Answer:
doppler shift's formula for source and receiver moving away from each other:
<em>λ'=λ°√(1+β/1-β)</em>
Explanation:
acceleration of spaceship=α=29.4m/s²
wavelength of sodium lamp=λ°=589nm
as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm
using doppler shift's formula:
<em>λ'=λ°√(1+β/1-β)</em>
putting the values:
700nm=589nm√(1+β/1-β)
after simplifying:
<em>β=0.17</em>
by this we can say that speed at that time is: v=0.17c
to calculate velocity at an acceleration of a=29.4m/s²
we suppose that spaceship started from rest so,
<em>v=v₀+at</em>
where v₀=0
so<em> v=at</em>
as we want to calculate t so:-
<em>t=v/a</em> v=0.17c ,c=3x10⁸ ,a=29.4m/s²
putting values:
=0.17(3x10⁸m/s)/29.4m/s²
<em>t=1.73x10⁶</em>
Answer:
Q= -6900 J
Explanation:
use the formula Q=mC(T_2 - T_1) and sub in givens
Q=mC(T_2 - T_1)
Q= (200 g)(0.444 J/g°C)(22-100)
Q= -6900 J
The negative sign means heat is lost, which agrees with the decrease in temperature
<h2>Right answer: Comets have very elliptical orbits that usually take them far beyond the orbit of Pluto, but also take them closer to the Sun than Earth</h2>
Comets are celestial bodies constituted by ice, dust and rocks that orbit around the Sun, after having been altered by the Oort cloud; following different trajectories that can be <u>highly eccentric elliptical</u><u> </u>(periodic trajectories), parabolic or hyperbolic.
One of the main characteristics of a comet is that it travels quite fast, on its way around the Sun and has a long tail. It should be noted that the tails of comets always go in the opposite direction to the Sun (due to the radiation pressure of sunlight).
Therefore, the correct option is C.
That's two different things it depends on:
-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.
Here's what I have in mind for an experiment to show those two dependencies:
-- a closed box with a wall down the middle, separating it into two closed sections;
-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.
-- a tiny fan that blows air through a tube into the hole in one outer wall.
<u>Experiment A:</u>
-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================
<u>Experiment B:</u>
-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
Answer:
W = 55.12 J
Explanation:
Given,
Natural length = 6 in
Force = 4 lb, stretched length = 8.4 in
We know,
F = k x
k is spring constant
4 = k (8.4-6)
k = 1.67 lb/in
Work done to stretch the spring to 10.1 in.

![W = \dfrac{k}{2}[x^2]_6^{10.1}](https://tex.z-dn.net/?f=W%20%3D%20%5Cdfrac%7Bk%7D%7B2%7D%5Bx%5E2%5D_6%5E%7B10.1%7D)

W = 55.12 J
Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.