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3 years ago

10

A periodic wave has a fixed pattern that is repeated. What is one completion of the wave pattern called?

1 answer:

Kaylis [27]3 years ago

7 0

It is called cycle. Letter D

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The dwarf planet Ceres contains over 50% of the mass of the main asteroid belt.

PolarNik [594]

False

Explanation:

Called an asteroid for many years, Ceres is so much bigger and so different from its rocky neighbors that scientists classified it as a dwarf planet in 2006. Even though Ceres comprises 25 percent of the asteroid belt's total mass, tiny Pluto is still 14 times more massive.

6 0

3 years ago

Bacteria vary somewhat in size, but a diameter of 2.9 μm is not unusual.

Sedbober [7]

Answer:

a) 6.4 x 10^-12 cm^3

b) 17 x 10^-6 mm^2

Explanation

a). The shape is assumed to be spherical The volume = volume of a sphere = \frac{4}{3} \pi r^3

3

4

πr

3

V = \frac{4}{3}*3.142* 1.15^3

3

4

∗3.142∗1.15

3

= 6.3715 μm^3

1 μm^3 = 10^-12 cm^3

6.3715 μm^3 = 6.3715 x 10^-12 cm^3

==> 6.4 x 10^-12 cm^3

8 0

2 years ago

Read 2 more answers

A house is wired so that one electrical source comes to a room but many outlets and lights work from that source . when one ligh

Misha Larkins [42]

The room is wired as a parallel circuit

5 0

4 years ago

What is the speed of sound in air with temperature of 355.8 k

nikklg [1K]

Answer:

Explanation:

20.05 √Tk = 20.05 √355.8 = 378.196... ≈ 378 m/s

7 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago