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-Dominant-
17 days ago
9

Which of the following equations correctly expresses the relation between vectors A⃗ , B⃗ , C⃗ , and D⃗ shown in the figure?

Physics
2 answers:
Sav [3.1K]17 days ago
7 0
C)  B ⃗ = A ⃗ + C ⃗  Pay attention to the direction of the arrows (vectors).
kicyunya [3.2K]17 days ago
5 0
The equation that accurately represents the relationship among vectors A⃗, B⃗, C⃗, and D⃗ depicted in the figure is: a. A⃗ + B⃗ − C⃗ − D⃗ = 0. Vectors are characterized by both a magnitude and direction, visually represented as directed line segments; a scalar, by contrast, is indicated by a single number. The question presents options including equations a. through d., with a. being correct.
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Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
kicyunya [3294]

Response:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Reasoning:

The laser power P = 1 mW = 1 x 10^-3 W

duration t = 250 ms = 250 x 10^-3 s

Taking a beam diameter of 2 mm = 2 x 10^-3 m

therefore

the beam's cross-sectional area A = \pi d^{2} /4 = (3.142 x (2*10^{-3} )^{2})/4

A = 3.142 x 10^-6 m^2

a) The intensity I = P/A

where P refers to the laser's power

and A represents the beam's cross-sectional area

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) The total energy delivered E =Pt

where P is the beam's power

and t is the exposure duration

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field can be computed as

E = \sqrt{2I/ce_{0} }

where I signifies the beam's intensity

and E is the electric field

c is the speed of light = 3 x 10^8 m/s

e_{0} = 8.85 x 10^9 m kg s^-2 A^-2

E = \sqrt{2*318.2/3*10^8*8.85*10^9} = 1.55 x 10^-8 v/m

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A firecracker breaks up into two pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/
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O momento total, p = 21.24 kg-m/s
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A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
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Response:

Q = 8,345 * v

Clarification:

We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:

Q = 8,345 * v

Where v signifies the number of gallons emptied from the tub.

Have a great day! Let me know if there's anything else I can assist with.

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The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
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J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
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