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Alex_Xolod [135]
3 years ago
11

An object is 39 cm away from a concave mirrors surface along the principles axis. If the mirrors focal length is 9.50 cm, how fa

r away is the corresponding image?
Physics
1 answer:
Tatiana [17]3 years ago
5 0

Answer:

12.6 cm

Explanation:

We can use the mirror equation to find the distance of the image from the mirror:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where here we have

f = 9.50 cm is the focal length

p = 39 cm is the distance of the object from the mirror

Solving the equation for q, we find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{9.50 cm}-\frac{1}{39 cm}=0.080 cm^{-1}\\q = \frac{1}{0.080 cm}=12.6 cm

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Answer:

 T = 5.36 s

Explanation:

given,

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t = \sqrt{\dfrac{2\times 122.5}{9.8}}

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A thermometer initially reading 212F is placed in a room where the temperature is 70F. After 2 minutes the thermometer reads 125
frez [133]

Answer:

91.3°F

Explanation:

Let T be the temperature of the thermometer at any time

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And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

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- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 70) = (212 - 70)e⁻ᵏᵗ

(T - 70) = 142 e⁻ᵏᵗ

At t = 2 minute, T = 125°F

125 - 70 = 142 e⁻ᵏᵗ

55/142 = e⁻ᵏᵗ

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kt = 0.4742 × 4 = 1.897

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T - 70 = 142 × 0.15 = 21.3

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