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2 years ago

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Q2 Solving MDPs 6 Points Consider the gridworld MDP for which \text{Left}Left and \text{Right}Right actions are 100% successful.

Specifically, the available actions in each state are to move to the neighboring grid squares. From state aa, there is also an exit action available, which results in going to the terminal state and collecting a reward of 10. Similarly, in state ee, the reward for the exit action is 1. Exit actions are successful 100% of the time.

1 answer:

trasher [3.6K]2 years ago

3 0

Answer:

pata nahi

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Effects of biological hazards are widespread. Select the answer options which describe potential effects of coming into contact

s2008m [1.1K]

Answer:

- Allergic Reactions

- Life Threatinging Events

Explanation:

Sources of biological hazards may include bacteria, viruses, insects, plants, birds, animals, and humans. These sources can cause a variety of health effects ranging from skin irritation and allergies to infections (e.g., tuberculosis, AIDS), cancer and so on.

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2 years ago

Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic j

mezya [45]

Answer:

A) flow depth after jump = 2.46 m, Froude number after jump = 0.464

B) head loss = 0.572 m

C) dissipation ratio = 0.173

Explanation:

Given data :

width of channel = 10-m

velocity of before jump (V1) = 7 m/ s

flow depth before jump (y1) = 0.8 m

A) determining the flow depth and the Froude's number after the jump

attached below is the solution

B) head loss

HL = Y1 -Y2 + $\frac{V_{1} ^2 - V_{2} ^2}{2g}$ = 0.8 - 2.46 + $\frac{49 - 5.1984}{19.62}$ = 0.572

C) dissipation ratio

HL / Es1 = 0.572 / 3.3 = 0.173

Es1 = 0.8 + $\frac{7^2}{2*9.81}$ = 0.173

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2 years ago

Kris and James are working at a construction site that has a significant amount of stagnant water. Which type of hazard are they

valkas [14]

Answer:

Biological

Explanation:

A biological hazard refers to a biological substance such as viruses, insects, fungus, etc. that can affect humans or other animals' health negatively. In the case of Kris and James, they are exposed to a biological hazard because stagnant water is commonly used by mosquitos to place eggs, this leads to a lot of mosquitos around stagnant waters and therefore a higher risk of mosquito-transmitted diseases such as malaria. Besides this, stagnant water is highly polluted and includes bacteria and parasites that are harmful.

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2 years ago

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A 30-ft tall drop tower is being used to study the shape of water droplets as they fall through air. A camera is to be carried b

sveta [45]

Answer:

The camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.

Explanation:

The time it takes for the drop to reach 7-ft and 9-ft at the acceleration of 32.2ft/s2:

$s = gt^2/2$

$t^2 = 2s/g$

$t = \sqrt{2s/g}$

When s = 7ft, $t_7 = \sqrt{2*7/32.2} = 0.66 s$

When s = 9ft, $t_9 = \sqrt{2*9/32.2} = 0.75 s$

So it would take Δt = 0.75 - 0.66 = 0.09 s for each drop to travel from 7ft to 9ft. As the drops are released every 0.5s, this means that after our cam follow the first drop to the end of the 7-9ft window, it has 0.5 - 0.09 = 0.41 s or less to reverse its acceleration and go back to the original spot to take care of the next drop.

So at 9ft, the drop and camera velocity would be:

$v_9 =gt_9 = 32.2*0.75 = 24.1 ft/s$

So if the camera reverses its direction to go back to original spot at the rate of a (ft/s2) and initial speed of 24.1 ft/s. Within the time span of 0.41s to catch the next drop at 7ft

$7 = 9 + 24.1*(0.41) - a0.41^2/2$

$0 = 2 + 9.881 - 0.085a$

$a = 11.881 / 0.085 = 140.54 m/s^2$

So the camera would flow the same velocity and acceleration of the droplet until 9ft. After that it would switch gear and accelerate back to its original spot at the rate of 140.54 m/s2 in order to catch the next droplet.

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2 years ago

Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe

Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

$\frac{p1}{p*}$ = 0.3636

$\frac{T1}{T*}$ = 0.5289

$\frac{T01}{T0*}$ = 0.7934

Isentropic Flow Chart: M1 = 2.0 , $\frac{T01}{T1}$ = 1.8

T1 = $\frac{1}{0.7934}$ (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= $\frac{T02}{T1}T1$ = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = $\frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K$ = 135.7* $10^{3}$ J/Kg.

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2 years ago

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