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jek_recluse [69]

3 years ago

15

How are light years used to measure distances in the universe

1 answer:

FinnZ [79.3K]3 years ago

4 0

A light year is a unit of distance. It is a distance that light can travel in a years time which is six trillion miles. It is used to measure the distances in space. To take one example, the distance to the next nearest big galaxy, the Andromeda Galaxy, from earth is 21,000,000,000,000,000,000 km.
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What are the formulas for Work, KE, GPE and Power?

Katyanochek1 [597]

Answer:

work=f×d

GPE =m×g×h

$kinetic energy = \frac{1}{2} mv {}^{2} \\ power = \frac{w}{t}$

6 0

3 years ago

The free-fall acceleration at the surface of planet 1 is 22 m/s^2. The radius and the mass of planet 2 are twice those of planet

algol13

Answer:

g₂ = 11 m/s²

Explanation:

The value of free-fall acceleration on the surface of a planet is given by the following formula:

$g = \frac{Gm}{r^2}$

where,

g = free-fall acceleration

G = Universal Gravitational Constant

m = mass of the planet

r = radius of planet

FOR PLANET 1:

$g_1 = \frac{Gm_1}{r_1^2}\\\\\frac{Gm_1}{r_1^2} = 22 m/s^2$ --------------------- equation (1)

FOR PLANET 2:

$g_2 = \frac{Gm_2}{r_2^2}\\\\g_2 = \frac{G(2m_1)}{(2r_1)^2}\\\\g_2 = \frac{1}{2}\frac{Gm_1}{r_1^2}\\\\$

using equation (1):

$g_2 = \frac{g_1}{2}\\\\g_2 = \frac{22\ m/s^2}{2}$

<u>g₂ = 11 m/s²</u>

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2 years ago

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Answer:

Semiconductor. sorry i'm late

Explanation:

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Read 2 more answers

A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

5 0

3 years ago