The temperature of a cup of coffee varies according, №18010330, 19.02.2021 13:33

Mathematics

The temperature of a cup of coffee varies according to Newton s Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. 74 67 60 42

Step-by-step answer

P Answered by Master

(b) 67

Step-by-step explanation:

A solution to the differential equation describing the temperature according to Newton's Law of Cooling could be written as ...

T = (final temp) + (initial difference)×(decay factor)^t

where the decay factor is the fraction of change during 1 unit of time period t.

__

Here, the initial difference of 100-25 = 75 degrees decays to 90-25 = 65 degrees in 1 minute. So, the units of t are minutes, the decay factor is 65/75, the initial difference is 75 degrees, and the final temperature is 25 degrees. That lets us write the equation as ...

T = 25 +75(65/75)^t

Then for t=4, the temperature is ...

T = 25 +75(13/15)^4 ≈ 67.3 . . . . degrees

After 4 minutes the temperature of the coffee is about 67 degrees.

Mathematics

The temperature of a cup of coffee varies according to Newton s Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. 74 67 60 42

Step-by-step answer

P Answered by Specialist

(b) 67

Step-by-step explanation:

A solution to the differential equation describing the temperature according to Newton's Law of Cooling could be written as ...

T = (final temp) + (initial difference)×(decay factor)^t

where the decay factor is the fraction of change during 1 unit of time period t.

__

Here, the initial difference of 100-25 = 75 degrees decays to 90-25 = 65 degrees in 1 minute. So, the units of t are minutes, the decay factor is 65/75, the initial difference is 75 degrees, and the final temperature is 25 degrees. That lets us write the equation as ...

T = 25 +75(65/75)^t

Then for t=4, the temperature is ...

T = 25 +75(13/15)^4 ≈ 67.3 . . . . degrees

After 4 minutes the temperature of the coffee is about 67 degrees.

Physics

The temperature of a cup of coffee varies according to Newton s Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. 74 67 60 42

Step-by-step answer

P Answered by Specialist

We can solve it mathematically also in simple way.

Common difference=100-90=-10°C=dInitial=a=100

$\\ \rm\Rrightarrow a_n=a+(n-1)d$

$\\ \rm\Rrightarrow a_5=100+4(-10)$

$\\ \rm\Rrightarrow a_5=100-40$

$\\ \rm\Rrightarrow a_5=60°C$

Mathematics

The temperature of a cup of coffee varies according to Newton s Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. 74 67 60 42

Step-by-step answer

P Answered by Specialist

67.3°C

Step-by-step explanation:

A = 25, T(0) = 100, T(1) = 90, find T(4)

dT/dt = -k(T - A)

d(T - A)/dt = -k(T - A)

d(T - A)/(T - A) = -k dt

ln(T - A) = lnC - kt

$T(t) -A=Ce^{-kt}\\T(t) = A + Ce^{-kt} = 25 + Ce^{-kt}\\T(0) = 25 + C = 100\\C=75$

$T(t) = 25 + 75e^{-kt}\\T(1) = 25 + 75e^{-k} = 90\\75e^{-k} = 65\\e^{-k} = \frac{65}{75}, k = ln\frac{75}{65} = 0.1431\\T(t) = 25+75e^{-0.1431t}\\T(4) = 25+75e^{-0.1431\times4} = 67.3$

Mathematics

The temperature of a cup of coffee varies according to Newton s Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes. 74 67 60 42

Step-by-step answer

P Answered by Specialist

67

Step-by-step explanation:

A = 25, T(0) = 100, T(1) = 90, find T(4)

dT/dt = -k(T - A)

d(T - A)/dt = -k(T - A)

d(T - A)/(T - A) = -k dt

ln(T - A) = lnC - kt

$T(t) -A=Ce^{-kt}\\T(t) = A + Ce^{-kt} = 25 + Ce^{-kt}\\T(0) = 25 + C = 100\\C=75$

$T(t) = 25 + 75e^{-kt}\\T(1) = 25 + 75e^{-k} = 90\\75e^{-k} = 65\\e^{-k} = \frac{65}{75}, k = ln\frac{75}{65} = 0.1431\\T(t) = 25+75e^{-0.1431t}\\T(4) = 25+75e^{-0.1431\times4} = 67.3$

Mathematics

The temperature of a cup of coffee varies according to newton s law of cooling: dt dt equals negative k times the quantity t minus a, where t is the water temperature, a is the room temperature, and k is a positive constant. if the coffee cools from 180°f to 100°f in 10 minutes at a room temperature of 75°f, how long (to the nearest minute) will it take the water to cool to 80°f?

