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Helga [31]
1 year ago
14

A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops

directly downward 1.4m to hover in front of the flower. What is the hummingbirds total displacement

Physics
2 answers:
polet [3.4K]1 year ago
7 0

Answer:

1.84m, 49.4⁰ South west

Explanation:

<u>Step 1:</u> construct a right angled triangle from the birds displacement

⇒From the diagram uploaded; we calculate the resultant/total displacement (R) of the bird using Pythagoras theorem.

R² = 1.2² + 1.4²

R² = 3.4

R = √3.4

R = 1.84 m

<u>Step 2:</u> calculate the direction of the bird's displacement

From vector diagram head-to-tail rule, we determine the cardinal direction;

From the diagram; tan(\alpha) = \frac{1.4}{1.2}

tan(\alpha) = 1.1667

\alpha =tan^{-1} 1.1667

\alpha = 49.4^o

Therefore the hummingbirds total displacement is 1.84m, 49.4⁰ South west

Anit [1.1K]1 year ago
4 0

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

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Answer:

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part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
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<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

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