Explain what happens to the atoms in the reactants during a chemical reaction

A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro

grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

Φ $._{E}$ = ∫ E. dA = $q_{int}$ / ε₀

For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

∫ E dA = $q_{int}$ / ε₀

The area of a sphere is

A = 4π r²

E 4π r² = $q_{int}$ / ε₀

E = (1 /4πε₀ ) q / r²

Having the solution of the problem let's analyze the points:

A ) r = 3R / 4 = 0.75 R.

In this case there is no charge inside the Gaussian surface therefore the electric field is zero

E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

E = (1 /4πε₀ ) Q / (1.25R)²

E = (1 /4πε₀ ) Q / R2 1 / 1.56²

E₀ = (1 /4π ε₀ ) Q / R²

$E_{B}$ = Eo /1.56 ²

$E_{B}$ = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

$E_{B}$ =(1 /4π ε₀ ) Q 1/(2R)²

$E_{B}$ = (1 /4π ε₀ ) q/R² 1/4

$E_{B}$ = Eo 1/4

$E_{B}$ = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0

3 years ago

You need to connect your doorbell to a battery. You should select wire that will

Harlamova29_29 [7]

Run electrity through or is postive to the circuit

6 0

3 years ago

If we want to describe work , we must have what ?

ivanzaharov [21]

An external force that is being applied in the direction of the displacement

4 0

3 years ago

A 3.5 kg object moving in two dimensions initially has a velocity v1 = (12.0 i^ + 22.0 j^) m/s. A net force F then acts on the o

lys-0071 [83]

Answer:

The work done by the force is 820.745 joules.

Explanation:

Let suppose that changes in potential energy can be neglected. According to the Work-Energy Theorem, an external conservative force generates a change in the state of motion of the object, that is a change in kinetic energy. This phenomenon is describe by the following mathematical model:

$K_{1} + W_{F} = K_{2}$

Where:

$W_{F}$ - Work done by the external force, measured in joules.

$K_{1}$ , $K_{2}$ - Translational potential energy, measured in joules.

The work done by the external force is now cleared within:

$W_{F} = K_{2} - K_{1}$

After using the definition of translational kinetic energy, the previous expression is now expanded as a function of mass and initial and final speeds of the object:

$W_{F} = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{1}^{2})$

Where:

$m$ - Mass of the object, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speeds of the object, measured in meters per second.

Now, each speed is the magnitude of respective velocity vector:

Initial velocity

$v_{1} = \sqrt{v_{1,x}^{2}+v_{1,y}^{2}}$

$v_{1} = \sqrt{\left(12\,\frac{m}{s} \right)^{2}+\left(22\,\frac{m}{s} \right)^{2}}$

$v_{1} \approx 25.060\,\frac{m}{s}$

Final velocity

$v_{2} = \sqrt{v_{2,x}^{2}+v_{2,y}^{2}}$

$v_{2} = \sqrt{\left(16\,\frac{m}{s} \right)^{2}+\left(29\,\frac{m}{s} \right)^{2}}$

$v_{2} \approx 33.121\,\frac{m}{s}$

Finally, if $m = 3.5\,kg$ , $v_{1} \approx 25.060\,\frac{m}{s}$ and $v_{2} \approx 33.121\,\frac{m}{s}$ , then the work done by the force is:

$W_{F} = \frac{1}{2}\cdot (3.5\,kg)\cdot \left[\left(33.121\,\frac{m}{s} \right)^{2}-\left(25.060\,\frac{m}{s} \right)^{2}\right]$

$W_{F} = 820.745\,J$

The work done by the force is 820.745 joules.

6 0

3 years ago

A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu

kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) = ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i) work required to pump the contents to the rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the cone's rim

Extra work required = 28,914.229 lb × 5ft = 144571.145 lb.ft

Total work required = (281,913.733 + 144571.145) lb.ft

= 426,484.878 lb.ft

5 0

3 years ago