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WITCHER [35]
2 years ago
11

Two diverging light rays, originating from the same point, have an angle of 5° between them. After the rays reflect from a plane

mirror, what is the angle between them? Construct one possible ray diagram that supports your answer.

Physics
2 answers:
Alexeev081 [22]2 years ago
4 0

Answer:

5°  

Explanation:

Two diverging light rays, originating from the same point, have an angle of 5° between them.

According to laws of reflection the angle of incidence is equal to the angle of reflection.

From the attached figure, it is clear that the after reflection from a plane mirror the angle between two rays will be 5°.

O be the originating point. m and n two diverging light rays. When they strikes at the plane mirror, the laws of reflection still fulfilled.

LuckyWell [14K]2 years ago
3 0
As a reference, consider the line from the point perpendicular to the mirror.
That direction is called 'normal' to the mirror.

The ray on the right leaves the point traveling 5° to the right of the normal,
and leaves the mirror on a path that's 10° to the right of the normal.

The ray on the left leaves the point traveling 5° to the left of the normal,
and leaves the mirror on a path that's 10° to the left of the normal.

The angle between the two rays after they leave the mirror is 20° .

Frankly, Charlotte, if there were more than 5 points available for this answer,
I'd seriously consider giving you a drawing too.
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3 years ago
The half-life of caffeine is 5 hours. If you ingested a 30 oz Big Gulp, how many oz of caffeine is left after one half life? * Y
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The amount of caffeine left after one half life of 5 hours is 15 oz.

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So that:

After one half life of 5 hours, the value of caffeine that would be left is;

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8 0
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A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arr
Paha777 [63]

Answer:

The question is incomplete, below is the complete question "A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.00 N)k and the corresponding torque about the origin is vector tau = (4 N · m)i hat + (10 N · m)j + (11N · m)k.

Determine Fx."

F_{x}=-1N.m

Explanation:

We asked to determine the "x" component of the applied force. To do this, we need to write out the expression for the torque in the in vector representation.

torque=cross product of force and position . mathematically this can be express as

T=r*F

Where

F=F_{x}i+(7N)j-(5N)k  and the position vector

r=(2m)i-(3m)j+(2m)k

using the determinant method to expand the cross product in order to determine the torque we have

\left[\begin{array}{ccc}i&j&k\\2&-3&2\\ F_{x} &7&-5\end{array}\right]\\\\

by expanding we arrive at

T=(18-14)i-(-12-2F_{x})j+(12+3F_{x})k\\T=4i-(-12-2F_{x})j+(12+3F_{x})k\\\\

since we have determine the vector value of the toque, we now compare with the torque value given in the question

(4Nm)i+(10Nm)j+(11Nm)k=4i-(-12-2F_{x})j+(12+3F_{x})k\\

if we directly compare the j coordinate we have

10=-(-12-2F_{x})\\10=12+2F_{x}\\ 10-12=2F_{x}\\ F_{x}=-1N.m

8 0
3 years ago
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