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3 years ago

Which type of galaxy is too small for gravity to form it into an easily describable shape?

Physics

2 answers:

oksano4ka [1.4K]3 years ago

6 0

If you think about it, irregular galaxies dont really have a describable shape, as you can tell by the name. So B.

ruslelena [56]3 years ago

4 0

The correct answer is B. irregular.

An irregular galaxy is a galaxy too small for gravity to form into a regular shape, or one shaped by collisions or the gravitational pull of other galaxies.

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19. State any 3 applications of capillary action (3mk)

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3 years ago

A skateboarder flies horizontally off a cement planter. After 3 seconds the skateboarder lands on the ground with a final veloci

evablogger [386]

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

$v = v_o + a \Delta t$

A skateboarder flies horizontally off a cement planter. After a time of 3 seconds (Δt), he lands with a final velocity (v) of −4.5 m/s. Assuming the acceleration is -9.8 m/s² (a), we can calculate the initial velocity of the skateboarder (v₀) using the kinematic equation A.

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$v = v_o + a \Delta t$

Learn more: brainly.com/question/4434106

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3 years ago

If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?

leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water ---> 1</span>

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m (ANSWER)</span>

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3 years ago