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3 years ago

Please help me with this i am so!! stuck

2 answers:

6 0

If small objects can be anything, basically anything goes! People could go, ants could go, they are really tiny, use your imagination, and you could think of many possibilities!

Alex17521 [72]3 years ago

4 0

The question seems kind of squishy.

The smallest objects are also the most numerous in the
whole universe. Those are the Hydrogen atoms.

The most numerous objects that are big enough for us to see
are the stars, but I'm not going to say that those are the answer.
Stick with me here, and follow my reasoning.

The Kepler space telescope is discovering so many planets
orbiting other stars outside our solar system, that it's beginning
to look like MOST stars have planets.

Stars are bigger. Planets are smaller. If most stars have planets,
then the most numerous objects that compose much of the universe
are the planets ! (plus maybe their moons too)

Does this make any sense to you ?

Now I have to ask YOU a question:
Are there any choices for answers to this question
that you're not telling us about ?

You might be interested in

If element x has 94 protons how many electrons does it have

Igoryamba

94 electrons. protons and electrons are always the same, but neutrons are different.

8 0

3 years ago

Read 2 more answers

A 3.0 pF capacitor consists of two parallel plates that have surface charge densities of 1.0 nC/mm2 . If the potential between t

Archy [21]

Answer:

$A=81mm^2$

Explanation:

We know that for a capacitor $Q=CV$ , where Q is the charge of one plate, C the capacitance and V the potential between the plates.

We also know that $Q=\sigma A$ , since $\sigma$ is the surface charge density and A the area of the plate (both equal in our case).

Putting all together:

$A=\frac{CV}{\sigma}$

Which for our values is:

$A=\frac{(3\times10^{-12}F)(27\times10^3V)}{1\times10^{-9}C/mm^2}=81mm^2$

Where we notice that the S.I. units combination FV/C must not have units (we can verify it directly from their definitions or we notice that $mm^2$ is enough to describe an area).

8 0

3 years ago

A 5000-lb wrecking ball hangs from a 50-ft cable of density 10 lb/ft attached to a crane. Calculate the work done if the crane l

aniked [119]

Answer:

total work is = 52450 J

Explanation:

given data

mass = 5000-lb

density = 10 lb/ft

height = 50 ft

solution

as we will treat here cable and ball are separate

and

here work need to lift cable is

w = (10Δy )(9.8 y ) j

and

now summing all segment of cable

so passing limit Δy to 0

so total work need

= $\int\limits^{10}_0 {98y} \, dy$

= $[49 y^2]^{50}_0$

= 2450J

so lifting 5000 lb wrcking 50 m required additional 5000 + 2450

so total work is = 52450 J

3 0

3 years ago

Find the current flowing through the following circuit

kap26 [50]

Answer:

0.5

Explanation:

calculating the total resistance for the both the parallel and the series resistors then later using the ohm's law formulae V=IR to calculate current.

4 0

3 years ago

A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the sp

SOVA2 [1]

solution;

$the expression for force applied on the spring due to the load is\\f=k\Delta x\\here,\Delta x is the extension in the spring due to appling force\\given three case as following\\110N=k(40-x_{o})----------1\\240N=k(60-x_{o})----------2\\w=k(30-x_{o})-------------3\\$ $To calculate the accrual length of the spring,solve th equation 1 and 2\\\frac{110N}{240N}=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458=\frac{k(40-x_{o})}{k(60-x_{o})}\\0.458(60mm-x_{o})=(40mm-x_{o})\\x_{o}(1-0.458)=(40-60(0.458))mm\\x_{o}\frac{12.52}{0.542}\\=23.1mm\\to calculate the force on the spring in case,\\solve the equation 1 and 2\\\frac{110}{w}=\frac{k(40-x_{o})}{k(60-x_{o})}\\\frac{110}{w}=\frac{(40mm-23.1mm)}{30mm-23.1mm}\\w=\frac{110}{2.45}=44.9N$

8 0

3 years ago