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Tasya [4]
2 years ago
12

A displacement vector points in a direction of θ = 23° left of the positive y-axis. The magnitude of this vector is D = 155 m. R

efer to the figure. Enter an expression for the x-component vector, Dx, in terms of D, θ, and the unit vectors i and j.

Physics
1 answer:
Lady bird [3.3K]2 years ago
3 0

Answer:

Dₓ = -155 sin 23° i + 0 j

Explanation:

The diagram showing the vector has been attached to this response.

As shown in the diagram,

The vector D has an x-component (also called horizontal component) of -D sinθ i. i.e

Dₓ = -D sin θ i   [The negative sign shows that D lies in the negative x direction]

Where;

D = magnitude of D = 155m

θ = direction of D = 23°

Therefore;

Dₓ = -155 sin 23° i

Since Dₓ represents the x component, its unit vector, j component has a value of 0.

Therefore, Dₓ can be written in terms of D, θ and the unit vectors i and j as follows;

Dₓ = -155 sin 23° i + 0 j

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Multiply the number 4.48E-8 by 5.2E-4 using Google. What is the correct answer in scientific notation?
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sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
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2 years ago
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