

First we need to determine the time it takes for the sound wave to travel from the stone to the bottom of the mine shaft and back up to our ears.

Let's start by finding the time it takes for the sound wave to reach the bottom of the mine shaft. We can use the formula:

time = distance / speed

The distance is the depth of the mine shaft, which is 15 meters. The speed of sound is 343 m/s, as given in the problem. Therefore, the time it takes for the sound wave to reach the bottom of the mine shaft is:

time = 15 m / 343 m/s

time = 0.0437 s

Now, we need to find the time it takes for the sound wave to travel back up to our ears. Since the sound wave travels at the same speed, 343 m/s, the distance it needs to cover is twice the depth of the mine shaft, or 30 meters. Therefore, the time it takes for the sound wave to travel back up to our ears is:

time = 30 m / 343 m/s

time = 0.0874 s

Finally, to find the total time it takes to hear the echo, we add the time it takes for the sound wave to reach the bottom of the mine shaft to the time it takes to travel back up to our ears:

total time = 0.0437 s + 0.0874 s

total time = 0.1311 s

Therefore, it takes 0.1311 seconds to hear the echo of the stone.

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Answer:

- 0.33 m/s

Explanation:

An illustration is shown above,

In this case, since the two objects move in opposite directions before collision, then move together, the formula to be used is,

m1u1 - m2u2 = (m1 + m2)v

Where,

m1 = mass of the first object

u1 = initial velocity of the first object

v1 = final velocity of the first object

m2 = mass of the second object

u2 = initial velocity of the second object

v2 = final velocity of the second object

Therefore,

(4.0 • 3.0) - (8.0 • 2.0) = (4.0 + 8.0)v

12 - 16 = 12v

-4 = 12v

Divide both sides by 12,

-4 / 12 = 12v / 12

-1 / 3 = v

v = -0.33 m/s

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Answer:

Explanation:

The time taken to travel from Fort Worth to Dallas is:

t = 3:23 pm - 2:41 pm = 42 minutes = 0.7 hours

The distance covered is:

d = 58 km

The average speed is:

v = d/t = 58 km / 0.7 hours = 82.86 km/h

To convert km/h to m/s, we can use the conversion factor:

1 km/h = 0.2778 m/s

Therefore, the average speed in m/s is:

v = 82.86 km/h × 0.2778 m/s/km = 23.06 m/s (rounded to two decimal places)

So the average speed is 23.06 m/s.

Answer: b. force per unit area.

Explanation: