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2 years ago

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Trace the path of a ray emitted from the tip of the object through the focal point of the mirror and then the reflected ray that

results. Start by extending the existing ray emitted from the tip of the object. Then create the reflected ray. Draw the vector for the reflected ray starting from the point where the incident focal ray hits the mirror. The location and orientation of the vector will be graded. The length of the vector will not be graded.

1 answer:

Valentin [98]2 years ago

5 0

Answer:

1 / f = 1 / p + 1 / q

Explanation:

Mirror problems can be solved analytically with the constructor equation

1 / f = 1 / p + 1 / q

An approximate solution can also be given using graphical methods, where two of three rays must be drawn,

1) A ray that passes through the focal length and bounces horizontally

2) A ray that passes through the center of curvature (c = 2f) of the mirror and does not deviate

3) A beam that comes horizontal and bounces through the focal

The attached shows the ray that passes through the focal length and bounces horizontally.

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A circular loop ( radius = 0.5 m) carries a current of 3.0 A and has unit normal vector of (2i - j +2k)/3 what is x component of

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Answer:

$T=9.42Nm$

Explanation:

From the question we are told that:

Radius $r= 0.5 m$

Current $I= 3.0 A$

Normal vector $n=\frac{(2i - j +2k)}{3}$

Magnetic field $B= (2i-6j) T$

Generally the equation for Area is mathematically given by

$A=\pi r^2$

$A=3.1415 *0.5^2$

$A=0.7853 m^2$

Generally the equation for Torque is mathematically given by

$T=A(i'*B)$

Where

$i'*B= \begin{bmatrix}2&-1&2\\2&-6&0\end{bmatrix}$

$X\ component\ of\ i'*B= [(-1 * 0)-(2*-6)]$

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Therefore

$T=0.7853*12$

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2 years ago

Summarize the weather conditions related to a warm front. Remember to include all data collected on warm fronts in this activity

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Answer:

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Explanation:

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3 years ago

A student standing on a cliff that is a vertical height d = 8.0 m above the level ground throws a stone with velocity v0 = 24 m/

PilotLPTM [1.2K]

Answer:

a) Vf = 27.13 m/s

b) It would have been the same

Explanation:

On the y-axis:

$Y=-Vo*sin\theta*t-1/2*g*t^2$

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Solving for t:

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Discarding the negative value and using the positive one to calculate the velocity:

$Vf_y = -Vo*sin\theta-g*t$

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So, the module of the velocity will be:

$Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}$

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If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.

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3 years ago

008 (part 1 of 3) 10.0 points A 3.2 kg object is subjected to two forces, F~ 1 = (1.9 N) ˆı + (−1.9 N) ˆ and F~ 2 = (3.8 N) ˆı

musickatia [10]

Answer:

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