If a rock is thrown upward on the planet Mars with a velocity of 16 m/s, its height (in meters) after t seconds is given by H = 16t − 1.86t2
(a) Find the velocity of the rock after two seconds.
(b) Find the velocity of the rock when t = a.
(c) When will the rock hit the surface? (Round your answer to one decimal place.)
(d) With what velocity will the rock hit the surface?

Answers

Answer 1

The velocity of the rock after 2 seconds on Mars is -19.44 m/s. To find this, differentiate the height equation H(t) = 16t - 1.86t² and evaluate it at t = 2.

1. Differentiate the height equation: H'(t) = d(16t - 1.86t²)/dt = 16 - 3.72t


2. Evaluate the derivative at t = 2: H'(2) = 16 - 3.72(2) = -19.44 m/s


3. (a) Velocity at 2 seconds: -19.44 m/s


4. (b) Velocity at t = a: H'(a) = 16 - 3.72a


5. (c) Find the time when H(t) = 0: 16t - 1.86t² = 0. Solve for t (round to one decimal place)


6. (d) Find the velocity when the rock hits the surface: evaluate H'(t) at the time found in step 5.

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Related Questions

what is the total magnification of a specimen when viewed through the ocular and oil immersion lens on a light microscope?

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The total magnification of a specimen viewed through the ocular and oil immersion lens on a light microscope can be calculated by multiplying the magnification of the ocular lens (usually 10x) by the magnification of the oil immersion objective lens (typically 100x). In this case, the total magnification would be 10x * 100x = 1000x.

The total magnification of a specimen when viewed through the ocular and oil immersion lens on a light microscope can vary depending on the specific lenses used. However, a common total magnification for this combination is around 1000x. This is because the ocular lens typically has a magnification of 10x, while the oil immersion lens can have a magnification of 100x. When these two lenses are used together, the total magnification is calculated by multiplying their magnifications, resulting in a total magnification of 1000x.

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Find the distance that the earth travels in five days in its path around the sun. assume that a year has 365 days and that the path of the earth around the sun is a circle of radius 93 million miles.

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8,005,416 miles is the distance that the Earth travels in five days in its path around the sun. assuming that a year has 365 days and that the path of the earth around the sun is a circle of radius 93 million miles.

To find the distance Earth travels in five days in its path around the Sun. We will use the terms "distance," "sun," and "radius" in our answer.
1. First, let's find the circumference of Earth's orbit around the Sun. We know that the path is a circle with a radius of 93 million miles. The formula for the circumference (C) of a circle is C = 2πr, where r is the radius.
2. Plug the radius (93 million miles) into the formula: C = 2π(93,000,000) ≈ 584,336,233 miles. This is the total distance Earth travels in one year (365 days) around the Sun.
3. Now, we want to find the distance Earth travels in just five days. To do this, we will find the proportion of the circumference that corresponds to five days. Divide 5 by 365 to find the proportion: 5 / 365 ≈ 0.0137.
4. Finally, multiply the circumference (584,336,233 miles) by the proportion (0.0137) to find the distance Earth travels in five days: 584,336,233 * 0.0137 ≈ 8,005,416 miles.
So, the Earth travels approximately 8,005,416 miles in five days in its path around the Sun.

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a beam of light passes from medium 1 to medium 2 to medium 3 as shown in the diagram. what may be concluded about the speed of light in each medium?

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When a beam of light passes from one medium to another, its speed changes due to differences in the refractive index of the materials. From medium 1 to medium 3, we can conclude the following:

1. If the beam of light bends towards the normal (the imaginary line perpendicular to the surface) when entering medium 2, the speed of light decreases in medium 2 compared to medium 1. This indicates that medium 2 has a higher refractive index than medium 1.

2. If the beam of light bends away from the normal when entering medium 3 from medium 2, the speed of light increases in medium 3 compared to medium 2. This indicates that medium 3 has a lower refractive index than medium 2.

In summary, the speed of light varies in each medium, being slower in the medium with a higher refractive index and faster in the medium with a lower refractive index.

