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3 years ago

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In the amusement park ride Mr. Freeze, riders are uniformly accelerated from rest by magnetic induction motors along a 70 meter

horizontal track in just 5 seconds. While accelerating, friction exerts 500N of force on the train. Then the train coasts through the loops and turns of the remainder of the ride. A train is loaded with passengers so that it has a total mass of 2500 kg. Determine the acceleration. Net Force, and all of the Forces acting upon the train during the ride knowing that forces are vectors.

1 answer:

pishuonlain [190]3 years ago

7 0

Answer:

Acceleration=5.6 m/s

Explanation:

Dunno the rest

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An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast

forsale [732]

Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, $h_{max}$ = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g· $h_{max}$

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, $v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }$

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, $v_e$ ≈ 3.13 m/s is therefore;

$3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}$

$r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45$

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

7 0

3 years ago

What are used for manufacturing paper along with wood and chemicals

Tju [1.3M]

Answer:

Paper and pulp are made from cellulosic fibers and other plant materials. Some synthetic materials may be used to impart special qualities to the finished product. Paper is made from wood fibers, but rags, flax, cotton linters, and bagasse are also used in some papers.

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3 years ago

The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re

lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

$\Delta x = \Delta y$

Where:

$\Delta x$ - Range of the bullet, measured in meters.

$\Delta y$ - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

$K_{1} = U_{2}$

Where:

$K_{1}$ - Kinetic energy at point 1, measured in joules.

$U_{1}$ - Gravitational potential energy at point 2, measured in joules, and:

$U_{2} = m\cdot g \cdot \Delta y$

Where:

$m$ - Mass of the bullet, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

$K_{1} = m\cdot g \cdot \Delta y$

$\Delta y = \frac{K_{1}}{m\cdot g}$

If $K_{1} = 1066\,J$ , $m = 0.25\,kg$ and $g = 9.81\,\frac{m}{s^{2}}$ , the maximum height is now computed:

$\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\Delta y = 434.791\,m$

$\Delta y = 0.435\,km$

Lastly, the range of the bullet is 0.435 kilometers.

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3 years ago

Sound is a disturbance that travels through a medium as a

Vinvika [58]

Answer:

Sound waves. Anything that vibrates is producing sound; soundis simply a longitudinal wave passing through a medium via the vibration of particles in themedium. Consider a sound wavetraveling in air

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3 years ago

How does the equivalence principle lead us to suspect that spacetime might be curved?

Dafna11 [192]

Answer:

To understand Einstein's thought processes,

imagine yourself in the sealed box, being accelerated through interplanetary

space at 9.8m/s^2. You grab the flashlight that you keep on the bedside

table and shine a beam of light perpendicular to the acceleration vector. Since the box is accelerating upward, the path of the light beam

will appear to you to be bent downward, as the floor of the box rushes up

to meet the photons. However, thanks to the equivalence principle, we can

replace the accelerated box with a stationary box experiencing a constant gravitational acceleration. Since there's no way to distinguish between these

two cases, we are led to the conclusion that the paths of photons will be

curved downward in the presence of a gravitational field. Gravity affects

photons, Einstein concluded, even though they have no mass. Contemplating the curved path of the light beam.

Fermat's principle, which states that

light travels between two points along a path which minimizes the travel time required. In a vacuum, where the speed of light is constant, this translates into the requirement that light takes the shortest path between two points. In Euclidean, or flat, space, the shortest path between two points is a straight line. However, in the presence of gravity, the path taken by light is not a straight line. Thus, Einstein concluded, space is not Euclidean. The presence of mass, in Einstein's view, causes space to be curved.

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3 years ago