which of the following best describes acceleration

A. Acceleration is the rate of change of position of an object.

B. Acceleration is the rate of change of the direction of an object.

C. Acceleration is the rate of change of force of an object.

D. Acceleration is the rate of change of velocity of an object.

Jumping, throwing, and weight lifting Olympic events would be drastically altered by the Moon's lower gravity in a hypothetical lunar Olympics. Athletes could jump higher and throw further due to gravity being 1/6th of earth's. Similarly, weights would feel lighter during weight lifting.

In the context of the lunar olympics, it's critical to understand how the Moon's lower gravity would drastically impact certain sports events. The Moon's gravity is about 1/6th of Earth's gravity, which would significantly influence events involving lifting, jumping and throwing.

Jumping events would drastically change because the lower gravity would allow participants to jump about six times higher than on Earth. Throwing events would also see remarkable enhancements as objects would travel a greater distance before dropping.

Moreover, Weight lifting would be greatly impacted as the same weights would feel much lighter on the Moon. Interestingly, sports such as bicycling, swimming on a level track, and running races on a level track would not be significantly affected because these activities are less dependent on gravity relative to the others mentioned. Nonetheless, athletes might perform differently due to the reduced atmospheric pressure and possible difficulties in maintaining balance because of lower gravity.

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Final answer:

The lunar olympics would see drastic changes in events such as jumping, throwing, and weight lifting due to the Moon's 1/6th Earth gravity, while bicycling would be less affected. Swimming would also change, but it assumes the presence of water in a lunar environment.

Explanation:

The "lunar olympics" concept highlights how sports would be drastically affected by the Moon's reduced gravity. Given that gravity on the Moon is about 1/6th of that on Earth, events such as jumping, throwing things, and weight lifting would be greatly altered. With lower gravity, athletes could jump much higher, throw objects much farther, and lift heavier weights with less effort, compared to Earth. Running races might see a difference in athletes' strides and suspension period off the ground, potentially leading to faster times. However, activities like bicycling would be less affected as they rely more on the rider's strength and endurance than on gravitational forces. Swimming is not practical without substantial water, which is currently not present on the Moon, but if somehow managed in a dome, the reduced gravity could affect water dynamics and a swimmer's buoyancy, thereby affecting the execution of strokes and turns.

White Dwarfs are less luminous than the sun, but with hotter surface temperatures. Therefore, option (B) is correct.

A white dwarf can be described as a stellar core remnant made mostly of electron-degenerate matter. A white dwarf is dense because its mass is comparable to the Sun's and volume is comparable to the Earth's.

A white dwarf has faint luminosity which comes from the emission of residual thermal energy and no fusion occurs in a white dwarf. There are currently eight white dwarfs among the 100 star systems nearest the Sun.

The material in a white dwarf has no fusion reactions, so there is no source of energy. It cannot support itself by the energy generated by fusion but is supported by electron degeneracy pressure.

A carbon-oxygen white dwarf approaches this mass limit by mass transfer from a companion star and may explode as a kind Ia supernova via a process called carbon detonation.

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the answer should be:

When the buoyant force is equal to the force of gravity

Answer:

The horizontal component of a projectile's acceleration is zero

Explanation:

When calculating all the different components of a projectile (acceleration, initial velocity, final velocity, etc.), you need to break it up into its X and Y directions.

During free fall, the only force acting on the projectile is gravity, and since gravity acts down (in the Y direction), there is no force acting in the X direction. Hence, the acceleration in the X direction is zero.

Since the two trains are passing in opposite directions, so this means that their relative velocities will be the sum of the two trains that is: 

relative velocity = (13 + 28) = 41m/s

a. The passengers aboard on train B will see that train A is moving at 41m/sec due east

b. The passengers aboard on train A will see that train B is moving at 41m/sec due west

Answer:

B and C

Explanation:

Answer:

12 m/s bye

Explanation:

The velocity of the object is 12 m/s. The correct answer is 12 m/s. The correct option is (e).

