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Free_Kalibri [48]
3 years ago
6

A faulty thermometer reads 2°C when dipped in ice at 0°C and 95°C when dipped in steam at 100°C. What would this thermometer rea

d if placed in water at room temperature at 18°C?​
Physics
1 answer:
kvasek [131]3 years ago
4 0

Answer:

The read will be 20.9[C]x=\frac{(y-y_{1} )}{(y_{2} -y_{1} )}*(x_{2}-x_{1})+y_{1}\\  x=\frac{(18-0 )}{(95 -0 )}*(100-2)+2\\\\x= 20.9[C]

Explanation:

This is a problem related to linear interpolation, Linear interpolation consists of tracing a line through two known points y = r (x) and calculating the intermediate values according to this line.

The equation of a known line two points (x1, y1)and (x2, y2) = (2,0) (100,95) is:

\frac{(y-y_{1} )}{(y_{2} -y_{1} )}=\frac{(x-y_{1} )}{(x_{2}-x_{1}  )}

If we clear y from the equation we have:

[tex]y=\frac{(x-y_{1})}{(x_{2}-x_{1} )}*(y_{2} -y_{1} )+y_{1}  \\replacing

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A student is running to catch the campus shuttle bus, which is stopped at the bus stop. The student is runnign at a constant spe
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Answer:

Part a)

t = 16.8 s

d = 100.8 m

Part b)

v_f = 2.86 m/s

Explanation:

Part a)

Constant speed by which the student will run is given as

v = 5 m/s

now after some time if student is going to overtake the position of bus

so here the final positions will be same

so we have

x_{bus} = x_{student}

0 + \frac{1}{2}at^2 + d = v_{student} t

\frac{1}{2}(0.170)t^2 + 60 = 5 t

0.085 t^2 - 5t + 60 = 0

so it is

t = 16.8 s

So student will run the total distance

d = vt

d = (6)(16.8)

d = 100.8 m

Part b)

Speed of bus when student reach the bus is given as

v_f = v_i + at

v_f = 0 + (0.170)(16.8)

v_f = 2.86 m/s

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