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3 years ago

Donde esta la orilla del universo?

Physics

1 answer:

irinina [24]3 years ago

5 0

Answer:

<em>El universo no tiene un borde finito</em>, continúa y se expande sin fin. La distancia adecuada desde la Tierra hasta el borde del universo observable es de aproximadamente 46.5 mil millones de años luz o 4.40 × 10 ^ 26 metros en cualquier dirección.

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Scott cuts a one meter long wire into five equal parts. What is the length of each part of the wire? (2 points) 0.02 centimeters

kompoz [17]

20 centimeters. 100 centimeters in a meter. 100/5= 20

4 0

3 years ago

Read 2 more answers

Please help! I especially need help with the second question but help with the first one would be most appreciated!

lara31 [8.8K]

Answer:

a) Team A will win.

b) The losing team will accelerate towards the middle line with 0.01 m/ $s^{2}$

Explanation:

Given that Team-A pulls with a force , $F_{1} = 50N$

and Team-B pulls with a force , $F_{1} = 45N$

∵ $F_{1} > F_{2}$

The rope will move in the direction of force $F_{1}$ .

∴ Team-A will win.

b) Considering both the teams as one system of total mass , $m = 246+253 = 499 kg$

Net force on the system , $F = F_{1} - F_{2}$ = 50-45 = 5N

Applying Newtons first law to the system ,

F = ma , where 'a' is the acceleration of the system.

Since , both the teams are connected by the same rope , their acceleration would be the same.

∴ 5 = 499×a

∴ a = 0.01 m/ $s^{2}$

4 0

4 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)