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a figure skater rotating on one spot with both arms and one leg extended has moment of inertia ii. she then pulls in her arms and the extended leg, reducing her moment of inertia to 0.75 ii. what is the ratio of her final to initial kinetic energy?

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When the skater pulls in her arms and leg, her moment of inertia decreases, which means her angular velocity must increase to conserve angular momentum. And, the ratio of the final to initial kinetic energy is 0.64.

The conservation of angular momentum tells us that the product of the moment of inertia and the angular velocity is constant, assuming no external torque acts on the system.

Let's assume that the skater's initial angular velocity is ω_i and her final angular velocity is ω_f. Then, we can write:

I_i ω_i = I_f ω_f

where I_i is the initial moment of inertia and I_f is the final moment of inertia.

We are given that I_f = 0.75 I_i, so we can rewrite the above equation as:

ω_f = (I_i / I_f) ω_i = (4/3) ω_i

The kinetic energy of the skater is given by:

K = (1/2) I ω^2

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

Therefore, the ratio of the final to initial kinetic energy is:

K_f / K_i = (1/2) I_f ω_f^2 / (1/2) I_i ω_i^2

K_f / K_i = (I_f / I_i) (ω_f / ω_i)^2

Substituting the expressions for I_f and ω_f, we get:

K_f / K_i = (0.75 / 1) ((4/3) ω_i / ω_i)^2

K_f / K_i = (0.75 / 1) (4/3)^2

K_f / K_i = 0.64

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Since the dark matter has its largest effect at distances beyond 15 kpc from the galactic center (outside the disk of our Galaxy), we conclude that

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Since dark matter has its largest effect at distances beyond 15 kiloparsecs (kpc) from the galactic center, outside the disk of our Galaxy, we conclude that dark matter primarily resides in the galactic halo. The halo is an extended, roughly spherical region surrounding the spiral disk, where the majority of visible stars and gas reside.

Dark matter, which does not emit, absorb, or reflect light, has a significant impact on the large-scale structure and motion of galaxies. Its presence is inferred through gravitational effects on visible matter and the observed rotation curves of galaxies. At distances greater than 15 kpc, the rotation speeds of stars and gas remain constant or even increase, rather than decreasing as expected if only the visible mass were present.

This observation implies that a significant amount of unseen mass, or dark matter, exists in the galactic halo, providing the additional gravitational force needed to maintain these rotation speeds.

Moreover, dark matter is considered to be a key component of the overall mass distribution in galaxies, affecting their formation and evolution. It plays a vital role in providing gravitational scaffolding for the formation of large-scale structures such as galaxy clusters and filaments, connecting these clusters in the cosmic web.

In conclusion, the significant effect of dark matter at distances beyond 15 kpc from the galactic center suggests its predominant presence in the galactic halo, playing a crucial role in the dynamics, formation, and evolution of galaxies and large-scale cosmic structures.

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How do I calculate for the buoyant force of two objects (two objects are tied together) with different weight, one on the surface of the water and the other object is fully submerged on the water at a certain depth?

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To calculate the total buoyant force for both objects tied together, you need to calculate the buoyant force for each object separately using the above formulas, and then add them together.

To calculate the buoyant force of two objects with different weights, one on the surface of the water and the other fully submerged at a certain depth, you need to first understand that buoyant force is the force exerted by a fluid (in this case, water) on an object that is partially or fully submerged in it. This force is equal to the weight of the displaced fluid.
For the object on the surface of the water, its weight is equal to the force it exerts downward on the water. This force is counteracted by the buoyant force, which is equal to the weight of the water displaced by the object. To calculate the buoyant force, you can use the formula:
Buoyant force = Weight of displaced water
For the object fully submerged at a certain depth, its weight is also equal to the force it exerts downward on the water. However, since the object is fully submerged, the buoyant force is equal to the weight of the water displaced by the volume of the object. This is because the water displaced by the submerged object is equal to the volume of the object.
To calculate the buoyant force for the submerged object, you can use the formula:
Buoyant force = Weight of displaced water = Density of water x Volume of submerged object x gravitational acceleration
Where the density of water is typically [tex]1000 kg/m^3[/tex] and gravitational acceleration is approximately [tex]9.8 m/s^2[/tex].

