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3 years ago

Pre-questioning identifies a purpose for reading.

Physics

2 answers:

nadya68 [22]3 years ago

8 0

Answer:

True

Explanation:

Pre-questioning may help a reader focus on information s/he hopes to find in the reading selection.

Lynna [10]3 years ago

7 0

Answer:

True!!

Explanation:

I took the practice thing on edge :))

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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,

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

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

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

3 0

3 years ago

4. True or false: Ionic<br> Compounds do not form<br> molecules.

Verizon [17]

Answer:

true

Explanation:

ionic compounds do not form molecules

7 0

3 years ago

Read 2 more answers

Does a falling rock have potential or kinetic energy

Neko [114]

depends t what stage in the fall it is. If it is at the peak, it is fully potential. If it is in the middle, it has both. If it is at the bottom of the fall, it is completely kinetic

3 0

3 years ago

Read 2 more answers

Two 10 kg pucks head straight towards each other with velocities of 10 m/s and -20 m/s. They collide and stick together. Calcula

RUDIKE [14]

The final velocity of the two pucks is -5 m/s

Explanation:

We can solve the problem by using the law of conservation of momentum.

In fact, in absence of external force, the total momentum of the two pucks before and after the collision must be conserved - so we can write:

$m_1 u_1 + m_2 u_2 = (m_1 +m_2)v$