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nata0808 [166]
4 years ago
14

What unit is similar to light year

Physics
1 answer:
Sidana [21]4 years ago
8 0
You could use miles
one light year equals 6 trillion miles
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A glacier can cause a change in landforms as it picks up rocks and moves the rock to new locations. How fast does the glacier ha
STatiana [176]

Answer:

30 meters/day.

Explanation:

Glacial motion can be fast up to 30 meters/day or slow up to 0.5 m/year on small glaciers moving the impact on local landforms. If the glacial motion goes up to 30 meters/day, more fast change occurs in the local landforms while on the other hand, if the glacial motion goes up to 0.5 m/year, very slow change occurs in the local landforms. So glacial motion has a direct affect on the changes occurs in the local landforms.

5 0
4 years ago
Learning good work ethics will help you develop _____.
Pavel [41]

Answer:

a sense of industry

Explanation:

I just had this question and got it right

3 0
3 years ago
You are given aqueous solutions of six different substances and asked to determine whether they are strong, weak, or nonelectrol
kogti [31]

Answer:

Answer is explained below;

Explanation:

Electrolytes are any substances that dissociate into charged particles called ions when dissolved in water. The positively charged ions called cations and the negatively charged ions called anions move toward the negative and positive terminals (cathode and anode) of an electric circuit.

When a substance dissolved in water completely dissociates into ions, it is called a strong electrolyte. The aqueous solutions containing strong electrolytes conduct electricity very well and the examples include strong acids and soluble ionic compounds such as barium chloride, sodium hydroxide, etc.  

When a substance dissolved in water does not completely dissociate into ions, it is called a weak electrolyte. Since the aqueous solutions containing weak electrolytes have relatively few ions, their electrical conductivity is very low compared to the solutions containing strong electrolytes. Examples of weak electrolytes include weak acids and bases like acetic acid, ammonia, etc.

When a substance does not dissociate into ions when dissolved in water, it is called a nonelectrolyte. Since the aqueous solutions containing nonelectrolytes do not contain any ions, such solutions do not conduct electricity. Examples of nonelectrolytes are ethanol, aldehydes, glucose, ketones, etc.

If a solution contains dissolved ions, it conducts electricity and as the ion concentration increases, the conductivity also increases. To determine whether the aqueous solutions of six different substances are strong, weak, or nonelectrolytes, we can test them by applying a voltage to electrodes immersed in the solutions and a light bulb. By observing the brightness of the light bulb or by measuring the flow of electrical current, we can find out which solution contains a strong electrolyte or weak electrolyte, or nonelectrolyte.

If the solution contains a nonelectrolyte, the current flow is nil and the light bulb does not glow. If the solution contains a strong electrolyte, the current flow is very strong and so the brightness of the light bulb is very high. If the solution contains a weak electrolyte, the current flow is much low compared to the strong electrolyte and the light bulb glows, but the brightness is very low.

3 0
4 years ago
What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingr
Alexxx [7]

Answer:

 E = 9.4 10⁶ N / C ,     The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

             Ф = E. dA = qint / ε₀

 

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of ​​the cylinder is the length of the circle along the length of the cable

         dA = 2π dr L

          A = 2π r L

They indicate that the distance at which we must calculate the field is

         r = 5 R₁

         r = 5 1.3

         r = 6.5 mm

The radius of the outer shell is

         r₂ = 10 R₁

         r₂ = 10 1.3

         r₂ = 13 mm

         r₂ > r

When comparing these two values ​​we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

         λ = q / L

         Qint = λ L

Let's replace

      E 2π r L = λ L /ε₀

       E = 1 / 2piε₀  λ / r

Let's calculate

         E = 1 / 2pi 8.85 10⁻¹²  3.4 10-12 / 6.5 10-3

         E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

5 0
3 years ago
You break a piece of Styro foam packing material, and it releases lots of little spheres whose electric charge makes them stick
stich3 [128]

Answer:

The magnitude of charge on each sphere is q=2.50\times 10^{-8}\ C.

Explanation:

Given that,

Force of repulsion between the charges, F = 22 mN

The distance between spheres, r = 16 mm = 0.016 m

It is mentioned that both the spheres carry equal charges. The force between charges is given by :

F=\dfrac{kq^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{22\times 10^{-3}\times (0.016)^2}{9\times 10^9}}

q=2.50\times 10^{-8}\ C

So, the magnitude of charge on each sphere is q=2.50\times 10^{-8}\ C. Hence, this is the required solution.

3 0
3 years ago
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