5

Physics

2 answers:

MakcuM [25]2 years ago

8 0

Answer:

I believe the answer is triangle

Hope it helps!

lana [24]2 years ago

4 0

C because your cutting from the side the the pointy part would make a triangle

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I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

3 0

2 years ago

A 6.3 g bullet leaves the muzzle of a rifle with a speed of 596.2 m/s. what constant force is exerted on the bullet while it is

ivanzaharov [21]

anwser will be

F = ma

where

F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet

Substituting appropriately,

F = 0.005a --- call this Equation 1

Next working equation is

Vf^2 - Vo^2 = 2as

where

Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel

Substituting appropriately,

326^2 - 0 = 2(a)(0.83)

a = 64,022 m/sec^2

the anwser will be
Substituting this into Equation 1,

F = 0.005(64,022)

F =320.11 Newtons

Hope this helps.

8 0

3 years ago

What is the Morgan-Keenan Spectral Classification system?

Angelina_Jolie [31]

Answer: A method of classifying stars by their temperatures and compositions.

Explanation:

4 0

2 years ago

A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0

Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is $5.74\times10^{13}\ m/s^2$