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vovangra [49]
3 years ago
7

How fast does light year travel

Physics
2 answers:
KIM [24]3 years ago
5 0
"Light year" is a distance, not a speed. It's the distance light travels in one year, at the speed of 299,792,458 meters per second.
Allisa [31]3 years ago
4 0
<span>299,792 kilometers per second im pretty sure


</span>
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Kevin is a major stockholder in Professional Transmission Services (PTS), a nationwide network of transmission repair shops foun
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Answer:

Kevin wants to make the firm private

Explanation:

Based on the information provided within the question in regards to the situation it seems that Kevin wants to make the firm private. Private Firms are companies that are owned by non-governmental entities or instead owned by  a single individual or a very small amount of shareholders. Which is what Kevin seems to be wanting to do since he wants to buy back all the shares of the company so that only his family owns the company.

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The average distance from the sun to Pluto is approximately 6.10 × 109 km. How long does it take light from 
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What are five types of organisms That can reproduce asexually
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PHYSICS QUESTION - CAN ANYONE HELP NOW?
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It's velocity when it strikes the ground is. D. 232.9 kg.m/s<span>.
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I hope this helps!!!







3 0
3 years ago
Read 2 more answers
A horizontal block-spring system with the block on a frictionless surface has total mechanical energy E = 50.9 J and a maximum d
Llana [10]

(a) 2446 N/m

When the spring is at its maximum displacement, the elastic potential energy of the system is equal to the total mechanical energy:

E=U=\frac{1}{2}kA^2

where

U is the elastic potential energy

k is the spring constant

A is the maximum displacement (the amplitude)

Here we have

U = E = 50.9 J

A = 0.204 m

Substituting and solving the formula for k,

k=\frac{2E}{A^2}=\frac{2(50.9)}{(0.204)^2}=2446 N/m

(b) 50.9 J

The total mechanical energy of the system at any time during the motion is given by:

E = K + U

where

K is the kinetic energy

U is the elastic potential energy

We know that the total mechanical energy is constant: E = 50.9 J

We also know that at the equilibrium point, the elastic potential energy is zero:

U=\frac{1}{2}kx^2=0 because x (the displacement) is zero

Therefore the kinetic energy at the equilibrium point is simply equal to the total mechanical energy:

K=E=50.9 J

(c) 8.55 kg

The maximum speed of the block is v = 3.45 m/s, and it occurs when the kinetic energy is maximum, so when

K = 50.9 J (at the equilibrium position)

Kinetic energy can be written as

K=\frac{1}{2}mv^2

where m is the mass

Solving the equation for m, we find the mass:

m=\frac{2K}{v^2}=\frac{2(50.9)}{(3.45)^2}=8.55 kg

(d) 2.14 m/s

When the displacement is

x = 0.160 m

The elastic potential energy is

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

So the kinetic energy is

K=E-U=50.9 J-31.3 J=19.6 J

And so we can find the speed through the formula of the kinetic energy:

K=\frac{1}{2}mv^2 \rightarrow v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(19.6)}{8.55}}=2.14 m/s

(e) 19.6 J

The elastic potential energy when the displacement is x = 0.160 m is given by

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

And since the total mechanical energy E is constant:

E = 50.9 J

the kinetic energy of the block at this point is

K=E-U=50.9 J-31.3 J=19.6 J

(f) 31.3 J

The elastic potential energy stored in the spring at any time is

U=\frac{1}{2}kx^2

where

k = 2446 N/m is the spring constant

x is the displacement

Substituting

x = 0.160 m

we find the elastic potential energy:

U=\frac{1}{2}kx^2=\frac{1}{2}(2446)(0.160)^2=31.3 J

(g) x = 0

The postion at that instant is x = 0, since it is given that at that instant  the system passes the equilibrium position, which is zero.

4 0
3 years ago
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