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Degger [83]
2 years ago
5

A cup of hot coffee can be cooled by placing a cold spoon in it. A spoon of which of the following materials would be most effec

tive for this purpose: aluminum, copper, iron, or glass? Assume all spoons have the same mass.
Physics
1 answer:
arsen [322]2 years ago
8 0

Answer: copper

Explanation: when a spoon is placed in a hot coffee, the coffee will cool off based on the concept of heat transfer.

The coffee is at a higher temperature compared to the spoon, hence there is conduction of heat by the spoon from the coffee.

Heat is now transfered from the coffee to the spoon, the rate of heat transfer ( for this case coffee to spoon) is dependent on some properties but for this case, the thermal conductivity is of the most important.

A material of high thermal conductivity have a very fast rate of heat transfer and a material of low thermal conductivity have a slow rate of heat transfer.

Below is the value of thermal conductivity for copper, aluminum, iron and glass.

Copper (385 w/m.K)

Aluminum (205 W/m.K)

Iron (79.5 W/m.K)

Glass (0.8 W/m.K)

It can been seen that copper has the highest value for thermal conductivity which implies that it will conduct heat faster compared to others thus cooling the coffee faster.

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Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What
likoan [24]
F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
F = 56.67N
4 0
2 years ago
What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at
Stolb23 [73]

Answer:

m = 0.111 kg

Explanation:

Heat required to release from the body of the person when his temperature cool down by 1 degree C is given as

Q = m s\Delta T

here we know that

m = 70 kg

s = 3840 J/kg K

\Delta T = 1.00^o C

now we know that

Q = (70 kg)(3840 J/kg ^oC)(1 ^o C)

Q = 268800 J

now the same heat is used to vaporize water of the body

so it is given as

Q = mL

268800 = m(2.42 \times 10^6)

m = 0.111 kg

7 0
2 years ago
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
A solid cylindrical bar conducts heat at a rate of 25 W from a hot to a cold reservoir under steady state conditions. If both th
expeople1 [14]

Answer:

Using the new cylinder the heat rate between the reservoirs would be 50 W

Explanation:

  1. Conduction could be described by the Law of Fourierin the form: Q=kA\frac{T_1-T_2}{L} where Q is the rate of heat transferred  by conduction, k is the thermal conductivity of the material, T_1 and T_2 are the temperatures of each heat deposit, A is the cross area to the flow of heat, and {L} is the distance that the flow of heat has to go.
  2. For the original cylinder the Fourier's law would be: kA_1\frac{T_1-T_2}{L_1}=25W, and if A_1=\frac{\pi D_{1}^{2}}{4}, then the expression would be:k\frac{\pi D_1^{2}}{4} \frac{T_1-T_2}{L_1}=25W where D_1 is the diameter of the original cylinder, and {L_1} is the length of the original cylinder.
  3. For the new cylinder, in the same fashion that for the first, Fourier's Law would be: Q_2=k\frac{\pi D_2^2}{4}\frac{T_1-T_2}{L_2},where Q_2 is the heat rate in the second case, D_2 and {L_2 are the new diameter and length.
  4. But, D_2=2D_1 and L_2=2L_1, substituting in the expression for Q_2: Q_2=k\frac{\pi (2D_1)^2}{4}\frac{T_1-T_2}{2L_1}.
  5. Rearranging: Q_2=\frac{2^2}{2}(k\frac{\pi D_1^2}{4}\frac{T_1-T_2}{L_1}).
  6. In the last declaration of  Q_2, it could be noted that the expressión inside the parenthesis is actually  Q_1, then:  Q_2=\frac{2^2}{2}(25W)=50W.
  7. <u>It should be noted, that the temperatures in the hot and cold reservoirs never change.</u>
7 0
2 years ago
Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc
xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is  equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2  v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0
2 years ago
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