All categories

2 years ago

Ice that formed thousands of years ago is often found to contain tiny bubbles of gas. This gas came from __________.

1 answer:

8 0

Answer:Ice that formed thousands of years ago is often found to contain tiny bubbles of gas. This gas came from the ATMOSPHERE.

Explanation:

Atmospheric gases which consists of carbondioxide, nitrogen oxides and rare gases can be trapped inside the water body as bubbles. If these bubbles are surrounded by ice when they form, they get trapped and remain in the ice cube. A typical example of ice that contains bubbles of gases includes glacier ice. These are mass of ice that moves slowly over the land. It preserve bits of atmosphere from thousands of years ago in tiny air bubbles, or, deeper within the core, trapped within the ice itself. This is one way scientists know that there have been several Ice Ages.

You might be interested in

Viewed moving objects from pond water under his microscope and named them “animalcules”.

galben [10]

Anton Von Leeuwenhoek, in the early 1600s, saw these tiny microbes and called them "animalcules" and "wee beasties".

5 0

2 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$

where in this problem:

$Q=15 \mu C = 15\cdot 10^{-6} C$ is the charge on the shell

$r=20 cm = 0.20 m$ is the distance from the centre of the shell

Substituting, we find:

$E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C$

4 0

3 years ago

The global positioning system (GPS) uses electromagnetic waves sent between computer devices on Earth and satellites in space to

zhenek [66]

Answer:

first plz mark brarinlest

Explanation:

5 0

2 years ago

Read 2 more answers

Why is pluto not a planet?

mihalych1998 [28]

Pluto revolves around a star instead of our SUN. So, it's an "Exo-planet" rather than a Planet. It has declared in 2006 by the scientists.

Hope this helps!

3 0

3 years ago

Read 2 more answers

A system had 150 kj of work done on it and its internal energy increased by 60 kj. How much energy did the system gain or lose a

mina [271]

Answer:

The system loses 90 kJ of heat

Explanation:

We can answer the question by using the 1st law of thermodynamics, which states that:

$\Delta U=Q-W$