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3 years ago

Where does the bulk of atmosphere oxygen come from

Physics

1 answer:

adell [148]3 years ago

4 0

The bulk of earths atmospheric oxygen comes from phytoplankton, small flower like things in the ocean.

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A 7.0 mm -diameter copper ball is charged to 40 nC. What fraction of its electrons have been removed? The density of copper is 8

mylen [45]

Answer:

$f = 2.6 \times 10^{-13}$

Explanation:

Let the mass of copper ball is "m" gram

now the total number of copper atom present in the ball is given as

$N = \frac{m}{29} \times 6.02 \times 10^{23}$

now the total number of electrons in one copper atom is 29

so total number of electrons in given sample of copper ball is

$N_e = m(6.02 \times 10^{29})$

now diameter of the ball is 7.0 mm

density of the ball = $8900 kg/m^3$

now we have

$m = (\frac{4}{3}\pi r^3)(8900)$

$m = (\frac{4}{3}\pi(\frac{0.007}{2})^3)(8900)$

$m = 1.6 gram$

now we have

$N_e = 9.63 \times 10^{23}$

now the charge on the copper ball is 40 nC

so the number of electrons removed

$Q = ne$

$40 \times 10^{-9} = n(1.6 \times 10^{-19}$

$n = 2.5 \times 10^{11}$

so the fraction of number of electrons removed is given as

$f = \frac{n}{N_e}$

$f = \frac{2.5 \times 10^{11}}{9.63 \times 10^{23}}$

$f = 2.6 \times 10^{-13}$

7 0

4 years ago

PLEASE HELP ON QUESTION

mina [271]

Answer:

fastest air slowest steel

Explanation:

4 0

2 years ago

As the distance of a sound wave from its source quadruples, how is the intensity changed?

Naddik [55]

d. it decreases by a factor of 16

3 0

4 years ago

Read 2 more answers

It is common problem to a physically active person who is experiencing dizziness, fatigue, dark-colored urine, and extreme thirs

choli [55]

Answer:

Look below.

Explanation:

All the symptoms provided points to the common problem of dehydration. Especially a physically active person because they are constantly sweating which leads to the loss of bodily fluids and dehydration.

4 0

3 years ago

A 2kg object is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete rev

Travka [436]

Answer:

a) 37.70 m/s

b)710.6 m/s²

Explanation:

Given that ;

Mass of object = 2 kg

Radius of the motion = 2m

Frequency of motion = 3 rev/s

The formula to apply is;

v= 2πrf where v is linear speed

v = 2×π×2×3 =12π = 37.70 m/s

Centripetal acceleration is given as;

a= 4×π²×r×f²

a= 4×π²×2×3²

a=710.6 m/s²

5 0

3 years ago