The process by which conclusions are determined is called

Why do the planets appear in different locations in the night sky while the pattern of stars in a constellation stays the same?

Goshia [24]

I'm sure you've noticed that an airplane high in the sky, far away
from you, looks like it's moving very slowly. At the same time,
somebody passing you on a skateboard whizzes past you at
high speed. The farther away something is from you, the slower
it appears to move.

The nearest star outside the solar system is almost 32 thousand times
as far away from us as the farthest visible planet (Saturn) is, and all of the
other stars are farther than that.

That's why you have to wait a few thousand years before you notice
that the shape of a constellation has changed.

To put it a slightly different way . . . Everything is in motion. The motion is
more noticeable for nearby things, and less noticeable for farther-away things.
Objects within our solar system are the only ones near enough so that a human
lifetime is a long enough period in which to notice the change in their position.
Even Pluto moves less then 1.5° against the 'background' stars in a whole year.

This all makes me feel small. How about you ?

7 0

4 years ago

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How are Babys Born?

My name is Ann [436]

Answer:

Baby are born due to the fertilization of Owen or egg present in vigina due to the fertilization from sperms

6 0

4 years ago

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An object has a mass of 30 grams and measures 3cmx2cmx1cm. What is the density of the object?

dimulka [17.4K]

The density of the object is the ratio of its mass and volume. From the given dimensions above, we determine the volume through the equation,

V = L x W x H

Substituting,

V = (3 cm)(2 cm)(1 cm) = 6 cm³

From the idea presented above,
d = m/V

Substituting the known values,

d = (30 g)/ (6 cm³) = 5 g/cm³

ANSWER: 5 g/cm³

4 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

Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has

a_sh-v [17]

Answer:

h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

Em₀ = U = m g h

final point. Lower, just before the crash

Em_f = K = ½ m $v_{b}^2$

energy is conserved

Em₀ = Em_f

m g h = ½ m v²

v_b = $\sqrt{2gh}$

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2m 0 + m v_b

final instant. Right after the crash

p_f = (2m + m) v

the moment is preserved

p₀ = p_f

m v_b = 3m v

v = v_b / 3

v = ⅓ $\sqrt{2gh}$

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

Em₀ = K = ½ 3m v²

Final point. Higher

Em_f = U = (3m) g h'

Em₀ = Em_f

½ 3m v² = 3m g h’

we substitute

h’= $\frac{v^2}{2g}$

h’ = $\frac{1}{3^2} \ \frac{ 2gh}{2g}$

h’ = 1/9 h

6 0

3 years ago