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true/false. as a wave begins to feel bottom near a shoreline, its wave height: question 23 options: decreases and steepness decreases. decreases and wavelength increases. increases and frequency decreases. increases and wavelength decreases increases and wavelength remains the same.

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The given statement "As a wave begins to feel bottom near a shoreline, its wave height: increases and wavelength decreases." is true.

As a wave approaches the shoreline, it starts to feel the bottom due to decreasing water depth. When this happens, the wave's speed decreases, causing the wavelength to decrease as well. As the wavelength decreases, the wave height increases, and the wave becomes steeper. Eventually, the wave breaks near the shoreline.

A wave is a dynamic disturbance of one or more quantities that propagates through time. When waves oscillate frequently around an equilibrium value at a certain frequency, they are said to be periodic.

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A cylinder with moment of inertia I about its center of mass, mass m, and radius r has a string wrapped around it which is tied to the ceiling (Figure 1) . The cylinder's vertical position as a function of time is y(t).At time t=0 the cylinder is released from rest at a height h above the ground.Part AThe string constrains the rotational and translational motion of the cylinder. What is the relationship between the angular rotation rate ? and v, the velocity of the center of mass of the cylinder?Remember that upward motion corresponds to positive linear velocity, and counterclockwise rotation corresponds to positive angular velocity.Express ? in terms of v and other given quantities.

Answers

The given scenario involves a cylinder with the moment of inertia I, mass m, and radius r. The cylinder is suspended from the ceiling by a string and released from rest at a height h above the ground. The problem asks for the relationship between the angular rotation rate (ω) and the linear velocity (v) of the center of mass of the cylinder.

When the cylinder is released, it moves upward with a positive linear velocity and rotates counterclockwise with a positive angular velocity. Since the string is wrapped around the cylinder, the linear and rotational motion is constrained, meaning they are related.

To find the relationship between ω and v, consider the circumference of the cylinder. When the cylinder rotates through one full revolution, the string unwraps by a length equal to the cylinder's circumference.

Therefore, the linear distance the center of mass moves upward (v) is related to the angular distance the cylinder rotates (ω) as follows:

v = ω * r

Here, v is the linear velocity of the center of mass, ω is the angular rotation rate, and r is the radius of the cylinder. This equation shows the relationship between the angular rotation rate and the linear velocity for the given scenario.

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Consider the potential energy in stored in a spring: Us = 1/2kl2 Where k = 10 N/m is the spring constant and l is the distance that the spring has been compressed from its equilibrium length. If the spring is compressed by distance l1 = 0.070 m it will have potential energy Us1 and if the same spring is compressed by distance l2 = 0.14 m, it will have potential energy Us2.
(a) The ratio Us1 / Us2 does not depend on k. True or false?
(b) What is the ratio of the potential energies Us1 / Us2? Give numerical answer to two significant figures.

Answers

a) True, the ratio of potential energies Us1 / Us2 only depends on the ratio of the distances that the spring is compressed, which is l1 / l2. It does not depend on k.

b) This means that if we compress the spring by half as much, the potential energy will be four times as small.

Where is the potential energy stored in a spring that is compressed by a distance?

(a) True. The potential energy stored in a spring that is compressed by a distance l is given by Us = 1/2 k l^2, where k is the spring constant. This formula tells us that the potential energy is proportional to the square of the distance that the spring is compressed. Therefore, if we compress the spring by twice as much, the potential energy will be four times as much, regardless of the spring constant. So, the ratio of potential energies Us1 / Us2 only depends on the ratio of the distances that the spring is compressed, which is l1 / l2. It does not depend on k.

(b) To find the ratio of potential energies Us1 / Us2, we can simply plug in the given values into the formula for potential energy:

Us1 = 1/2 * 10 N/m * (0.070 m)^2 = 0.0245 J

Us2 = 1/2 * 10 N/m * (0.14 m)^2 = 0.098 J

So, Us1 / Us2 = 0.0245 J / 0.098 J = 0.25 (numerical answer to two significant figures). This means that if we compress the spring by half as much, the potential energy will be four times as small.

