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love history [14]
3 years ago
5

A small metal bar, whose initial temperature was 30° C, is dropped into a large container of boiling water. How long will it tak

e the bar to reach 80° C if it is known that its temperature increases 2° during the first second? (The boiling temperature for water is 100° C. Round your answer to one decimal place.)

Physics
2 answers:
Law Incorporation [45]3 years ago
8 0

Answer:

t = 25s

Explanation:

The required temperature change is 50°C. (80 –30). The temperature of the bar increased 2°C in the first second. Assuming uniformity in the properties of the bar, in 2s the temperature would now be 2+2 = 4°C, in 3s it will be 2+2+2 =6°C.

So for gain a temperature change of 50°C, it will take 50/2 = 25s

ΔT /t = 2

t = ΔT/2 = 50/2 = 25seconds

kirill [66]3 years ago
7 0

Answer: It will take 43.2 seconds for the bar to reach 80 °C

Explanation: Please see the attachments below

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3 years ago
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A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction betw
umka2103 [35]

Answer:

<h3>0.69</h3>

Explanation:

Using the Newtons law of motion;

\sum Fx = ma_x\\Fm - Ff = ma_x

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69

Hence the coefficient of kinetic friction between the box and floor is 0.69

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3 years ago
Five liters of air at 50 c is warmed to 100c what is the new volume if the pressure remain constant
KonstantinChe [14]
To solve the problem, we can use Charle's law, which states that for an ideal gas at constant pressure the ratio between absolute temperature T and volume V remains constant:
\frac{T}{V}=k
For a gas transformation, this law can be rewritten as
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Before applying the law, we must convert the temperatures in Kelvin:
T_1 = 50^{\circ}C + 273 = 323 K
T_2 = 100^{\circ}C+273=373 K
The initial volume of the gas is V_1 = 5 L, so if we re-arrange (1) we find the new volume of the gas:
V_2 = V_1  \frac{T_2}{T_1}=(5 L) \frac{373 K}{323 K}=5.77 L
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The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher. 
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Answer:

Explanation:

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