A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele

Result:

,

Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

while the mass of the electron is

and the charge is

By plugging in the values into the formula, we achieve

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=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:

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

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.

if you want the short reply, the answer is B

Answer:

x = v₀ cos θ   t,   y = y₀ + v₀ sin θ t - ½ g t2

Explanation:

This pertains to a projectile motion scenario. Here, we will express the equations for both the x and y dimensions.

Now, we will apply trigonometry to determine the initial velocity components.

              sin θ = / v₀

              cos θ = v₀ₓ / v₀

              v_{y} = v_{oy} sin θ

              v₀ₓ = v₀ cos θ

Next, let's formulate the equations of motion.

X axis

         x = v₀ₓ t

         x = v₀ cos θ   t

        vₓ = v₀ cos θ

Y axis

        y = y₀ + t - ½ g t2

        y = y₀ + v₀ sin θ t - ½ g t2

        v_{y} = v₀ - g t

       v_{y} = v₀  sin θ - gt

        = v_{oy}^2 sin² θ - 2 g y

It is evident that the major distinction lies in the fact that in an inclined launch compared to a horizontal one, the velocity comprises different components

B) 14.0 N

To address this inquiry, we need to evaluate the kinetic energy of the box before and after crossing the rough section. The kinetic energy is given by the formula:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Now, utilizing the known data, we compute the energy prior and post.

Before:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (2.25 m/s)^2

E = 6.75 kg * 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (1.2 m/s)^2

E = 6.75 kg * 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

Hence, the box consumed energy equal to 34.17188 J - 9.72 J = 24.451875 J over a length of 1.75 meters. Next, we will calculate the loss per meter by dividing the energy loss by the distance traversed.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

When we round to one decimal point, we arrive at 14.0 N, which corresponds with option “B.”