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3 years ago

8

If you place a cup of 210 degree Fahrenheit coffee on a table in a room that is 68 degrees fareinheit, and ten minutes later, it

is 200 degrees fareinheit. How long will it be before the coffee is 180 degrees fareinheit? Use newtons law of cooling

1 answer:

Lana71 [14]3 years ago

8 0

Newton's Law of Cooling

Tf=Ts+(Ti-Ts)e^(-kt) where Tf is temp at time t, Ts is temp of surroundings, Ti is temp of object/fluid. So we need to find k first.

200=68+(210-68)e^(-10k)

132=142e^(-10k)

132/142=e^(-10k)

ln(132/142)=-10k

k=-ln(132/142)/10

k≈0.0073 so

T(t)=68+142e^(-0.0073t) so how long until it reaches 180°?

180=68+142e^(-0.0073t)

112=142e^(-0.0073t)

112/142=e^(-0.0073t)

ln(112/142)=-0.0073t

t= -ln(112/142)/(0.0073)

t≈32.51 minutes

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Fred begins walking toward John house at 3 mi/h. John leaves his house at the same time and walks toward Fred's house on the sam

Tcecarenko [31]

Let us assume as at when they meet, the time elapsed is x hours.

Fred 3 mi/h
John 2 mi/h

Distance = speed * time

Fred = 3x
John = 2x

Houses are 4miles apart.
So: 3x + 2x = 4 5x = 4

x = 4/5 hour = 0.8 hour or 0.8*60 = 48 minutes

Time elapsed is 0.8 hour or 48 minutes.

7 0

3 years ago

A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the po

astra-53 [7]

Answer:

a. See Attachment 1

b. $PT = 12.3\ m$

c. $HT = 31.1\ m$

d. $OH = 28.4\ m$

Step-by-step explanation:

Calculating PT

To calculate PT, we need to get distance OT and OP

Calculating OT;

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OT}{20}$

Multiply both sides by 20

$20 * tan50 = \frac{OT}{20} * 20$

$20 * tan50 = OT$

$20 * 1.1918 = OT$

$23.836 = OT$

$OT = 23.836$

Calculating OP;

We have to consider angle 30, distance OH and distance OP

The relationship between these parameters is;

$tan30 = \frac{OP}{20}$

Multiply both sides by 20

$20 * tan30 = \frac{OP}{20} * 20$

$20 * tan30 = OP$

$20 * 0.5774= OP$

$11.548 = OP$

$OP = 11.548$

$PT = OT - OP$

$PT = 23.836 - 11.548$

$PT = 12.288$

$PT = 12.3\ m$ (Approximated)

--------------------------------------------------------

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

$HT^2 = OT^2 + OH^2$

Substitute 20 for OH and $OT = 23.836$

$HT^2 = 20^2 + 23.836^2$

$HT^2 = 400 + 568.154896$

$HT^2 = 968.154896$

Take Square Root of both sides

$HT = \sqrt{968.154896}$

$HT = 31.1\ m$ <em>(Approximated)</em>

--------------------------------------------------------

Calculating the position of H

This is represented by OH

See Attachment 2

We have to consider angle 50, distance OH and distance OT

The relationship between these parameters is;

$tan50 = \frac{OH}{OT}$

Multiply both sides by OT

$OT * tan50 = \frac{OH}{OT} * OT$

$OT * tan50 = {OH$

$OT * 1.1918 = OH$

Substitute $OT = 23.836$

$23.836 * 1.1918 = OH$

$28.4= OH$

$OH = 28.4\ m$ <em> (Approximated)</em>

5 0

3 years ago

Please help me <br>Due in an hour

vladimir1956 [14]

131.88 or jus do radius times 2 so 21+ 21 = 42.

42 x pi(3.14) = 131.88

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3 years ago

If k is a constant, what is the value of k such that the polynomial k^2x^3 - 6kz+9 is

Tresset [83]

$x-1$ divides $p(x)=k^2x^3-6kx+9$ if the remainder is $p(1)=0$ . (This is the remainder theorem)

Then we must have

$p(1)=k^2-6k+9=(k-3)^2=0\implies\boxed{k=3}$

3 0

3 years ago

29+5x=4<br> 21+3x=51<br> 13+5x=33<br> 47-3x=32<br> 157+-11x=69<br> 1x+72=82

saul85 [17]

1. x=-5
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2 years ago

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