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amm1812
3 years ago
15

During an earthquake, you should do all of the following EXCEPT _________________________.

Physics
2 answers:
Nady [450]3 years ago
6 0
<span>The key word is "electric". If the power goes out during an earthquake, items that run on electricity won't help you any. You need battery-operated radios and flashlights, as well as extra batteries.
What are your answer choices?
</span>
frosja888 [35]3 years ago
4 0
Run from the area in a zig-zag formation or keep hands visible. if these are wrong then sorry.
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I help much help on 18!!!
ra1l [238]
I think it's something like electrons don't attract, cuz you know the saying "Opposites attract." Cause electrons are negative... Ahaha... sorry, I don't know the answer.
4 0
3 years ago
A scientist is designing a device that will mimic Earth's atmosphere by
mafiozo [28]

Answer:

1. Ultraviolet light (UV)

2. X-rays

3. Gamma-rays

Explanation:

Though there are different types of energy or electromagnetic waves with varying wavelengths, including the likes of Gamma X-rays, ultraviolet light, visible light, infrared radiation, and microwave radiation.

What is more certain is that the atmosphere blocked the high-energy waves from getting to the earth surface or biosphere such as Ultraviolet light (UV), X-rays and Gamma-rays

8 0
3 years ago
I'm not sure if the answer is A or B... someone help
Talja [164]
Its b i literally have had this exact question
8 0
3 years ago
1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

4 0
3 years ago
The distance between Earth and Mars is 225 million km. When converted using the conversion factor 1 AU = 1.5 × 108 km, the dista
noname [10]
To convert km to AU, we divide 225,000,000 km by the factor of 1.5 x 10^8 = 150,000,000 km. This gives us 225,000,000 / 150,000,000 = 1.5 AU. Therefore, the distance between Earth and Mars in AU is 1.5 AU.
The AU is not equivalent to a light-year. A light-year is equivalent to around 9.5 x 10^12 kilometers.

4 0
3 years ago
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