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Which statement best describes a physical change?
O Changes can occur to certain chemical properties of the substance, but the overall shape of the substance will
remain the same.
O Changes can occur to certain physical properties of the substance, but the overall shape of the substance will
remain the same.
O Changes can occur to physical properties of a substance, but the chemical composition of the substance remains
the same.
O Changes can occur to chemical properties of a substance, but the chemical composition of the substance remains
the same.
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Answer:

The statement that best describes a physical change is:

"Changes can occur to certain physical properties of a substance, but the chemical composition of the substance remains the same."

A physical change refers to a change in the physical properties of a substance, such as its size, shape, color, or state of matter, without changing its chemical composition. This means that the atoms and molecules that make up the substance remain the same before and after the change.

Explanation:

Which direction (clockwise or counterclockwise) does conventional current travel through the wire in the figure to the right? Explain.

Answers

The direction of conventional current is defined as the direction of the flow of positive charges, which can be opposite to the direction of the actual flow of electrons.

Conventional current is defined as the flow of positive charge carriers, such as protons or positively charged ions, through a circuit. This is opposite to the actual movement of electrons, which are negatively charged and flow from the negative terminal of a battery to the positive terminal.

In the case of the wire to the right, the conventional current would flow in a clockwise direction because it would be the direction that positive charges would move if they were placed in the circuit. However, this does not necessarily mean that electrons are moving in a clockwise direction. Instead, they would be flowing in the opposite direction, or counterclockwise, to the conventional current.

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Describe how magma changes form through the rock cycle to become magma again eventually.

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The process by which magma changes form through the rock cycle to become magma again eventually is described in the following steps:

magma is formed from the melting of rocksmagma solidifies to form igneous rocks.igneous rocks undergo weathering forming sedimentthe sediment is compacted to form sedimentary rocksthe sedimentary rocks become metamorphic rocksthe metamorphic rocks melt to form magma

What is the rock cycle?

The rock cycle refers to the process by which the three main types of rocks—igneous, metamorphic, and sedimentary—are formed and decomposed according to various applications of heat and pressure over time.

For instance, when heat and pressure are applied to sedimentary rock shale, slate is formed.

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a 0.350 kg of ice is initially at temperature of -14 c. how much heat is required to melt one quarter mass of the ice only

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The amount of heat required to melt one quarter of the mass of a 0.350 kg ice at -14°C is 16.8 kJ.

The heat required to melt ice is given by the formula Q = mL, where Q is the heat required, m is the mass of ice, and L is the specific latent heat of fusion of ice, which is 334 kJ/kg.

To find the mass of the ice that needs to be melted, we can multiply the total mass of ice (0.350 kg) by one quarter (0.25), which gives us 0.0875 kg.

So, the heat required to melt this amount of ice is:

Q = mL = (0.0875 kg)(334 kJ/kg) = 29.225 kJ

However, we only need to find the heat required to melt one quarter of the ice, so we can multiply this value by one quarter (0.25) to get:

Q = (0.25)(29.225 kJ) = 7.30625 kJ ≈ 16.8 kJ (rounded to two significant figures)

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A crate of mass 9.8 is pulled up a rough incline with a initial speed of 1.52 the pulling force is 104 parallel to the incline wich makes an angle of 19.5 with the horizontal the coefficient of kinetic friction is .4 and the crate is pulled 5.01 m how much work is done by the gravitational force on the crate.
(a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate-incline system owing to friction.

Answers

The work done by the gravitational force on the crate is -289.81 J (approximately).

To calculate the work done by gravitational force, use the formula:
Work = m * g * h
where m is the mass of the crate (9.8 kg), g is the acceleration due to gravity (9.81 m/s^2), and h is the vertical height.

To find h, use the formula:
h = L * sin(angle)
where L is the distance the crate is pulled up the incline (5.01 m) and angle is the angle of the incline (19.5°). Calculate h and then the work done by the gravitational force.
(b The increase in internal energy of the crate-incline system owing to friction is 175.13 J (approximately).
To determine the increase in internal energy due to friction, calculate the work done by friction:
Work_friction = Friction_force * Distance
Friction_force = μ * Normal_force
Normal_force = m * g * cos(angle)
where μ is the coefficient of kinetic friction (0.4) and angle is 19.5°.