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 0.5 × m × v²

Where:

KE is the kinetic energy

m is the mass of the object

v is the velocity of the object

The object has a mass of 11 kilograms and kinetic energy of 792 joules, we can rearrange the formula to solve for velocity:

v² = (2 × KE) / m

v = √((2 × KE) / m)

v = √((2 × 792) / 11)

v = 12 m/s

So, the correct answer is 12 m/s. The correct option is (e).

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To determine Morgan's position on the bed (whether standing, sitting, or lying), one must calculate the area of contact she has with the bed using her mass and the pressure she exerts. By comparing the calculated contact area with typical areas for different positions, one can infer her position.

The question asks to determine whether Morgan is standing, sitting, or lying on the bed based on the pressure she exerts on the mattress and her mass of 85 kg. Pressure is defined as the force applied per unit area; in this case, force is due to Morgan's weight. Given that Morgan's weight (W) is the product of her mass (m) and Earth's gravity (g), we can write W = m * g = 85 kg * 9.81 m/s2. The weight finds us the force Morgan exerts downward.

To find the area of contact (A), we rearrange the pressure formula P = F/A, to A = F/P. Inserting the given pressure (P = 9530 N/m2) and her weight calculated earlier, we can find the area. A smaller contact area would suggest standing, while a larger area would suggest sitting or lying down. For example, if calculated area is around the size of feet, she's likely standing; if it's larger - similar to a body's contact area when sitting or much larger - she's sitting or lying down, respectively.

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1) Magnitude

Let's take south as positive y-direction and east as positive x-direction. Then we have to resolve both displacements into their respective components:

[tex]d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km[/tex]

[tex]d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km[/tex]

[tex]d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km[/tex]

[tex]d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km[/tex]

So, the components of the total displacement are

[tex]d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km[/tex] east (so, 1.26 km west)

[tex]d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km[/tex] south

So, the magnitude of the resultant displacement is

[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km[/tex]

2) Direction

the direction of the hiker's displacement is

[tex]\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ}[/tex] south of west.

The magnitude of the hiker's resultant displacement is 5.8 km south of west.

To find the magnitude of the hiker's resultant displacement, we need to first find the x and y components of the displacement vectors.

Path 1, the hiker walks 3.5 km in a direction of 55° south of west. The x component of this displacement is -3.5 * cos(55°) = -2.503 km, and the y component is -3.5 * sin(55°) = -2.845 km.

Path 2, the hiker walks 2.7 km in a direction of 16° east of south. The x component of this displacement is 2.7 * sin(16°) = 0.732 km, and the y component is -2.7 * cos(16°) = -2.606 km.

The total x component is -2.503 + 0.732 = -1.771 km, and the total y component is -2.845 - 2.606 = -5.451 km. Using the Pythagorean theorem, the magnitude of the resultant displacement is sqrt((-1.771)^2 + (-5.451)^2) = 5.797 km.

The direction can be found using the inverse tangent function, atan((-5.451)/(-1.771)) = 73°. Since the displacement is south of west, the direction is 180 - 73 = 107° south of west.

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The distance from the center of the earth that you will get, given that you leave the surface of the earth with a velocity of 5534 m/s, is [tex]6.5099\times10^6\ m[/tex]

How to calculate the distance from the earth you will get to?

The distance from the center of the earth that you will get, knowing that you leave the surface of the earth with a velocity of 5534 m/s can be calculated as shown below:

[tex]d = \frac{GM}{2v^2} \\\\d = \frac{6.67428\times10^{-11}\ \times\ 5.9742\times10^{24}}{2\ \times\ 5534^2} \\\\d = 6.5099\times10^6\ m[/tex]

From the above calculation, we can conclude that the distance you will get is [tex]6.5099\times10^6\ m[/tex]

The pressure at a hang glider's stagnation point is the sum of atmospheric pressure and dynamic pressure due to airflow collision, resulting in higher than atmospheric pressure. Using Bernoulli's equation with provided values, the stagnation pressure is 95430 Pa, and at 150 m/s, the pressure is approximately 82920 Pa.

