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4 years ago

15

Is radioactivity on the Earth something relatively new?

1 answer:

luda_lava [24]4 years ago

3 0

naturally occurring radioactive elements have been around since the Earth's formation

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The largest and brightest galaxies are those that contain _____.

user100 [1]

They contain quasars

6 0

3 years ago

Read 2 more answers

What is the minimum size copper equipment grounding conductor required to serve a multi section motor control center equipped wi

Yakvenalex [24]

The minimum size copper equipment grounding conductor that is required is ; #4 copper

<h3>Equipment grounding conductor </h3>

An Equipment grounding conductor is a conductive part of ground fault current path which connects the non-conducting metallic parts of the euipment together and also connects the system grounded conductor.

For a copper euipment grounding conductor required to serve a control center connected to a 300 amp over current device the minimum size of copper is #4 copper

Hence we can conclude that The minimum size copper equipment grounding conductor that is required is ; #4 copper.

Learn more about Equipment grounding conductor : brainly.com/question/14124204

4 0

2 years ago

1. Fill in the blanks. (3 pts)

Usimov [2.4K]

Answer:

a) mass

b) Newtons

c) momentum formula where p stands for momentum, m stands for mass, and v stands for velocity

Hope this helps!

4 0

3 years ago

Read 2 more answers

By applying a force of one Newton, one can hold a body of mass _

SVEN [57.7K]

Answer:

By applying a force of one Newton, one can hold a body of mass of 102 gram.

Explanation:

Force is the pull or push of an object. It can be mathematically measured as, F= m* g.

where, F= force in newton

m= mass in kg

g= acceleration due to gravity ( $meter/second^{2}$ )

For 1 newton force,

F= m* 9.8

or, m= $\frac{1}{9.8}$ = 0.102 kg

or, m= 102 gram.

Hence, 102 gram mass can be hold by one Newton force.

3 0

4 years ago

On an asteroid, the density of dust particles at a height of 3 cm is ~30% of its value just above the surface of the asteroid. A

Anvisha [2.4K]

From the law of atmosphere

$N_v(y) = n_0*e^{-\frac{mgy}{Kb*T}}$

Where

$n_0$ = constant and is number density where the height y = 0cm

$n_V$ = Number density at height y=3cm

Kb = Boltzmann constant $= 1.38*10^{-23}J/K$

$T=20K$

$m = 10^{-19}kg$

Re-arranging the equation to have the value of the gravity,

$\frac{N_v(y)}{n_0} = e^{-\frac{mghy}{KbT}}$

$ln(\frac{N_v(y)}{n_0}) = -\frac{mgy}{KbT}$

Since it is 30% of value above surface, therefore $N_v = 0.3n_0$

$ln(\frac{0.3n_0}{n_0}) = -\frac{mgy}{KbT}$

$g = -\frac{KbT ln(0.3)}{my}$

$g = -\frac{(1.38*10^{-23}J/K)(20K)(Ln(0.3))}{10^{-19}(3*10^{-2})}$

$g = \frac{1.38*2*ln(0.3)*10^{-22}}{3*10^{-4}}$

$g = 1.104*10^{-1}m/s^2$

$g = 0.1m/s^2$

Therefore the correct answer is C.

4 0

4 years ago