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aalyn [17]
2 years ago
10

We know today that atoms cannot be divided into smaller parts true or false

Physics
1 answer:
djverab [1.8K]2 years ago
3 0
Hi , the answer is false ,atoms can be divided into smaller parts , electrons , protons and neutrons.
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Atoms in a molecule held together through sharing ______. (points 3)
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You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving
BabaBlast [244]

Y₀ = initial position of the balloon at the top of the building = 44 m

Y = final position of the balloon at halfway down the building = 44/2 = 22 m

a = acceleration of the balloon = - 9.8 m/s²

v₀ = initial velocity of the balloon = 0 m/s

v = final velocity of the balloon = ?

using the kinematics equation

v² = v₀² + 2 a (Y - Y₀)

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4 0
3 years ago
To view an enlarged upright image of an object through a simple magnifier, where must the object be located?.
Andre45 [30]

Answer:within the focal length of the lens, provided the focal length is shorter than the near point distance.

Explanation:Hope it helps

6 0
2 years ago
R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d
padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

#SPJ10

3 0
2 years ago
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