The calculation of the gage pressure at a stagnation point involves understanding the flow dynamics around the structure of a hang glider as it moves through air. The stagnation point is where the airflow comes to a complete stop at the leading edge of the wing, causing the pressure to be higher than the atmospheric pressure due to the dynamic pressure of the colliding air mass. Using the provided stagnation pressure of 95430 Pa and the density of air at sea-level conditions (1.14 kg/m³), and given that the airspeed is 11.8 m/s, we would apply Bernoulli's equation to confirm the dynamic pressure at the stagnation point. But for the second part of the question, we adjust the equation to account for the higher speed of 150 m/s over the wing and determine that the pressure is 82920 Pa, acknowledging that turbulent flow makes these results approximate.

The stagnation pressure on a hang glider's stagnation point at sea-level conditions and an airspeed of 11.8 m/s can be calculated as 95430 Pa using the sea-level air density (1.14 kg/m³) and standard atmospheric pressure (8.89 × 10⁴ N/m²).

The concept being discussed in the question is the stagnation pressure which is the pressure at a point on a structure where the fluid (air) velocity is zero because the fluid there has been brought to rest abruptly. This pressure is higher than the atmospheric pressure because it includes both the atmospheric pressure and the additional pressure resulting from the air hitting the structure. The stagnation pressure (Pstagnation) at a hang glider's stagnation point can be calculated using Bernoulli's equation.

Given that the airspeed is 11.8 m/s, the sea-level air density is 1.14 kg/m³, and standard atmospheric pressure is 8.89 × 10⁴ N/m², we can find the stagnation pressure on the structure. Using the provided values:

Pstagnation = Po + (1/2) ρV²
Pstagnation = 8.89 × 10⁴ Pa + (1/2)(1.14 kg/m³)(11.8 m/s)²
Pstagnation = 8.89 × 10⁴ Pa + (1/2)(1.14)(139.24)
Pstagnation = 8.89 × 10⁴ Pa + 79.37 Pa
Pstagnation = 95430 Pa

Fracture toughness, minimum crack length calculation for fracture under stress.

Fracture toughness is the ability of a material to resist brittle fracture when a crack is present. In this scenario, we are given a fracture toughness value of 82.4 MPa√m.

To determine the minimum length of a surface crack that will lead to fracture when exposed to a tensile stress of 345 MPa, we can use the plane strain fracture toughness formula.

By plugging in the given values and solving the equation, we can find the minimum length of the surface crack that will cause fracture.

To find the minimum crack length that could lead to fracture in a steel plate, the plane strain fracture toughness equation is used, incorporating the values for fracture toughness, applied stress, and the geometric factor Y.

To determine the minimum length of a surface crack that will lead to fracture in a steel plate subjected to a tensile stress during service, we can use the plane strain fracture toughness equation:

\[a = \left(\frac{1}{\pi} \cdot \left(\frac{K_{IC}}{\sigma \cdot Y}\right)^{2}\right)
\]

Where:

Plugging in the values, the minimum crack length a can be calculated as follows:

\[a = \left(\frac{1}{\pi} \cdot \left(\frac{82.4 \, \text{MPa}\sqrt{m}}{345 \, \text{MPa} \cdot 1.0}\right)^{2}\right)
\]

After calculating the expression inside the parenthetical first and then squaring it, you would divide by π to get the minimum crack length a.

The answer is watershed. It is an area of land that drains all the rainfall and streams to a shared passage for instance, mouth of a bay, the outflow of a reservoir, or any point alongside a stream network. The word watershed is occasionally used interchangeably with catchment or drainage basin. This contains surface water like reservoirs, streams, lakes, and wetlands.

Answer:

THe answer is watershed

Explanation: