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cestrela7
1 month ago
14

Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic ener

gy kb of particle
b. remember that both particles rotate about the y axis.
Physics
1 answer:
Keith_Richards [3.2K]1 month ago
5 0
<span>Response: KE = (11/2)mω²r², particle B should have a mass of 2m, while A consists of mass m. Thus, the moment of inertia for the system is I = Σ md² = m*(3r)² + 2m*r² = 11mr² and subsequently KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2 Therefore, I will assume this as my basis. For particle A, translational KEa = ½mv² but v = ω*d = ω*3r, thus KEa = ½m(3ωr)² = (9/2)mω²r² For particle B, translational KEb = ½(2m)v² but v = ω*r, so KEb = ½(2m)ω²r² total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2 which equals our rotational KE.</span>
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True or False: Molecules in a gas resist crowding and get as far apart as possible. Free electrons also resist crowding and get
ValentinkaMS [3465]

Answer:

This assertion is inaccurate.

Explanation:

The random nature of gas molecules results in their erratic motion and occasional collisions. While it is true that they tend to avoid being tightly packed, achieving the maximum separation from each other is not always feasible due to their lack of fixed positions. Consequently, gas molecules in a container cannot consistently maintain the furthest distance from their neighboring molecules.

In contrast, the separation among electrons is primarily influenced by repulsive forces, not random movement as in gases. Electrons maintain distance as a result of repulsion between similarly charged particles. Therefore, the arrangement of electrons on a charged copper sphere occurs not from a random distribution but rather due to repulsion, establishing a set distance between them.

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2 months ago
Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle
Ostrovityanka [3204]

Answer:

Explanation:

Let us denote the launch angle as \theta _1=30^{\circ}.

\theta _2=60^{\circ}, \theta _1, and \theta _2 are complementary angles which means their ranges are identical.

H_{max}=\frac{u^2(\sin \theta )^2}{2g}

H_{30}=\frac{u^2(\sin 30)^2}{2g}

H_{30}=\frac{u^2}{8g}

Time of flight is indicated as =\frac{2u\sin \theta }{g}.

t_{30}=\frac{u}{g}

For \theta =60^{\circ}

H_{60}=\frac{u^2(\sin 60)^2}{2g}

H_{60}=\frac{3u^2}{8g}

t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}

H_{60}=3\times H_{30}

t_{60}=\sqrt{3}\times t_{30}

7 0
2 months ago
A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the
ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

  • 52.92% metal: 52.92 g M
  • 47.80% oxygen: 47.80 g O

 By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

moles of oxygen=\frac{47.8g*1mol}{16g}

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M

Given 52.92 g of metal in the sample, the molar mass of the metal is:

molar mass=\frac{52.92 g}{5.975 mol}

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0
2 months ago
The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m
Sav [3153]

Answer:

The density comes out to be 10^{6} Mg/µL

Explanation:

Given data:

The density of nuclear matter is approximately 10^{18} kg/m³

1 ml corresponds to 1 cm³

To determine:

The density of nuclear matter in Mg/µL

Solution:

We recognize that:

1 Mg equals 1000 kg

Thus, 1 m³ is equal to 10^{6} cm³

Moreover, 1 cm³ is equivalent to 1 mL

Thus, we can conclude that 1 mL is equal to 10³ µL

With this, we convert the density as follows:

Density = 10^{18} kg/m³

Density = 10^{18} kg/m³ × \frac{10^{-3} }{10^{6} }  Mg/µL

Density = 10^{6} Mg/µL

8 0
2 months ago
Read 2 more answers
A solid uniform disk of diameter 3.20 m and mass 42 kg rolls without slipping to the ) bottom of a hill, starting from rest. If
kicyunya [3294]

Answer:

(A) = 3.57 m

Explanation:

According to the question, the information provided is:

diameter (d) = 3.2 m

mass (m) = 42 kg

angular speed (ω) = 4.27 rad/s

Using the conservation of energy principle, we have

mgh = 0.5 mv² + 0.5Iω²...equation 1

where

Inertia (I) = 0.5mr²

ω = v/r

Revising equation 1, it turns into

mgh = 0.5 mv² + 0.5(0.5mr²)(v/r)²

resulting in gh = 0.5 v² + 0.5(0.5)v²

This simplifies to 4gh = 2v² + v²

thus h = 3v² ÷ 4g... equation 2

Given ω = v/r, we find v = ωr = 4.27 × (3.2 ÷ 2)

which yields v = 6.8 m/s

Next, substituting the value of v into equation 2 gives us

h = 3v² ÷ 4g

h = 3 × (6.8)² ÷ (4 × 9.8)

h = 3.57 m

8 0
2 months ago
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