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amid [387]
2 years ago
11

Jack's model includes one object consisting of two molecules and one object consisting of six molecules. He put 16 energy cubes

in each object and stated that his model showed the two objects at equilibrium.
Explain what is wrong with Jack’s model.
Physics
1 answer:
Novay_Z [31]2 years ago
5 0

Answer:

the molecules have different energy and the system is not in equilibrium

Explanation:

The model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.

 So when you book them together the molecules have different energy and the system is not in equilibrium

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Which of the following describes a physical change? A. A marshmallow burns over an open flame. B. Icicles melt from a rooftop. C
AnnZ [28]

Answer:

B

Explanation:

Icicles melt from a rooftop

5 0
2 years ago
Two identical metal balls are rolling without slipping along a horizontal surface with speed V. Each ball encounters a hill ('tw
Anastaziya [24]

Answer:

The angular velocity of Ball A will be greater than the angular velocity of Ball B when they reach the top of the hill.

Explanation:

Angular velocity can be defined as how fast an object rotates relative to a given point or frame of reference.

The question said the hill encountered by Ball A is frictionless, so Ball A will continue to rotate at the same rate it started with even when it reached the top of the hill.

Ball B on the other hand rolls without slipping over its hill, i.e there's friction to slow down its rotational motion which thus reduces how fast Ball B will rotate at the top of the hill

3 0
2 years ago
A 100 cm3 block of lead weighs 11N is carefully submerged in water. One cm3 of water weighs 0.0098 N.
Pie

#1

Volume of lead = 100 cm^3

density of lead = 11.34 g/cm^3

mass of the lead piece = density * volume

m = 100 * 11.34 = 1134 g

m = 1.134 kg

so its weight in air will be given as

W = mg = 1.134* 9.8 = 11.11 N

now the buoyant force on the lead is given by

F_B = W - F_{net}

F_B = 11.11 - 11 = 0.11 N

now as we know that

F_B = \rho V g

0.11 = 1000* V * 9.8

so by solving it we got

V = 11.22 cm^3

(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N

(iii) Buoyant force = 0.11 N

(iv)since the density of lead block is more than density of water so it will sink inside the water


#2

buoyant force on the lead block is balancing the weight of it

F_B = W

\rho V g = W

13* 10^3 * V * 9.8 = 11.11

V = 87.2 cm^3

(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight =  11.11 N

(iii) Buoyant force = 11.11 N

(iv) since the density of lead is less than the density of mercury so it will float inside mercury


#3

Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid

3 0
2 years ago
Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H
kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

we substitute the value of B = 2

             U = 1 / r²

5 0
2 years ago
A 5.00-pF, parallel-plate, air-filled capacitor with circular plates is to be used in a circuit in which it will be subjected to
Ne4ueva [31]

Answer:

a) r=4.24cm d=1 cm

b) Q=5x10^{-10} C

Explanation:

The capacitance depends only of the geometry of the capacitor so to design in this case knowing the Voltage and the electric field

V=1.00x10^{2}v\\E=1.00x10^{4} \frac{N}{C}

V=E*d\\d=\frac{V}{E}\\d=\frac{1.0x10^{2}}{1.0x10^{4}}\\d=0.01m

The distance must be the separation the r distance can be find also using

C=\frac{Q}{V_{ab}}

But now don't know the charge these plates can hold yet so

a).

d=0.01m

C=E_{o}*\frac{A}{d}\\A=\frac{C*d}{E_{o}}

A=\frac{5pF*0.01m}{8.85x10^{-12}\frac{F}{m}}\\A=5.69x10^{-3}m^{2}

A=\pi *r^{2}\\r=\sqrt{\frac{A}{r}}\\r=\sqrt{\frac{5.64x10^{-3}m^{2} }{\pi } }  \\r=42.55x^{-3}m

b).

C=\frac{Q}{V_{ab}}

Q=C*V\\Q=5x10^{-12} F*1x10^{2}\\Q=5x10^{-10}C

8 0
2 years ago
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