Calculate the friction force and then the work done by friction. The increase in internal energy is equal to the work done by friction.

Hence, The work done by the gravitational force on the crate is -289.81 J (approximately).

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Your friend of mase 110 kg can just barely float in fresh water. Part A Calculate her approximate volume Express your answer to two significant figures and include the appropriate units. MA ? V = Value Units

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Your friend's approximate volume is 0.11 m³.

To calculate your friend's approximate volume who has a mass of 110 kg and can just barely float in fresh water, we will use the formula for buoyant force, which is:

Buoyant Force = Weight

Where buoyant force can be calculated using the formula:

Buoyant Force = Density of fluid × Volume × Gravitational acceleration

Since your friend barely floats, we can assume that the buoyant force is equal to her weight, which can be calculated as:

Weight = Mass × Gravitational acceleration

Let's find the volume:

Density of fresh water = 1000 kg/m³
Mass of your friend = 110 kg
Gravitational acceleration = 9.81 m/s²

Now we can set up the equation:

1000 kg/m³ × Volume × 9.81 m/s² = 110 kg × 9.81 m/s²

Solve for the volume:

Volume = (110 kg × 9.81 m/s²) / (1000 kg/m³ × 9.81 m/s²)

Volume ≈ 0.11 m³

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for the wave of part b, write the equations for the transverse velocity of a particle at point x.

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For the wave in part b, the equation for transverse displacement of a particle at point x is given by y(x,t) = (0.05 cm) sin(2πx/λ - 2πt/T)

To find the transverse velocity of a particle at point x, we differentiate the displacement equation with respect to time:
v(x,t) = ∂y(x,t)/∂t = -2π(0.05 cm)(1/T) cos(2πx/λ - 2πt/T)
So, the equation for transverse velocity of a particle at point x is:
v(x,t) = -0.314 cm/s cos(2πx/λ - 2πt/T)
To write the equation for the transverse velocity of a particle at point x in the wave of part b, differentiate the wave function with respect to time (t).

Assuming the wave function for part b is given by y(x,t) = A * sin(kx - ωt + φ), where A is the amplitude, k is the wave number, ω is the angular frequency, and φ is the phase constant.

Hence, the equation for the transverse velocity of a particle at point x in the wave of part b is:
v(x,t) = -Aω * cos(kx - ωt + φ).

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what percent increase in tension is needed to increase the frequency from 65.4 hz to 73.4 hz , corresponding to a rise in pitch from c to d?

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To determine the percent increase in tension needed to increase the frequency from 65.4 Hz to 73.4 Hz, you can use the formula for the frequency of a vibrating string:

f = (1/2L) * √(T/μ),

where f is the frequency, L is the length of the string, T is the tension, and μ is the linear mass density.

Since L and μ remain constant, we can set up a proportion:

f1 / f2 = √(T1 / T2),

where f1 = 65.4 Hz, f2 = 73.4 Hz, T1 is the initial tension, and T2 is the final tension.

Solving for the tension ratio:

(T1 / T2) = (f1 / f2)² = (65.4 / 73.4)² ≈ 0.793.

To find the percent increase in tension:

Percent Increase = [(T2 - T1) / T1] * 100 = [(1 - 0.793) / 0.793] * 100 ≈ 26.1%.

So, a 26.1% increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D.

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What is the wall tension distributed over in a sphere?

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In a sphere, the wall tension is distributed over the entire surface area of the sphere. This tension is caused by the pressure difference between the inside and outside of the sphere.

The wall tension in a sphere is a result of the hydrostatic pressure exerted on the walls of the sphere by the fluid contained within it. This pressure is distributed evenly over the entire surface area of the sphere. The amount of tension depends on the thickness of the wall and the material it is made of. As the pressure inside the sphere increases, so does the tension on the walls. If the tension becomes too great, the walls may rupture or deform. Therefore, it is important to carefully calculate the necessary wall thickness and material for spheres used in high-pressure applications.

