g A boat is anchored 2000 ft from shore anddirects its searchlight towards an automobile travelingdown the straight road. At the particular moment whenthe distance rfrom the searchlight to the automobile is3000 ft, the automobile has speed 80 ft/s and increasesits speed at a rate of 15 ft/s2 down the road.Find the required angular velocity and angularacceleration of the boat's searchlight to track the automobile at this instant.(Answers: 0.018 rad/s CCW, 0.004 rad/s2 CCW)

Explanation:

The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v  . We are required to calculate  angular velocity ω .

v = 80 ft /s

R = 3000 ft

ω = v / R

= 80  / 3000 = .027 rad / s

For angular acceleration the formula is

angular acceleration α = a / R

a is linear acceleration = 15 ft / s²

α = 15 / 3000 = .005 rad / s².

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