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4 years ago

What is the name for a star and planets held together by gravity? A. solar system B. galaxy C. black hole D. supernova

Physics

1 answer:

DaniilM [7]4 years ago

5 0

These answers aren´t valid .

The correct answer will be:

Planetary system.

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In which way are the cathode rays deflected in the x-plates of the Cathode Ray Oscilloscope

MrRissso [65]

Answer:

The cathode ray is deflected vertically to the fluorescent screen

Explanation:

$.$

4 0

3 years ago

Read 2 more answers

PLEASE HELP!! 9TH K12 GCA LIGHT TEST REVIEW SHEET FOR PHYSICS! 100 POINTS!!

Svetllana [295]

I got a 100 on this. I took a bunch of snips of it for you. Mine is a little different though. I only have 6 questions and most of yours are the same as mine. And some of yours that I have are in a different order. I hope I helped.

6 0

4 years ago

the brightest , hottest, and most massive stars are the brilliant blue stars designated as spectral class O. if a class O star w

4vir4ik [10]

The speed is 0.956 m / s.

<u>Explanation</u>:

The kinetic energy is equal to the product of half of an object's mass, and the square of the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

where K.E represents the kinetic energy,

m represents the mass,

v represents the velocity.

K.E = 1/2 $\times$ m $\times$ $v^{2}$

1.10 $\times$ 10^42 = 1/2 $\times$ 3.26 $\times$ 10^31 $\times$ $v^{2}$

$v^{2}$ = (1.10 $\times$ 10^42 $\times$ 2) / (3.26 $\times$ 10^31)

v = 0.956 m / s.

6 0

4 years ago

A young lady can paddle a canoe in a lake 7.7 m/s. She paddles downstream in a river whose current is 12.4 m/s. What is the comb

GenaCL600 [577]

Answer:20.1m/s

Explanation:

6 0

4 years ago

You are explaining why astronauts feel weightless while orbiting in the space shuttle. Your friends respond that they thought gr

Yuri [45]

Answer:

The answer is "83.1%".

Explanation:

Given:

$\text{Mass of the earth}\ (M_E) = 5.97 \times 10^{24}\ \ kg\\\\\text{Radius of the earth}\ (R_E) = 6380 \ km = 6.38 \times 10^{6}\ \ m\\\\\text{acceleration of gravity}\ (g_E) = 608 \ \ km= 608,000 \ \ m \\\\G= 6.67 \times 10^{-11} \ \ \frac{N \cdot n^2}{kg^2}$

Using formula:

$\to g_E = G \frac{M_E}{(R_E +h)^2}$

$\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6.38 \times 10^{6}+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km} =6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,988,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{4.88 \times 10^{13}}\\\\$

$\to g_{608 \ km}=6.67\times10^{-24}\frac{5.97 \times 10^{24}}{4.88}\\\\\to g_{608 \ km}=6.67\times \frac{5.97}{4.88}\\\\\to g_{608 \ km}=6.67\times 1.22336066\\\\\to g_{608 \ km}= 8.15 \ \frac{m}{s^2}$

Calculating the gravity on the Earth’s surface:

$\to \frac{g_{608 \ km} }{ g_{\ earth \ surface}}$ $= \frac{8.15}{9.8} \times 100=83.1 \%$

8 0

3 years ago