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insens350 [35]
2 years ago
9

In general it is best to conceptualize vectors as arrows in space, and then to make calculations with them using their component

s. (You must first specify a coordinate system in order to find the components of each arrow.) This problem gives you some practice with the components. Let vectors A⃗ =(1,0,−3), B⃗ =(−2,5,1), and C⃗ =(3,1,1). Calculate and express your answers as ordered triplets of values separated by commas

Physics
1 answer:
Dimas [21]2 years ago
3 0

(1) A - B

(2) B - C

(3) - A + B - C

(4) 3A - 2C

(5) - 2A + 3B - C

(6) 2A - 3 (B - C)

Answer:

(1)  (3,-5,-4)

(2) (-5, 4, 0)

(3) (-6, 4, 3)

(4) (-3, -2, -11)

(5) (-11, 14, 8)

(6) (17, -12, -6)

Explanation:

A⃗ =(1,0,−3)

B⃗ =(−2,5,1)

C⃗ =(3,1,1)

Vector additions and subtraction are done on a component by component basis, that is, only data from component î can be added to or subtracted from another Vector's component î. And so on for components j and k.

1) (A - B) = (1,0,−3) - (−2,5,1) = (1-(-2), 0-5, -3-1) = (3,-5,-4)

2)  (B - C) = (−2,5,1) - (3,1,1) = (-2-3, 5-1, 1-1) = (-5, 4, 0)

3) -A + B - C = -(1,0,−3) + (−2,5,1) - (3,1,1) = (-1-2-3, 0+5-1, 3+1-1) = (-6, 4, 3)

4) 3A - 2C = 3(1,0,−3) - 2(3,1,1) = (3,0,-9) - (6,2,2) = (3-6, 0-2, -9-2) = (-3, -2, -11)

5) -2A + 3B - C = -2(1,0,−3) + 3(−2,5,1) - (3,1,1) = (-2,0,6) + (-6,15,3) - (3,1,1) = (-2-6-3, 0+15-1, 6+3-1) = (-11, 14, 8)

6) 2A - 3 (B - C) = 2(1,0,−3) - 3[(−2,5,1) - (3,1,1)] = (2,0,-6) - 3(-5,4,0) = (2+15, 0-12, -6-0) = (17, -12, -6)

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We know that

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\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g

\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt

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2 years ago
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Writing this expression for both discs

          I₁ A₁ = I₂ A₂

          I₂ = I₁ A₁ / A₂

The area of ​​a sphere is

          A = 4π r²

           I₂ = I₁ (r₁ / r₂)²

          r₂ = r₁ ± 5

          I₁ = I₂ ( (r₁ ± 5)/r₁)²

.

        Let's write the Stefan equation

         P / A = σ e T⁴

          I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

         σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values ​​of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

        e₁ T₁⁴ = T₂⁴   (r1 + 5)²/r₁²

       T₁ = T₂  {(e₂/e₁)}^{1/4}  √(1 ± 1/ r₁)  

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

           (1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

          T₁ = T₂ {(e₂/e₁)}^{1/4}  (1 ± ½   5/r1)

           T₁ = T₂ {(e₂)}^{1/4}   (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

              T₁ = T₂  {(e₂)}^{1/4} ¼

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2 years ago
A field researcher uses the slow-motion feature on her phones camera to shoot a video of an eel spinning at its maximum rate. Th
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Answer:

\alpha=(1/120)*14*360=42degrees

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Eels have been recorded to spin at up to 14 revolutions per second when feeding in this way

So the angle the eel rotate is going to be

Camera record 120 frames per second

time taken for 1 frame record

t_t= 1/120 seconds

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Xander's friend Corey has a skateboard that he rides at Newton's Skate Park.
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Data obtained from the question.

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Net force = 30N

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Step 2

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30 = 60 x Acceleration

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2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

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