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vazorg [7]
2 years ago
6

700 miles away, to try out his luck. The ride there with the wind takes four hours. A few months later almost broke he heads bac

k, on the same plane and the same wind. The flight then, against the wind, takes five hours. What is the speed of the plane and what is the speed of the wind?
Physics
1 answer:
Black_prince [1.1K]2 years ago
8 0

Answer:

speed of plane = 157.5 mph

speed o wind = 17.5 mph

Explanation:

Distance traveled to move from one position to other is given as

d = 700 miles

time taken by it

t = 4 hr

so here the speed is given as

v_p + v_w = \frac{700}{4}

v_p + v_w = 175 mph

during his return journey we have

v_p - v_w = \frac{700}{5}

v_p - v_w = 140 mph

now from above two equations we have

v_p = 157.5 mph

v_w = 17.5 mph

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A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be​
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The question is incomplete, the complete question is;

A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?

Answer:

See explanation

Explanation:

From Newton's law of cooling;

θ1 - θ2/t = K(θ1 + θ2/2 - θo]

Where;

θ1 and θ2 are initial and final temperatures

θo is the temperature of the surroundings

K is the constant

t is the time taken

Hence;

100 - 60/5 = K(100 + 60/2 - θo)

100 - 40/10 = K(100 + 40/2 - θo)

8= (80 - θo)K -----(1)

6= (70 - θo)K -----(2)

Diving (1) by (2)

8/6 = (80 - θo)/(70 - θo)

8(70 - θo) = 6(80 - θo)

560 - 8θo = 480 - θo

560 - 480 = -θo + 8θo

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Again from Newton's law of cooling;

θ = θo + Ce^-kt

Where;

t= 0, θ = 60° and θo = 11.4°

60 = 11.4 + C e^-K(0)

60 - 11.4 = C

C= 48.6°

To obtain K

40 = 11.4 + 48.6e^-10k

40 -11.4 = 48.6e^-10k

28.6/48.6 = e^-10k

0.5585 = e^-10k

-10k = ln0.5585

k= ln0.5585/-10

K= 0.0583

Hence, the temperature in 15 minutes;

θ= 11.4 + 48.6e^(-0.0583 × 15)

θ= 31.7°

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