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stiks02 [169]
2 years ago
6

3. Two fans blow at 5 ms^-1 in a easterly direction and 8ms^-1 in a Northerly

Physics
1 answer:
Anvisha [2.4K]2 years ago
3 0

Addition of vectors is done by adding the components of the vectors

The total speed of the fan  is approximately <u>9.43 m/s</u>,

Direction of total wind 58° counterclockwise from the positive x-axis

<u />

The reason the value is correct is as follows:

The given parameters are:

The direction at which the fan blowing at 5 m/s is blowing = Easterly

The direction in which the fan blowing 8 m/s is blowing = Northerly

The total speed of the van, |v| ≈ <u>9.43 m/s</u>

<u />

Required:

To find the total wind speed of the fans

Solution:

Taken the easterly direction as the ith component, and the northerly direction as the jth component,  the vector representing the speed of the fan is presented as follows

v = 5·i + 8·j

The magnitude of the total wind speed of the fan, |v| = √(5² + 8²) = √(89) ≈ 9.43

The total wind speed of the fan , |v| ≈ <u>9.43 m/s</u>

The direction of the total speed with respect to the x-axis, θ, is given as follows;

θ = arctan(8/5) ≈ <u>58°</u>

<u />

Learn more about addition of vectors here:

brainly.com/question/17691926

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The force exerted on the board by the karate master given the data is -4500 N

<h3>Data obtained from the question </h3>
  • Initial velocity (u) = 10 m/s
  • Final velocity (v) = 1 m/s
  • Time (t) = 0.002 s
  • Mass (m) = 1 Kg
  • Force (F) = ?
<h3>How to determine the force</h3>

The force exerted can be obtained as illustrated below:

F = m(v - u) / t

F = 1 (1 - 10) / 0.002

F = (1 × -9) / 0.002

F = -4500 N

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What is one basic need that both monkey and a tree have in common​
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both a monkey and a tree need water to survive.

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Sugar molecules can be oxidized in the cell of animals . This oxidation releases energy for the animal to use. From what soured
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Answer: through digested food

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What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
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Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

7 0
3 years ago
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