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alina1380
1 month ago
9

Anthony and Maelynn are watching a football game outside on a sunny day. Anthony is wearing a black shirt and Maelynn is wearing

a white shirt. Which person will be warmer after the game? Explain your answer.
Physics
2 answers:
Ostrovityanka [3.2K]1 month ago
7 0

Answer: After the game, Anthony will feel warmer.

Explanation:

Outside on a sunny day, both Anthony and Maelynn are spectating a football match. Anthony sports a black shirt while Maelynn has a white one on. Anthony will indeed be warmer post-game. The black hue is effective at absorbing radiation but not at reflecting it.

This means that the black color gathers heat until balancing out thermally. Consequently, it is better to opt for cotton fabrics in the summer months rather than dark clothing.

kicyunya [3.2K]1 month ago
3 0

Answer:

Anthony will be warmer after the game due to his black shirt. Dark shades absorb more heat energy compared to lighter shades.

Explanation:

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Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Consequently, the temperature difference across the material will be \Delta T = 375 K

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

\Delta T is the temperature difference we need to determine

represents the thickness of the material\Delta x=2.5 cm =0.025 m

To isolate \Delta T from equation (1), we obtain:

\Delta T =\frac{Q \Delta x}{Ak}

Initially, we convert 3KW to W, resulting in:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

With all variables accounted for, we can substitute and calculate:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

Thus, the temperature difference across the material will be \Delta T = 375 K

5 0
14 days ago
A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

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initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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1 month ago
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For the object you've mentioned, it translates to

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