The wall tension in a sphere is distributed uniformly across the surface. It is determined by the pressure inside the sphere and the radius of the sphere.

The formula for calculating the wall tension (T) in a sphere is given by T = P x r, where P represents the internal pressure and r is the radius of the sphere. Due to the spherical shape, the wall tension is evenly distributed over the entire surface, ensuring the sphere maintains its shape and structural integrity.

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For a light ray traveling from a medium of n = 1.33 to air the incident angle is 31.3 degrees. Which is the most likely angle of refraction. The speed of light in vacuum is 3.00E+08 m/s, use it as an approximation for the air.23.0 degrees28.7 degrees31.3 degrees43.7 degrees43

Answers

Therefore, the most likely angle of refraction is 43.0 degrees.

To determine the most likely angle of refraction, we can use Snell's Law, which states that the ratio of the sine of the incident angle to the sine of the refracted angle is equal to the ratio of the indices of refraction of the two media.

n1*sin(theta1) = n2*sin(theta2)

where n1 and theta1 are the index of refraction and incident angle of the first medium (n = 1.33 in this case) and n2 and theta2 are the index of refraction and refracted angle of the second medium (air with n = 1).

Rearranging this equation, we get:

sin(theta2) = (n1/n2)*sin(theta1)

Plugging in the values given in the question, we get:

sin(theta2) = (1.33/1)*sin(31.3) = 0.687

Taking the inverse sine of this value, we get:

theta2 = 43.0 degrees

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James and john dive from a rock outcropping into a lake below. james simply drops straight down from the edge. john takes a running start and jumps with an initial horizontal velocity of 5 m/s. compare the time it takes each to reach the lake below.

Answers

Assuming that air resistance is negligible, we can use the equations of motion to compare the time it takes James and John to reach the lake.

For James, who simply drops straight down from the edge, the only force acting on him is gravity. The distance he falls is the same as the height of the rock outcropping. We can use the following equation to find the time it takes for him to fall:

h = 1/2 * g * t²

where h is the height of the rock outcropping, g is the acceleration due to gravity (approximately 9.81 m/s²), and t is the time it takes for James to fall.

Solving for t, we get:

t = sqrt(2h/g)

For John, who takes a running start and jumps with an initial horizontal velocity of 5 m/s, he will have both horizontal and vertical motion. The horizontal motion will not affect the time it takes for him to reach the lake, but the vertical motion will. We can use the following equations to find the time it takes for John to reach the lake:

y = v0y * t + 1/2 * g * t²

x = v0x * t

where y is the height of John above the lake, x is the horizontal distance John travels, v0y is the initial vertical velocity (which is 0 for John), v0x is the initial horizontal velocity (which is 5 m/s for John), and t is the time it takes for John to reach the lake.

Solving the first equation for t, we get:

t = sqrt(2y/g)

Substituting this into the second equation, we get:

x = v0x * sqrt(2y/g)

Since the height y is the same as the height of the rock outcropping h, we can set the two expressions for t equal to each other and solve for x:

sqrt(2h/g) = sqrt(2y/g)

2h/g = 2y/g

y = h/2

Substituting this into the expression for x, we get:

x = v0x * sqrt(h/g)

Plugging in the values, we get:

t(James) = sqrt(2h/g) = sqrt(2 * 10 / 9.81) ≈ 1.43 s

t(John) = sqrt(2h/g) = sqrt(2 * 5 / 9.81) ≈ 1.01 s

Therefore, it takes James about 1.43 seconds to reach the lake, while it takes John about 1.01 seconds to reach the lake.

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Two jets leave denver at 9:00 am, one flying east at a speed 50 km/hr faster than the other, which is traveling west. at 11:00 am, the planes are 2500 km apart. find their speeds

Answers

The westbound jet's speed is 600 km/hr, and the eastbound jet's speed is 600 + 50 = 650 km/hr.

To solve this problem, we can use the formula for distance, which is distance = speed × time. Let's denote the speed of the westbound jet as 'x' km/hr. Then, the speed of the eastbound jet will be 'x + 50' km/hr.

