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sertanlavr [38]
1 year ago
13

Joanna claims that a large block of ice will cool a substance more than a small block of ice will at the same temperature. To su

pport her claim, Joanna places two blocks of ice, one larger than the other, into separate beakers each containing some water. She compares the final water temperatures of the two beakers after each block of ice has melted.
Physics
1 answer:
Levart [38]1 year ago
3 0

She puts each block of ice in the same 3000 mL beaker, each with 2000 mL of water at room temperature, and measures the temperature before and after adding ice. Therefore, small blocks of ice will have the same temperature.

Joanna puts two blocks of ice (one larger than the other) into separate cups and fills each with water. She compares the final water temperature of the two cups after each block of ice melts.

Put each block of ice in the same 3000 mL beaker, each at room temperature, put 2000 mL of water in it, and measure the temperature before and after adding ice. This way you keep the water at the same temperature in the beginning, then the temperature changes after you add the ice, giving you a better idea of ​​the final temperature reading.

Learn more about Temperature here brainly.com/question/24746268

#SPJ9

                 

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3 years ago
A doppler effect occurs when a source of sound moves. True or False
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2 years ago
What is the relationship between force in velocity selector in a bain bridge.​
kifflom [539]

Answer:

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» Electric field Intensity

» Magnetic field density

<u>Relationship</u><u>:</u>

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3 years ago
A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag
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Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

R_s = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = \frac{150}{3.6}\ m/s

L = Length of pole

Density

\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3

Drag force

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N

Force on the circular sign is 147.7 N

M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm

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<span>the overload principle hope this helps

</span>
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