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valentina_108 [34]

3 years ago

5

Is carbon a metal, nonmetal, or semi-metal

1 answer:

Vlada [557]3 years ago

5 0

Answer: Non metal.

Explanation: Carbon has 4 electrons in it’s valence shell which makes it a metalloid but commonly it is considered as a non metal.

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Geosynchronous satellites orbit the Earth at a distance of 42 000 km from the Earth's center. Their angular speed at this height

frozen [14]

Answer:

Force will be 7.97 N

Explanation:

We have given that earth is at s distance of 42000 km from the earth center

So r = 42000 km = 42000000 m

Mass m = 1500 kg

Angular velocity of earth $\omega =7.29\times 10^{-5}rad/sec$

We have to find the force

We know that force is given by

$F=m\omega ^2r=1500\times (7.29\times 10^{-5})^2\times 42000000=797.1615\times 10^{-2}=7.97N$

7 0

3 years ago

Someone please help. science <3 thx<br> ill give 15 pts for it.

STALIN [3.7K]

Answer:

put these numbers in the boxes from up to down. hope this helps! :)

Explanation:

7

6

3

1

8

4

2

5

5 0

2 years ago

Hai điện tích +q và -q (q>0) đặt tại hai điểm a và b với AB=a, M là một điểm nằm trên đường trung trực của AB cách AB một đoạ

stepladder [879]

I de momento que no what now what the que no se puede el gobierno del presidente y el gobierno

6 0

2 years ago

If some raps 3words per second then how many words would he say in 9,999,999,990 seconds

wolverine [178]

Answer:

29,999,999,970 words

Explanation:

9,999,999,990x3

3 0

2 years ago

Read 2 more answers

Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30∘ above the

Pavlova-9 [17]

Answer:1.71 m/s

Explanation:

Given

mass of Susan $m=12 kg$

Inclination $\theta =30^{\circ}$

Tension $T=29 N$

coefficient of Friction $\mu =0.18$

Resolving Forces Along x axis

$F_x=T\cos \theta -f_r$

where $f_r=friction\ Force$

$F_y=mg-N-T\sin \theta$

since there is no movement in Y direction therefore

$N=mg-T\sin \theta$

and $f_r=\mu N$

Thus $F_x=T\cos \theta -\mu N$

$F_x=29\cos (30)-\0.18\times (12\times 9.8-29\sin (30))$

$F_x=25.114-18.558$

$F_x=6.556 N$

Work done by applied Force is equal to change to kinetic Energy

$F_x\cdot x=\frac{1}{2}\cdot mv_f^2-\frac{1}{2}\cdot mv_i^2$

$6.556\times 2.7=\frac{1}{2}\cdot 12\times v_f^2$

$v_f^2=\frac{6.556\times 2.7\times 2}{12}$

$v_f^2=2.95$

$v_f=1.717 m/s$

8 0

3 years ago