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Wewaii [24]
3 years ago
10

How are red dwarf stars and yellow dwarf stars alike?

Physics
1 answer:
valkas [14]3 years ago
8 0
Yellow and red dwarf stars are normal stars - they burn hydrogen in their cores and live on the main sequence of stellar lifetimes.
You might be interested in
a heavy jar sits on top of a 3.4 m shelf with a gravitational potential energy of 180 j. What is the mass of the jar?
sp2606 [1]

Answer:

5.2941176471 kg or 5294.1 grams

Explanation:

g.p.e= mgh

g.p.e/gh=m

180j/10×3.4= m

180/34= m

5.2941 kg= m

6 0
3 years ago
you know that there are 1609 meter in a mile. the number of feet in a mile is 5280. what is the speed snail from problem 7 per m
Zanzabum

Answer:

We know that 1 meter = 100 centimeters, and 1 foot = 12 inches.

So  (1,609 meters) x (100 centimeters/meter)  =  (5,280 feet) x (12 inches/foot)

The second fraction on each side of the equation is equal to ' 1 ', because

the numerator is equal to the denominator, so sticking it in there doesn't

change the value of that side of the equation.  But now we can cancel some

units,and wind up with the units we need.

  (1,609 meters) x (100 centimeters/meter)  =  (5,280 feet) x (12 inches/foot)

        (1,609 x100) centimeters  =  (5,280 x 12) inches

            160,900 centimeters  =  63,360 inches

Divide each side by  63,360 :        2.54 centimeters =  1 inch

Explanation:

5 0
2 years ago
If you weigh 882 N on EARTH (HINT: What number do we ALWAYS use for gravity on Earth), what is your mass? ​
Softa [21]
882 divided by 9.81 (this is acceleration due to gravity) it equals 89.91
4 0
3 years ago
three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0
3 years ago
What is the energy difference between the second excited state and first excited state of an electron in the "box" of size L=1nm
jolli1 [7]
I thinks it’s A, tell me if you get it right
4 0
3 years ago
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