3 years ago

The value of a scientific variable _____.

Physics

2 answers:

Bas_tet [7]3 years ago

8 0

Answer:

Explanation:

Sloan [31]3 years ago

3 0

The value of a scientific variable can change. (A)

Hint hint: That's why it's called a "variable". If it always remained the same, then it would've been called a "constant".

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An asteroid is discovered in a nearly circular orbit around the Sun, with an orbital radius that is 3.570 times Earth's. What is

inysia [295]

Explanation:

Given:

$r_a = 3.570R_E$

$R_E = 1.499×10^{11}\:\text{m}$

$M_S = 1.989×10^{30}\:\text{kg}$

$G = 6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2$

Let $m_s$ = mass of the asteroid and $r_a$ = orbital radius of the asteroid around the sun. The centripetal force $F_c$ is equal to the gravitational force $F_G:$

$F_c = F_G \Rightarrow m_a\dfrac{v_a^2}{r_a} = G\dfrac{m_aM_S}{r_a^2}$

$\dfrac{4\pi^2 r_a}{T^2} = G\dfrac{M_S}{r_a^2}$

where

$v = \dfrac{2\pi r_a}{T}$

with T = period of orbit. Rearranging the variables, we get

$T^2 = \dfrac{4\pi^2 r_a^3}{GM_S}$

Taking the square root,

$T = 2\pi \sqrt{\dfrac{r_a^3}{GM_S}}$

$\:\:\:\:=2\pi \sqrt{\dfrac{(3.57(1.499×10^{11}\:\text{m}))^3}{(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)(1.989×10^{30}\:\text{kg})}}$

$\:\:\:\:= 2.13×10^8\:\text{s} = 6.75\:\text{years}$

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What is released when rocks shift or move along a tectonic plate boundary?

Phoenix [80]

Answer:

juuhu

Explanation:

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When an object is lifted 12 meters off the ground, it gains a certain amount of potential energy. If the same object is lifted 2

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I think the answer is twice as much

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F~1/r²
weakens by 1/4

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true or false Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer

salantis [7]

Hello Micu212006

Question: <span> Both the large loose rocks and the small loose rocks used to be part of earth's solid rock layer
</span><span>
Answer: True

Hope This Helps!
-Chris </span>

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3 years ago