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Alekssandra [29.7K]
3 years ago
9

A packet weighs 40kg in air but when it is totally submerged into a 1mx1m square tank the weight of the packet is only 18kg. How

much does the water rise in the tank?
Engineering
1 answer:
Irina18 [472]3 years ago
5 0

Answer:

water  rise = 22 mm

Explanation:

weight of packet IN AIR = 40 *9.81 =392.4 N

weight of packet  IN WATER= 18 *9.81 =176.58 N

by Archimedi's principle

difference in weight = weight of displaced water

w_a - w_w = \rho_w v_d g

392.4 - 176.58 = 1000* v_d* 9.81

v_d = 0.022 m^3

v_d = A*H_rise

0.022 =1*H_rise

H_rise = 0.022 m = 22 mm

water  rise = 22 mm

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The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet
Crank

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, I_s = 2 A

power supplied, P_s = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A

The number of turns on the input side is calculated as;

\frac{N_p}{N_s} = \frac{I_s}{I_p}  \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns

4 0
2 years ago
Initialize the tuple team_names with the strings 'Rockets', 'Raptors', 'Warriors', and 'Celtics' (The top-4 2018 NBA teams at th
Drupady [299]

Answer:

#Initialise a tuple

team_names = ('Rockets','Raptors','Warriors','Celtics')

print(team_names[0])

print(team_names[1])

print(team_names[2])

print(team_names[3])

Explanation:

The Python code illustrates or printed out the tuple team names at the end of a season.

The code displayed is a function that will display these teams as an output from the program.

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3 years ago
The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find
Nataly_w [17]

The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25

<h3>How to find the Instantaneous rate of change?</h3>

The annual inventory cost C for a manufacturer is given as;

C = (1012000/Q) + 7.5Q

where Q is the order size when the inventory is replenished.

Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341

Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23

Instantaneous rate of change in C is first order derivative C':

C'(Q) = -1,012,000/(Q²) + 7.5

C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25

Read more about Instantaneous rate of change at; brainly.com/question/14666106

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8 0
1 year ago
Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following
torisob [31]

Given:

Assuming the transition to turbulence for flow over a flat plate happens at a Reynolds number of 5x105, determine the following for air at 300 K and engine oil at 380 K. Assume the free stream velocity is 3 m/s.

To Find:

a. The distance from the leading edge at which the transition will occur.

b. Expressions for the momentum and thermal boundary layer thicknesses as a function of x for a laminar boundary layer

c. Which fluid has a higher heat transfer

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The transition from the lamina to turbulent begins when the critical Reynolds

number reaches 5\times 10^5

(a).  \;\text{Rex}_{cr}=5 \times 10^5\\\\\frac{\rho\;vx}{\mu}=5 \times 10^5\\\text{density of of air at}\;300K=1.16  \frac{kg}{m\cdot s}\\\text{viscosity of of air at}\;300K=1.846 \times 10^{-5} \frac{kg}{m\cdot s} \\v=3m/s\\\Rightarrow x=\frac{5\times 10^5 \times 1.846 \times 10^{-5} }{1.16 \times 3} =2.652 \;m \;\text{for air}\\(\text{similarly for engine oil at 380 K for given}\; \rho \;\text{and} \;\mu)\\

(b).\; \text{For the lamina boundary layer momentum boundary layer thickness is given by}:\\\frac{\delta}{x} =\frac{5}{\sqrt{R_e}}\;\;\;\;\quad\text{for}\; R_e(c). \frac{\delta}{\delta_t}={P_r}^{\frac{r}{3}}\\\text{For air} \;P_r \;\text{equivalent 1 hence both momentum and heat dissipate with the same rate for oil}\; \\P_r >>1 \text{heat diffuse very slowly}\\\text{So heat transfer rate will be high for air.}\\\text{Convective heat transfer coefficient will be high for engine oil.}

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Answer:

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Explanation:

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7 0
3 years ago
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