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My name is Ann [436]
3 years ago
14

How did early scientists attempt to organize the elements

Physics
1 answer:
Nookie1986 [14]3 years ago
8 0
They organized them by color and size
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A satellite circles the Earth in an orbit whose radius is twice the Earth’s radius. The Earth’s mass is 5.98 x 1024 kg, and its
gavmur [86]

Hello!

Recall the period of an orbit is how long it takes the satellite to make a complete orbit around the earth. Essentially, this is the same as 'time' in the distance = speed * time equation. For an orbit, we can define these quantities:

d = 2\pi r ← The circumference of the orbit

speed = orbital speed, we will solve for this later

time = period

Therefore:

T = \frac{2\pi r}{v}

Where 'r' is the orbital radius of the satellite.

First, let's solve for 'v' assuming a uniform orbit using the equation:
v = \sqrt{\frac{Gm}{r}}

G = Gravitational Constant (6.67 × 10⁻¹¹ Nm²/kg²)

m = mass of the earth (5.98 × 10²⁴ kg)

r = radius of orbit (1.276 × 10⁷ m)

Plug in the givens:
v = \sqrt{\frac{(6.67*10^{-11})(5.98*10^{24})}{(1.276*10^7)}} = 5590.983 m/s

Now, we can solve for the period:

T = \frac{2\pi (1.276*10^7)}{5590.983} =\boxed{ 14339.776 s}

7 0
2 years ago
The modern periodic table is arranged in order of increasing atomic
tankabanditka [31]
The periodic table is in increasing atomic mass. Hope this helped.
7 0
3 years ago
Read 2 more answers
A dog of mass 18 kg runs at a speed of 4 m/s. What is the momentum of the
Andrej [43]

Answer:

A, 72 kg•m/s

Explanation:

p=mv

p=18x4

p=72

6 0
3 years ago
Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1
hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

\sum F=ma

We know that the force between two bodies due to gravity is given by the following equation:

F_{g} = G\frac{m_{1}m_{2}}{r^{2}}

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

F_{g} =G\frac{Mm}{r^{2}}

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

F_{c}=ma_{c}

where the centripetal acceleration is given by:

a_{c}=\omega ^{2}r

where

\omega = \frac{2\pi}{T}

Where T is the period, and \omega is the angular speed of the planet, so:

a_{c} = ( \frac{2\pi}{T})^{2}r

or:

a_{c}=\frac{4\pi^{2}r}{T^{2}}

so:

F_{c}=m(\frac{4\pi^{2}r}{T^{2}})

so now we can do the sum of forces:

\sum F=ma

F_{g}=ma_{c}

G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})

in this case we can get rid of the mass of the planet, so we get:

G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})

we can now solve this for T^{2} so we get:

T^{2} = \frac{4\pi ^{2}r^{3}}{GM}

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

T_{1}^{2}>T_{2}^{2}

So let's see what's going on there, we'll call:

M_{1}= mass of Star 1

M_{2}= mass of Star 2

So:

\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}

we can get rid of all the constants so we end up with:

\frac{1}{M_{1}}>\frac{1}{M_{2}}

and let's flip the inequality, so we get:

M_{2}>M_{1}

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0
3 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
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