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larisa [96]
3 years ago
12

Which factors allow for life on Earth​

Physics
2 answers:
Elan Coil [88]3 years ago
6 0

Are there supposed to be multiple choices for this question?

Vlad1618 [11]3 years ago
5 0

Answer: water some food ALOT of FUNNN!!! and you'll live you also need the big guy up there

^

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A dog of mass 18 kg runs at a speed of 4 m/s. What is the momentum of the
Andrej [43]

Answer:

A, 72 kg•m/s

Explanation:

p=mv

p=18x4

p=72

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3 years ago
How do u answer this?
gayaneshka [121]

Answer:

food

Explanation:

did you get a chance to look at the maximum number of devices allowed by

4 0
3 years ago
the primary functionn of the bias circuit is to hold the circuit stable at the destinated Q-point. true or false
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Shake r risks e wish sis false
3 0
2 years ago
Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might
Tanya [424]
<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
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
2 years ago
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