Step-by-step answer

P Answered by PhD

13

This is a doozy. I actually have another form of the equation for cooling, and it's in Celcius (so I convert) and seconds (we can switch back to minutes in the end). The Celcius and Fahrenheit don't matter because it's the time we are interested in and as long as we use the C and F equivalents, we're fine. Promise. The formula is as follows: $T(t)= T_{s}+( T_{0}- T_{s})e ^{-kt}$ . It's really easy to use. T(t) is the ending temperature; $T_{s}$ is the room temp; $T_{0}$ is our initial temperature of the liquid; e is Euler's number, k is the constant of variation, and t is the time in seconds. In Celcius, our temps are as follows: 180 F = 82.2 C; 100 F = 37.8 C; 75 F = 23.9 C; 80 F = 26.7 C. 10 minutes is 600 seconds. Filling in our formula with all of that, we will solve for k. Then we will use that k value to solve for the time in the question we are told to answer. Let's do this. $37.8=23.9+(82.2-23.9)e ^{-600k}$ . Simplifying a bit we have $37.8=23.9+(58.3)e ^{-600k}$ . Subtracting 23.9 from both sides we have $13.9=58.3e ^{-600k}$ . divide both sides by 58.3 to get $.2384219=e ^{-600k}$ . At this point in your mathematical career, I'm assuming that since you're doing this, you're studying natural and common logs. The base of a natural log is e, therefore, if we take the natural log of e, they both "undo" each other and cancel each other out completely. So let's do that. ln(.2384219)=-600k

. Simplifying on the left gives us -1.433713=-600k

. Dividing both sides by -600 gives us a k value of .002389. Now we will do the equation again, using that k value, to solve for how long it takes for the coffee to cool to 80 F (26.7 C). $26.7=23.9+(82.2-23.9)e ^{-.002389t}$ . Simplifying a bit again gives us $26.7=23.9+(58.32)e ^{-.002389t}$ . Subtracting 23.9 from both sides gives us $2.8=(58.32)e ^{-.002389t}$ . We will now divide both sides by 58.32 to get $.0480109=e ^{-.002389t}$ . Again we will take the natural log of both sides to "undo" the e: ln(.0480109)=-.002389t

. The natural log of .0480109 is -3.036325671. So, dividing both sides by the negative decimal on the right gives us $\frac{-3.036325671}{-.002389}=t$ and t = 1270.9 seconds. Divide that by 60 to get the time in minutes. t = 21.1826...so 21 minutes. Phew!!!

Mathematics

The temperature of a cup of coffee varies according to newton s law of cooling: dt dt equals negative k times the quantity t minus a, where t is the water temperature, a is the room temperature, and k is a positive constant. if the coffee cools from 180°f to 100°f in 10 minutes at a room temperature of 75°f, how long (to the nearest minute) will it take the water to cool to 80°f?

Step-by-step answer

P Answered by PhD

13

This is a doozy. I actually have another form of the equation for cooling, and it's in Celcius (so I convert) and seconds (we can switch back to minutes in the end). The Celcius and Fahrenheit don't matter because it's the time we are interested in and as long as we use the C and F equivalents, we're fine. Promise. The formula is as follows: $T(t)= T_{s}+( T_{0}- T_{s})e ^{-kt}$ . It's really easy to use. T(t) is the ending temperature; $T_{s}$ is the room temp; $T_{0}$ is our initial temperature of the liquid; e is Euler's number, k is the constant of variation, and t is the time in seconds. In Celcius, our temps are as follows: 180 F = 82.2 C; 100 F = 37.8 C; 75 F = 23.9 C; 80 F = 26.7 C. 10 minutes is 600 seconds. Filling in our formula with all of that, we will solve for k. Then we will use that k value to solve for the time in the question we are told to answer. Let's do this. $37.8=23.9+(82.2-23.9)e ^{-600k}$ . Simplifying a bit we have $37.8=23.9+(58.3)e ^{-600k}$ . Subtracting 23.9 from both sides we have $13.9=58.3e ^{-600k}$ . divide both sides by 58.3 to get $.2384219=e ^{-600k}$ . At this point in your mathematical career, I'm assuming that since you're doing this, you're studying natural and common logs. The base of a natural log is e, therefore, if we take the natural log of e, they both "undo" each other and cancel each other out completely. So let's do that. ln(.2384219)=-600k

. Simplifying on the left gives us -1.433713=-600k

. Dividing both sides by -600 gives us a k value of .002389. Now we will do the equation again, using that k value, to solve for how long it takes for the coffee to cool to 80 F (26.7 C). $26.7=23.9+(82.2-23.9)e ^{-.002389t}$ . Simplifying a bit again gives us $26.7=23.9+(58.32)e ^{-.002389t}$ . Subtracting 23.9 from both sides gives us $2.8=(58.32)e ^{-.002389t}$ . We will now divide both sides by 58.32 to get $.0480109=e ^{-.002389t}$ . Again we will take the natural log of both sides to "undo" the e: ln(.0480109)=-.002389t

. The natural log of .0480109 is -3.036325671. So, dividing both sides by the negative decimal on the right gives us $\frac{-3.036325671}{-.002389}=t$ and t = 1270.9 seconds. Divide that by 60 to get the time in minutes. t = 21.1826...so 21 minutes. Phew!!!

