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qwelly [4]
3 years ago
10

The Earth is instantly replaced in its orbit by a speck of dust. Which statement best describes the subsequent orbital motion of

that piece of dust? The Earth is instantly replaced in its orbit by a speck of dust. Which statement best describes the subsequent orbital motion of that piece of dust? The dust particle will continue in the same orbit as the Earth did, orbiting the Sun in 1 year. The dust particle will move to a smaller orbit and orbit the Sun in less than 1 year. The dust particle will move to a larger orbit and orbit the Sun in more than 1 year. The dust particle will spiral into the Sun. The dust particle will be ejected from the Solar System.
Physics
1 answer:
Alik [6]3 years ago
5 0

The correct statement is:

The dust particle will move to a larger orbit and orbit the Sun in more than 1 year.

In fact, the dust particle has smaller mass than the Earth, therefore the gravitational attraction exerted by the Sun on the dust will be smaller, according to the formula of the gravitational force:

F=G\frac{M_S m}{r^2}

where G is the gravitational constant, MS is the Sun mass, m is the mass of the Earth (at first) and then of the dust particle (later), and r is the distance from the Sun. When we replace the Earth with the dust particle we see that m decreases, so the gravitational force F decreases, and the particle will move to a larger orbit. Then its orbital period will increase, according to the third law of Kepler, which states that the square of the orbital period is proportional to the cube of the orbital radius:

T^2 \sim r^3

therefore, since the orbit is larger, the orbital period is greater.

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If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?
Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

A=100 mm^2

multiplied by the height of rain that fell, which is

h=5.0 mm

Therefore, the volume is

V=hA=(5 mm)(100 mm^2)=500 mm^3

7 0
4 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

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3 years ago
8x = -6. What does x equal?
allsm [11]

Answer:

x= -3/4

Explanation:

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uppose two train cars are moving toward one another, the first with a mass of 150,000 kg and a velocity of 0.300 m/s; the second
kondaur [170]

The value was determined to be 0.122 m/s. The velocity of a body or object determines its direction of motion. Speed is a scalar quantity in its most fundamental form.

Velocity is essentially a vector quantity. It is the rate of change in distance. The initial speed of the first train, which has a mass of 150,000 kg, is 0.3 m/s. The second train has an initial speed of -0.120 m/s and a mass of 110,000 kg.

Let v represent the post-collision speed of the connected mass.

Utilize the idea of momentum.

The speed of the trains is constant both before and after a collision.

150.000 + 110.000v 45.000 - 13200 = 260.000 v 31800 = 260.000 v v = 0.122 m/s 150000 x 0.3 - 110000 x 0.120

After colliding, they move at a speed of 0.122 m/s towards the direction of the right.

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1 year ago
Explain the physical and chemical changes that occur from when you eat food until it is digested.
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The physical changes are you are breaking down the food u eat and the chemical changes are when the enzymes in your stomach break down the food that you have already eaten
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