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Reil [10]
2 years ago
8

Please help im being timed on this and need to get my grade up

Physics
1 answer:
nadya68 [22]2 years ago
8 0

Answer:

a I think hope this helps

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True or False: The average atomic mass is always closer to the isotope with the largest mass
svlad2 [7]
The awnser would be True. Hope I helped!
3 0
3 years ago
A book sits on a bookshelf without moving until a student picks it up. Which law best explains why the book remains at rest unti
svetoff [14.1K]

Answer:

Newtons first law

Explanation:

object in rest stays at rest

object in motion stays in motion

8 0
3 years ago
The potential energy an object has due to it position Is called _______ potential energy
Romashka-Z-Leto [24]
Gravitational potential energy :)
4 0
3 years ago
A balloon is filled with helium at a pressure of 2.4 x 105 Pa. The balloon is at a
garri49 [273]
1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L

Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol

Avogadro number = 1 mol = 6.022 * 10^23 atoms

Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms

2) Double atoms => double volume

V2 / V1 = r2 ^3 / r1/3

2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3

r2 = [∛2]r1

The factor is ∛2
5 0
3 years ago
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

3 0
3 years ago
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