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zheka24 [161]
3 years ago
12

An 8.0 cm object is 40.0 cm from a concave mirror that has a focal length of 10.0 cm. Its image is 16.0 cm in front of the mirro

r.
Physics
1 answer:
svet-max [94.6K]3 years ago
6 0
 We can rearrange the mirror equation before plugging our values in. 
1/p = 1/f - 1/q. 
1/p = 1/10cm - 1/40cm
1/p = 4/40cm - 1/40cm = 3/40cm
40cm=3p  <-- cross multiplication
13.33cm = p

Now that we have the value of p, we can plug it into the magnification equation.

M=-16/13.33=1.2
1.2=h'/8cm
9.6=h'

So the height of the image produced by the mirror is 9.6cm.
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2) The speed with which the gazelle needs to run, to beat the cheater by 2 seconds is approximately 29.79 m/s

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1) The distance of the gazelle, from the gazelle safe zone = 275 m

The distance of the cheetah from the gazelle = 455 m

The speed of the gazelle = 25 m/s

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Therefore, we have;

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t = 275 m/(25 m/s) = 11

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Let the time the cheetah reaches the gazelle = t₁, we have;

455 + 25 × t₁ = 65 × t₁

t₁ = 455/40 = 11.375

t₁ = 11.375 seconds

The gazelle reaches the gazelle safe zone before the cheetah reaches the gazelle

Therefore, the gazelle gets to safety

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The time for the cheetah to reach the safe zone = (275 + 455)/65 = 11.23 seconds

Therefore, we have;

In order for the gazelle to beat the gazelle by 2 seconds the time the cheetah reaches the safe zone = 11.23 - 2 = 9.23 s

The speed of the gazelle is then 275/9.23 ≈ 29.79 m/s

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