Physics

An amateur player is about to throw a dart with an initial velocity of 15 meters/second onto a dartboard that is at a distance of 2.7 meters. Calculate the vertical distance by which the player will miss the target if he throws the dart horizontally, in line with the dartboard.

Step-by-step answer

P Answered by Specialist

Options:
a. 0.08 meters
b. 0.16 meters
c. 0.32 meters
d. 1.8 meters

Answer:
b. 0.16 meters

Explanation:
In the picture
Options:
a. 0.08 meters
b. 0.16 meters
c. 0.32 meters
d. 1.8 meters

Answer:
b. 0.16 meters

Physics

The system shown above consists of two identical blocks that are suspended using four cords, each of a different length. Which of the following claims are true about the magnitudes of the tensions in the cords?

Step-by-step answer

P Answered by Specialist

Answer: Option B and C are True.

Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂:
w = T₁/cos 60° -----(1);
w = T₂/cos 30° ----(2);
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°;
T₁ cos 30° = T₂ cos 60°;
T₂/T₁ = cos 30°/cos 60°;
T₂/T₁ =1.73.
Therefore, option a is false since T₂ > T₁.
Option B is true since T₁ cos 30° = T₂ cos 60°.
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
Answer: Option B and C are True.

Explanation:
The weight of the two blocks acts downwards.
Le

Physics

A proton (1.6726 ? 10-27 kg) and a neutron (1.6749 ? 10-27 kg) at rest combine to form a deuteron, the nucleus of deuterium or “heavy hydrogen.” In this process, a gamma ray (high-energy photon) is emitted, and its energy is measured to be 2.2 MeV (2.2 ? 10 6 eV).

Step-by-step answer

P Answered by Master

Answer:

see below.

Step-by-step explanation:

To solve this problem, we can use the conservation of energy and conservation of momentum principles.

Conservation of energy:

The total initial energy is the rest energy of the proton and neutron, which is given by:

Ei = (mp + mn)c^2

where mp and mn are the masses of the proton and neutron, respectively, and c is the speed of light.

The total final energy is the rest energy of the deuteron plus the energy of the gamma ray, which is given by:

Ef = (md)c^2 + Eg

where md is the mass of the deuteron and Eg is the energy of the gamma ray.

According to the conservation of energy principle, the initial energy and final energy must be equal, so we have:

Ei = Ef

(mp + mn)c^2 = (md)c^2 + Eg

Conservation of momentum:

The total initial momentum is zero because the proton and neutron are at rest. The total final momentum is the momentum of the deuteron and the momentum of the gamma ray. Since the gamma ray is massless, its momentum is given by:

pg = Eg/c

where pg is the momentum of the gamma ray.

According to the conservation of momentum principle, the total final momentum must be equal to zero, so we have:

0 = pd + pg

where pd is the momentum of the deuteron.

Solving for md and pd:

From the conservation of energy equation, we can solve for md:

md = (mp + mn - Eg/c^2)/c^2

Substituting this expression into the conservation of momentum equation, we get:

pd = -pg = -Eg/c

Substituting the given values, we have:

mp = 1.6726 × 10^-27 kg mn = 1.6749 × 10^-27 kg Eg = 2.2 × 10^6 eV = 3.52 × 10^-13 J

Using c = 2.998 × 10^8 m/s, we get:

md = (1.6726 × 10^-27 kg + 1.6749 × 10^-27 kg - 3.52 × 10^-13 J/(2.998 × 10^8 m/s)^2)/(2.998 × 10^8 m/s)^2 = 3.3435 × 10^-27 kg

pd = -Eg/c = -(3.52 × 10^-13 J)/(2.998 × 10^8 m/s) = -1.1723 × 10^-21 kg·m/s

Therefore, the mass of the deuteron is 3.3435 × 10^-27 kg, and its momentum is -1.1723 × 10^-21 kg·m/s.

Step-by-step answer

Faq

Try asking the Studen AI a question.