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Sladkaya [172]
3 years ago
11

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.

Physics
1 answer:
Valentin [98]3 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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At baseball practice, Mason and Alfredo both picked up the same bat and neither would let go until one of them had it for himsel
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Answer:

Option B. O because the net force was 5 N in Alfredo's direction

Explanation:

To know the the correct answer to the question given above, we shall determine the net force acting on the bat. This can be obtained as follow:

Force of pull by Mason (Fₘ) = 15 N

Force of pull by Alfredo (Fₐ) = 20 N

Net force (Fₙ) =?

Fₙ = 20 – 15

Fₙ = 5 N in Alfredo's direction

From the calculation made above, we can see that the net force is 5N in Alfredo's direction. This is the reason why Alfredo end up having the bat.

7 0
3 years ago
A student initially 10.0 m East of his school walks 17.5 m West. The magnitude of the student's displacement, relative to the sc
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Answer:

1. 7.5 m

2. towards west side

explanation:

I hope it will help you

3 0
2 years ago
A body is at aheight of 81m and is ascending upwards with a velocity of 12m/s .a body of 2 kg weight is dropped from it.if g=10m
butalik [34]
First the velocity drops to zero in 1.2 secs. In those seconds it went upwards for 7.2 m, then it went from 87.2 to 0m. x-x0=v0*t+1/2*g*t^2 ergo t=sqrt(2x/g) that is 4.1761 s. Finally the total time required is 5.3761 s
7 0
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An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remai
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<span>79.75m/s  .................................</span>
5 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
1 year ago
Read 2 more answers
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