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Ket [755]
3 years ago
12

In the sum →A+→B=→C, vector →A has a magnitude of 12.0 m and is angled 40.0° counterclockwise from the +x direction, and vector

→C has a magnitude of 15.0 m and is angled 20.0° counterclockwise from the −x direction. What are (a) the magnitude and (b) the angle (relative to +x) of →B?
Physics
1 answer:
icang [17]3 years ago
7 0

Answer:

Explanation: Ok, first caracterize the two vectors that we know.

A = ax + ay = (12*cos(40°)*i + 12*sin(40°)*j) m

now, see that C is angled 20° from -x, -x is angled 180° counterclockwise from +x, so C is angled 200° counterclockwise from +x

C = cx + cy = (15*cos(200°)*i + 15*sin(200°)*j) m

where i and j refers to the versors associated to te x axis and the y axis respectively.

in a sum of vectors, we must decompose in components, so: ax + bx  = cx and ay + by = cy. From this two equations we can obtain B.

bx= (15*cos(200°) - 12*cos(40°)) m = -23.288 m

by = (15*sin(200°) - 12*sin(40°)) m = -12.843 m

Now with te value of both components of B, we proceed to see his magnitude an angle relative to +x.

Lets call a to the angle between -x and B, from trigonometry we know that tg(a) = by/bx, that means a = arctg(12.843/23.288) = 28.8°

So the total angle will be 180° + 28.8° = 208.8°.

For the magnitude of B, lets call it B', we can use the angle that we just obtained.

bx = B'*cos(208.8°) so B' = (-23.288 m)/cos(208.8°) =  26.58 m.

So the magnitude of B is 26.58 m.

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Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

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A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

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now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

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(c). surface area A of the sun on mars is = 4πR²

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A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

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The statement is true: velocity and acceleration have opposite directions in the interval of braking.

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a=\frac{dv}{dt}

v decreases ⇒\frac{dv}{dt}

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