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patriot [66]
3 years ago
15

In an experiment, Mary compares a collision of two 1-kilogram steel balls with a collision of two 1-kilogram foam rubber balls.

She makes sure that, except for the material, the two collisions are the same. Mary finds that the force of impact of the steel ball collision is greater than the force of impact of the foam rubber ball collision.
From her force measurements, Mary concludes that the foam rubber balls’ acceleration during the collision was (BLANK) the steel balls’ acceleration. She concludes that the collision between the foam rubber balls is an collision.

BLANK 1
Greater than
Less than
The same as
Physics
2 answers:
IgorC [24]3 years ago
8 0
1. Greater then
<span>2. Inelastic</span>
Lynna [10]3 years ago
4 0

Complete sentence:

Mary concludes that the foam rubber balls’ acceleration during the collision was less than the steel balls’ acceleration. She concludes that the collision between the foam rubber balls is an inelastic collision.

Explanation:

Mary find that the force of impact on the steel ball is greater than the force on the foam rubber ball. According to Newton's second law, the acceleration of each ball is proportional to the force applied:

a=\frac{F}{m}

where F is the force applied and m is the mass of the ball. The two balls have same mass (1 kg), so since the force on the steel ball is greater, the acceleration of the steel ball is also greater.

The collision between the foam rubber balls is inelastic: it means that only the total momentum is conserved, while the total kinetic energy is not conserved.

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kumpel [21]

Answer:

a)mass = 300kg

b) no of atoms = 1.81 x 10^26atoms.

c)Fission Activity = 207Bq

d) no of fissions = 1.79 × 107 d^-1

Explanation:

a)mass= (3×10-4)(10^6kg) = 300kg

b) no of atoms = (300kg) X (6.022×10^23/kg) = 1.81 x 10^26atoms.

c)Fission Activity = (0.69)(300kg) = 207Bq

d) no of fissions = (207)(86400) = 1.79 × 107 d^-1

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3 years ago
Calculate the value of ∆g for these initial partial pressures. (the partial pressure of each gas is set a 5. 0 atm at 25°c) (use
Sati [7]

∆g for these initial partial pressures is 10,403.31 KJ.

   

ΔG gets increasingly positive as a product gas's partial pressure is raised. ΔG becomes more negative as the partial pressure of a reactant gas increases.

                                  ∆g = RT ln (q/k)

In this equation: R = 8.314 J mol⁻¹ K⁻¹ or 0.008314 kJ mol⁻¹ K⁻¹

                            K = 325

If ΔG < 0, then K > Q, and the reaction must proceed to the right to reach equilibrium.

∴∆g = RT ln (q/k)

        = 8.314 × 298 ln ( 5 / 325)

        = 2477.57 ln 0.015

        = 2477.57 × (-4.199)

        = 10,403.31 KJ

Products are preferred over reactants at equilibrium if G° 0 and both the products and reactants are in their standard states. When reactants are preferred above products in equilibrium, however, if G° > 0, K 1. At equilibrium, neither reactants nor products are preferred if G° = 0, hence K = 1.

Therefore, ∆g for these initial partial pressures is 10,403.31 KJ.

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That information describes the plane's speed. 
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