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3 years ago

What is the biggest planet in our galaxy

1 answer:

4 0

We won't be able to answer that until we've seen all of them and
we know that there aren't any more that we haven't seen yet.

There are estimated to be between 200 billion and 400 billion stars
in our galaxy, and a large part of the stars in the galaxy aren't visible
to us. So you can see that it's going to take some time before we're
able to answer your question. Check back with us next Tuesday.

You might be interested in

Assume that segment r exerts a force of magnitude t on segment l. what is the magnitude flr of the force exerted on segment r by

mrs_skeptik [129]

If we are talking on the force being exerted by a segment of a rope of lenght R on the right on a point M which is being also pulled from the Left by a segment of rope R as shown in the figure attached. Then we invoke Newton's Third Law:
"Any force exerted by an object (in this case a segment of the rope) also suffers a equal and opposite force".
If we pick $T_R=T$ whis is the tension exerted by the right segment then the left segment will also exert an equal and opposite force so we have that $T_L=-T$

8 0

3 years ago

Par 1/2

BaLLatris [955]

part 1

mass = ρ x V

mass = 1739 kg/m³ x 3.8 km³ = 6608.2 kg

PE (potential energy)= mgh

PE = 6608.2 kg x 9.81 x 403

PE = 2.61 x 10⁷ J

part 2

megaton of TNT (Mt) =4.2 x 10¹⁵ J

convert PE to Mt:

2.61 x 10⁷ J : 4.2 x 10¹⁵ J = 6.21 x 10⁻⁹ Mt

4 0

3 years ago

A solid cylinder of mass 12.0 kgkg and radius 0.250 mm is free to rotate without friction around its central axis. If you do 75.

dybincka [34]

Answer:

The final angular velocity is 20rad/s

Explanation:

We are given;

mass, m = 12 kg

radius, r = 0.25 m

Work done;W = 75 J

Moment of inertia of cylinder, I = (1/2) mr²

Thus,

I = (1/2) x 12 x 0.25² = 0.375 kg.m²

Now, from work energy theorem,

Work done = Change in kinetic energy

So, W = KE_f - KE_i

Now, Initial Kinetic Energy (KE_i) = 0

Final Kinetic Energy; KE_f = (1/2)Iω²

So, KE_f = (1/2) x 0.375 x ω²

KE_f = 0.1875 ω²

Now, W = 75 J

Thus,

From, W = KE_f - KE_i, we have;

75 = 0.1875 ω² - 0

75 = 0.1875 ω²

ω² = 75/0.1875

ω² = 400

ω = √400

ω = 20 rad/s

5 0

3 years ago

PLEASE HELP ASAP.

kifflom [539]

Volume of the original block (V) of length (l), width (w) and height (h) is given by the following equation

$V = l*b*h$

hence volume of original block, V = (6)(4)(10) = $240 cm^{3}$

Now say we cut this block in two equal parts, than the length of new block will be $6/2 = 3 cm$ ; where as its width and height will be same as of original block i.e. 4 cm and 10 cm respectively.