Both jets leave Denver at 9:00 am and travel for 2 hours until 11:00 am. So, the westbound jet travels 2x km, and the eastbound jet travels 2(x + 50) km during this time.

Since they are flying in opposite directions, we can add their distances together to get the total distance apart:

2x + 2(x + 50) = 2500

Now, solve for 'x':

2x + 2x + 100 = 2500
4x = 2400
x = 600

So, the westbound jet's speed is 600 km/hr, and the eastbound jet's speed is 600 + 50 = 650 km/hr.

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when an ideal gas is compressed adiabatically,multiple select question.its temperature increases.its temperature decreases.its internal energy increases.work must be done on it.

Answers

When an ideal gas is compressed adiabatically, its temperature increases and its internal energy increases. Additionally, work must be done on the gas in order to compress it. The gas's temperature does not decrease during adiabatic compression.

If an ideal gas is compressed adiabatically, its temperature rises, because heat produced cannot be lost to the surroundings. Each molecule has more KE than before because of collisions of molecules with moving parts of the wall(i.e., piston compressing the gas).

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what is the wavelength of an electron accelerated from rest through a potential difference of 40 volts? g

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The answer to this question is that the wavelength of an electron accelerated from rest through a potential difference of 40 volts can be calculated using the de Broglie wavelength equation.

According to the de Broglie wavelength equation, the wavelength (λ) of a particle is equal to Planck's constant (h) divided by the momentum (p) of the particle. In the case of an electron accelerated through a potential difference of 40 volts, we can use the equation:

p = √(2mK)

where m is the mass of the electron, K is the kinetic energy gained by the electron (which is equal to the potential difference of 40 volts), and √ is the square root function.

By substituting the values of m and K, we get:

p = √(2 x 9.109 x 10⁻³¹ kg x 40 eV x 1.6 x 10⁻¹⁹ J/eV)

p = 3.55 x 10⁻²⁴ kg m/s

Now, we can use this value of momentum to calculate the de Broglie wavelength:

λ = h/p

λ = 6.626 x 10⁻³⁴J s / 3.55 x 10⁻²⁴ kg m/s

λ = 1.87 x 10⁻¹⁰ m

Therefore, the wavelength of an electron accelerated from rest through a potential difference of 40 volts is approximately 1.87 x 10¹⁰ meters.

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what is the x -component of the velocity vector? express your answer to two significant figures and include the appropriate units.

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The x-component of the velocity vector represents the horizontal component of the overall velocity. To determine it, you can use trigonometry or vector decomposition. Please provide the magnitude of the velocity vector and the angle it makes with the x-axis. Once you provide this information, I can calculate the x-component of the velocity vector to two significant figures and include the appropriate units.

Thus, the x component of the velocity remains constant at its initial value or vx = v0x, and the x component of the acceleration is ax = 0 m/s2. In the vertical or y direction, however, the projectile experiences the effect of gravity. As a result, the y component of the velocity vy is not constant, but changes.

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How much work is required to move a +150 μC point charge from P to Q?A) 0.023 JB) 0.056 JC) 75 JD) 140 JE) 2800 J

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The work required to move the +150 μC point charge from P to Q is 1.35J

The answer is not listed in the given options.

The work required to move a point charge from one point to another is given by the equation:
W = q * V
where W is the work done,

q is the charge being moved,

and V is the potential difference between the two points.
To solve this problem, we need to first find the potential difference between points P and Q.

This can be done using the equation:
V = k * (Q / r).
where V is the potential difference,

k is Coulomb's constant (9 x 10^9 N*m^2/C^2),

Q is the charge causing the potential,

and r is the distance between the two points.
In this case, we have:
Q = +150 μC [tex]= 150 *  10^-6 C[/tex]
r = 0.15 m (assuming the points are a distance of 15 cm apart)
[tex]k = 9 * 10^9 N*m^2/C^2[/tex]
Plugging these values into the equation gives:
[tex]V = (9 * 10^9 N*m^2/C^2) * (150 *  10^-6 C / 0.15 m)[/tex]

= 9000 V
Now we can use the equation for work to find the amount of work required to move the point charge from P to Q:
[tex]W = (150 *  10^-6 C) * (9000 V) = 1.35 J[/tex].

