A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of charge distribution is inside the surface? a positively charged plane parallel to the end faces of the cylinder a positive line charge situated on and parallel to the axis of the cylinder a collection of positive point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a collection of negative point charges arranged in a line at the center of the cylinder and perpendicular to the axis of the cylinder a negatively charged plane parallel to the end faces of the cylinder (b) If the radius of the cylinder is 0.66 m and the magnitude of the electric field is 300 N/C, what is the net electric flux through the closed surface? How is the electric flux related to the electric field vector and the normal to the surface? What is the orientation of the electric field relative to the curved surface? N · m2/C (c) What is the net charge inside the cylinder?

Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

Magnitude of electric field, E=300 N/C

We have to find the net flux through the closed surface.

Net electric flux,

c.

Net charge,

Where

Where

Independent variable: the best method to get rid of them.

Dependent variable: washing with soap and water.

Hypothesis: Organic oils

Control group: water

Answer:

Step-by-step explanation:

D.
Heat each cube to the same temperature, place each cube into different containers with 500 grams of water at the same temperature, and measure the temperature of the water.

Answer:

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

The question specifies the diameter of the screw, therefore the IMA of this screw is 0.812? / 0.318 = 8.02