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Maru [420]
3 years ago
8

Light year is a unit of?

Physics
1 answer:
IrinaVladis [17]3 years ago
3 0
Light year is the unit of distance. It is the distance that an object travels in one year with the speed of light.

In short, Your Answer would be "Distance"

Hope this helps!
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A sample of chloroform is found to contain 12.0 g of carbon, 106.4 g of chlorine, and 1.01 g of hydrogen. If a second sample of
GREYUIT [131]

Given:

Sample 1:

Chloroform is CHCl_{3}

12 g Carbon

1.01 g Hydrogen

106.4 g Cl

Sample 2:

30.0 g of Carbon

Solution:

mass of chloroform from sample 1:

12 + 1.01 +106.4 =119.41 g

Now, for the total mass of chloroform in sample 2:

mass of chloroform \times\frac{given mass of carbon}{mass of carbon atom}

mass of chloroform = 119.41 \times\frac{30}{12} = 298.53 g

6 0
3 years ago
ANSWER THE 3 QUESTION WILL MARK AS BRAINLIEST THIS IS 20 POINTS :)))
PtichkaEL [24]
15) a 
16) b
17) a

Hope this helps
3 0
3 years ago
Read 2 more answers
A baby elephant is stuck in a mud hole. to help pull it out, game keepers use a rope to apply a force f with arrowa, as part a o
Julli [10]

The two forces should be equal therefore:

2.10 * Fa = Fa + 2 * F * cos 18

simplifying the right side:

2.10 * Fa = Fa + 1.902 * F

1.10 Fa = 1.902 F

<span>F / Fa = 0.578</span>

3 0
2 years ago
Help please!
Talja [164]

The correct expression for the maximum speed of the object during its motion is \frac{FT}{m}.

<h3>Maximum speed of the object</h3>

The maximum speed of the object is determined using the following formulas;

v(max) = Aω

where;

  • A is the amplitude of the motion
  • ω is angular speed

The maximum speed of the object can also be obtained from the maximum net force on the object,

F = ma

where;

  • F is the maximum net force
  • a is the acceleration
  • m is mass of the object

F = m(v/t)

mv = Ft

v = Ft/m

Thus, the correct expression for the maximum speed of the object during its motion is \frac{FT}{m}.

Learn more about maximum speed here: brainly.com/question/4931057

3 0
2 years ago
A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th
Shalnov [3]
Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.= 
<span>Wt. of skier. </span>

<span>Fp=588*sin35 = 337 N.=Force parallel to </span>
<span>incline. </span>
<span>Fv = 588*cos35 = 482 N. = Force perpendicular to incline. </span>

<span>Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force </span>
<span>of kinetic friction. </span>
<span>d =h/sinA = 2.5/sin35 = 4.36 m. </span>

<span>Ek + Ep = Ekmax - Fk*d </span>
<span>Ek = Ekmax-Ep-Fk*d </span>
<span>Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. </span>
<span>Ek = 0.5m*V^2 = 2682 J. </span>
<span>30*V^2 = 2682 </span>
<span>V^2 = 89.4 </span>
<span>V = 9.5 m/s = Final velocity.</span>
4 0
3 years ago
Read 2 more answers
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