A block is pushed so that it moves up a ramp at constant speed. Identify from choices (a)-(e) below the appropriate description for the work done by the specified force while the block moves from point A to point B. (a) is zero. (b) is less than zero. (c) is greater than zero. (d) could be positive or negative depending on the choice of coordinate systems. (e) cannot be determined.

*The work of the Normal (N) y Wy are zero  answer a

*The work of the applied force (F1)  is positive answer c

*The work of the friction force (fr) is negative, answer b

*The work of the Wy isnegative, answer d

Explanation:

In this exercise it is asked to identify the type of work, unfortunately the diagram cannot be seen, but in the attached we can see the diagram of a body moving upward on an inclined plane, the existing forces are shown.

As the body moves at constant speed the accelerations are zero. Let's look for the job that is defined

           W = F. d

            W = F d cos θ

where the dot represents the dot product and the bold letters are vectors.

* The work of the Normal (N) and the y component of the weight (Wy) are zero because they are perpendicular to the motion

    answer a

* The work of the applied force (F1) is positive because it is in the same direction of motion

        W = F1 Δx

answer c

* The work of the friction force (fr) is negative because the force in the displacement have opposite directions

        W = -fr Δx

answer b

* the work the x component of the weight (Wx) in this case is negative

answer d

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:

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.

Answer:

Step-by-step explanation:

Time taken by the tomatoes to each the ground

using h = 1/2 g t^2 

t^2 = 2h/g = 2 x 50/ 9.8 = 10.2

t = 3.2 sec 

horizontal ditance = speed x time = 3 x 3.2 = 9.6 meters

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