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Naddik [55]
3 years ago
6

Match each item to its best description.

Physics
2 answers:
Mashcka [7]3 years ago
5 0
A solar system is a group of planets which orbit a star. Our solar system is comprised of planets like Mercury, Earth, and Jupiter. A solar nebula is a large cloud of gas and dust. An accretion disk occurs when gas and dust spin, typically the main body of this disk is a star.
andre [41]3 years ago
5 0

Solar nebula --> "gas and dust spins"

Accretion disk --> "large cloud of gas and dust"

Solar system --> "planets orbit a star"

(Apex class ;)

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Match these items.
viva [34]
-- 'Ca' (Calcium) is an element.
-- The proton has a positive charge.
-- Nuclear fusion results in the synthesis of atoms of new elements.
-- H₂O (water) is a chemical compound. 
-- Nuclear fission is a decay of the nucleus.
-- The atomic number of an element is the number of protons 
      in each atom of it.
-- I suppose you're using the Greek letter  <span>η  ('eta', not 'nu') 
      to represent the neutron.
-- I suppose you're using ' e ' to represent the electron.
</span>
6 0
3 years ago
Read 2 more answers
Does decomposing water require high activation energy
skelet666 [1.2K]
decomposing water does not require High activation energy
6 0
4 years ago
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
4 years ago
A heavy boy and a lightweight girl are balanced on a mass-less seesaw. If they both move forward so that they are one-half their
Rom4ik [11]

Answer:

b) Nothing will happen,  the sea saw will still be balanced.

Explanation:

b) Nothing will happen,  the sea saw will still be balanced.

Reason:-

When two kids are balanced, the sum of torques on the seesaw will be zero.

if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.

Therefore the seesaw is balanced

4 0
3 years ago
Read 2 more answers
A rocket ship starts from rest and turns on its forward booster rockets causing it to have a constant acceleration of 4 meters p
bagirrra123 [75]

Complete question is;

A rocket ship starts from rest and turns on its forward booster rockets, causing it to have a constant acceleration of 4 m/s² rightward. After 3s, what will be the velocity of the rocket ship?

Answer:

v = 12 m/s

Explanation:

We are given;

Initial velocity; u = 0 m/s (because ship starts from rest)

Acceleration; a = 4 m/s²

Time; t = 3 s

To find velocity after 3 s, we will use Newton's first equation of motion;

v = u + at

v = 0 + (4 × 3)

v = 12 m/s

6 0
3 years ago
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