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solmaris [256]
2 years ago
15

HELP ASAP!!!

Physics
1 answer:
miss Akunina [59]2 years ago
4 0

Write each force in component form:

<em>v </em>₁ : 50 N due east   →   (50 N) <em>i</em>

<em>v</em> ₂ : 80 N at N 45° E   →   (80 N) (cos(45°) <em>i</em> + sin(45°) <em>j</em> ) ≈ (56.5 N) (<em>i</em> + <em>j</em> )

The resultant force is the sum of these two vectors:

<em>r</em> = <em>v </em>₁ + <em>v</em> ₂ ≈ (106.5 N) <em>i</em> + (56.5 N) <em>j</em>

Its magnitude is

|| <em>r</em> || = √[(106.5 N)² + (56.5 N)²] ≈ 121 N

and has direction <em>θ</em> such that

tan(<em>θ</em>) = (56.5 N) / (106.5 N)   →   <em>θ</em> ≈ 28.0°

i.e. a direction of about E 28.0° N. (Just to clear up any confusion, I mean 28.0° north of east, or 28.0° relative to the positive <em>x</em>-axis.)

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(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

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P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

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