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Anit [1.1K]
3 years ago
13

According to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the differen

ce in temperature between the object and the surrounding medium. If T(t) represents the temperature of the object (CC) at time t (in hours), and T5 represents the constant temperature of the surrounding medium, then the differential equation best describing the rate of change in the temperature of the object is:
Physics
2 answers:
adelina 88 [10]3 years ago
7 0

Answer:

dT/dt = k[T5 - T]

Explanation:

DT/dt represents rate of change in temperature for Celsius degrees per hour. Its proportional to the difference in temperature between the object and the Surrounding medium. This means either dT/Dt =k (T-TS) or dT/dt=k(TS-T) with k being some positive constant of proportionality which depends on the object.

We can see that dT/dt = k(T-T5) has temperature increasing when the temperature of the object T is greater than the surrounding medium (T5).

The equation dT/dt=k(T5-T) has the temperature increasing when the object (T) is less than the temperature of the surrounding medium. Therefore the differential equation best describing the rate of change in temperature of the object is dT/dt=k(T5-T) for some positive constant of proportionality k.

Kryger [21]3 years ago
4 0

Answer:

dT(t)/dt = k[T5 - T(t)]

Explanation:

Since T(t) represents the temperature of the object and T5 represents the temperature of the surroundings, according to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the difference in temperature between the object and the surrounding medium, that is dT(t)/dt ∝ T5 - T(t)

Introducing the constant of proportionality

dT(t)/dt = k[T5 - T(t)]

which is the desired differential equation

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Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
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Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

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