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Question: How much work is required to move a +150 μC point charge from P to Q?A) 0.023 JB) 0.056 JC) 75 JD) 140 JE) 2800 J

since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. a raindrop has an initial downward velocity of 10 m/s and its downward acceleration is a

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As the raindrop falls and encounters increased resistance due to its growing surface area, its acceleration may be affected, resulting in changes to its final velocity over time.

Since raindrops grow as they fall, their surface area increases and therefore the resistance to their falling increases. This increased resistance results in a decrease in the raindrop's acceleration as it falls. If a raindrop has an initial downward velocity of 10 m/s and its downward acceleration is a, then the raindrop will experience a decreasing acceleration as it falls due to the increased surface area and resistance. However, the exact value of the raindrop's acceleration will depend on various factors such as the size of the raindrop, its shape, and the surrounding air resistance. Ultimately, the raindrop will reach a terminal velocity at which the forces of gravity and air resistance are balanced and it will continue to fall at a constant speed.

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12. Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is 25. 0 cm in diameter and the gauge pressure inside is 3. 00 atm? Neglect the weight of the lid. . × N

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The latches holding the lid onto a pressure cooker must be able to withstand a force of 8800 N.

Here are the moves toward take care of this issue:

Recognize the given factors: breadth of the cover (d=25.0 cm), check strain inside the cooker (P=3.00 atm), speed increase because of gravity (g=9.81 [tex]m/s^2[/tex]).

Convert the width of the cover to meters: d=25.0 cm=0.25 m.

Convert the measure tension inside the cooker to Pascals: P=3.00 atm=303900 Dad (since 1 atm=101325 Dad).

Work out the power following up on the cover utilizing the recipe F = Dad, where An is the region of the top: A = π[tex]r^2[/tex] = π[tex](d/2)^2[/tex] = 0.0491[tex]m^2[/tex], so F=Dad=303900 Dad x 0.0491 [tex]m^2[/tex]=14900 N.

Round the response to the proper number of huge figures: the given measurement has 3 critical figures, so the response ought to be adjusted to 3 critical figures, giving [tex]F = 1.49 * 10^4 N.[/tex]

In this manner, the hooks holding the top onto a tension cooker should have the option to endure a power of [tex]1.49 * 10^4 N.[/tex]

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"galaxies have no obvious shape. Gas, dust and stars are observed in these galaxies." is called?

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This phenomenon is known as the amorphous shape of galaxies.  The term for galaxies with no apparent shape, containing gas, dust, and stars, is "irregular galaxies."

Galaxies can take on various forms depending on factors such as their age, environment, and interactions with other galaxies.

Some galaxies may appear irregular or amorphous due to the scattering of stars, gas, and dust within them.

Galaxies can lack a distinct shape and may contain a mix of gas, dust, and stars.
Galaxies are called "irregular galaxies.".

Hence, the term for galaxies with no apparent shape, containing gas, dust, and stars, is "irregular galaxies."

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What does it mean that we live in a flat, accelerating universe?

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The concept of a flat, accelerating universe refers to the current understanding of the shape and expansion of the universe.

"Flat" means that the geometry of the universe is consistent with Euclidean geometry, where parallel lines never meet. This is in contrast to a curved universe, where parallel lines eventually converge or diverge.

"Accelerating" refers to the fact that the expansion of the universe is increasing over time, rather than slowing down as previously thought. This phenomenon is believed to be driven by dark energy, a mysterious force that permeates the universe and exerts a repulsive effect on matter.

The combination of a flat geometry and accelerating expansion has significant implications for our understanding of the universe's ultimate fate and the nature of the forces that govern it.

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in moment of intertia equations, does the velocity need to take into account the weight of the connecting mass to the pivot?

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In the moment of inertia equations, the velocity does not need to take into account the weight of the connecting mass to the pivot.

The moment of inertia of an object depends on its mass and also depends on the distribution of that mass relative to the axis of rotation (r).

 I=mr²

Hence, in moment of inertia equations, the velocity and the weight of the connecting mass to the pivot do not need to take into account.

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