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koban [17]
3 years ago
8

A thermometer is taken from a room where the temperature is 24oc to the outdoors, where the temperature is −15oc. After one minu

te the thermometer reads 14oc. (a) what will the reading on the thermometer be after 3 more minutes?
Mathematics
1 answer:
kap26 [50]3 years ago
7 0

Solution:

Use Newton's Law of Cooling.  


T = T_s + (T_0 - T_s)*e^(-kt)  


where  

T = temperature at any instant  

T_s = temperature of surroundings  

T_0 = original temperature  

t = elapsed time  

k = constant  


Now, we need to find this constant. We are given that after one hour, the temperature drops to 13° C in a 7°C Environment.  

T = 14, T_0 = 24, T_s = -15, t = 1, k = ?  

T = T_s + (T_0 - T_s)*e^(-kt)  

==> 14 = -15 + (24 - 7)*e^(-k)  

==> 14 = 7 + 17*e^(-k)  

==> 7 = 17*e^(-k)  

==> 7/13 = e^(-k)  

==> -k = ln(7/17)  

==> k = -ln(7/17) ≈ 0.774  

Now,


Let's calculate temperatures!  

T = ?, T_0 = 24, T_s = -15, k = 0.773, t = 3  

T = T_s + (T_0 - T_s)*e^(-kt)  

==> T = -15 + (24 –(-15))*e^[ -(0.774)(2) ]  

==> T = -15 + 39*e^(-1.548)  

==> T ≈ 15.72° C  

This the required answer.


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Answer:

x = 4\sqrt{5}

Step-by-step explanation:

Since the triangle is right, solve for x using Pythagoras' identity, that is

x² = 8² + 4² = 64 + 16 = 80

Take the square root of both sides

x = \sqrt{80} = \sqrt{16(5)} = \sqrt{16} × \sqrt{5} = 4\sqrt{5} ← exact value



3 0
3 years ago
Amy is pulling a wagon with a force of 30 pounds up a hill at an angle of 25°. Give the force exerted on the wagon as a vector a
Fofino [41]

Answer:

Vector (ordered pair - rectangular form)

\vec F = (27.189,12.679)\,[lbf]

Vector (ordered pair - polar form)

\vec F = (30\,lbf, 25^{\circ})

Sum of vectorial components (linear combination)

\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]

Step-by-step explanation:

From statement we know that force exerted on the wagon has a magnitude of 30 pounds-force and an angle of 25° above the horizontal, which corresponds to the +x semiaxis, whereas the vertical is represented by the +y semiaxis.

The force (\vec F), in pounds-force, can be modelled in two forms:

Vector (ordered pair - rectangular form)

\vec F =  \left(\|\vec F\|\cdot \cos \theta, \|\vec F\|\cdot \sin \theta\right) (1)

Vector (ordered pair - polar form)

\vec F = \left(\|\vec F\|, \theta\right)

Sum of vectorial components (linear combination)

\vec {F} = \left(\|\vec F\|\cdot \cos \theta\right)\cdot \hat{i} + \left(\|\vec F\|\cdot \sin \theta \right)\cdot \hat{j} (2)

Where:

\|\vec F\| - Norm of the vector force, in newtons.

\theta - Direction of the vector force with regard to the horizontal, in sexagesimal degrees.

\hat{i}, \hat{j} - Orthogonal axes, no unit.

If we know that \|\vec F\| = 30\,lbf and \theta = 25^{\circ}, then the force exerted on the wagon is:

Vector (ordered pair - rectangular form)

\vec F= \left(30\cdot \cos 25^{\circ}, 30\cdot \sin 25^{\circ}\right)\,[lbf]

\vec F = (27.189,12.679)\,[lbf]

Vector (ordered pair - polar form)

\vec F = (30\,lbf, 25^{\circ})

Sum of vectorial components (linear combination)

\vec F = (30\cdot \cos 25^{\circ})\cdot \hat{i} + (30\cdot \sin 25^{\circ})\cdot \hat{j}\,[N]

\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]

8 0
2 years ago
Select the expression that makes the equation true.
svetoff [14.1K]

Answer:

a.) -18 x 10^6

Step-by-step explanation:

multiply 9 x-2

add exponents 9 and -3

4 0
3 years ago
Solve for a. 1/5(25−5a)=4−a
aalyn [17]
1/5(25-5a)=4-a
⇔5-a=4-a
<span>NO SOLUTION</span>
3 0
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7) A photo booth charges a $500 fee for two hours at a party, plus $50 per additional hour. Cindy doesn't want to spend more tha
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Since the photo booth charges a $500 fee for two hours at the party and an additional 50 dollars per hour, we have:

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50x+500=700 \implies\\ 50x=200\implies\\ x=4

So she spent an additional 4 hours on the rental.

3 0
3 years ago
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