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AlladinOne
1 month ago
14

Two thermometers are calibrated, one in degrees Celsius and the other in degrees Fahrenheit.

Physics
1 answer:
Keith_Richards [3.2K]1 month ago
8 0

Answer:

The temperature is 233.15 K

Explanation:

Recall the formula for converting Celsius (C) to Fahrenheit (F):

\frac{9}{5} C+32=F

To make degree C equal to degree F, we set it up as C = F

By substituting F with C on the right side of the equation above, we find C:

\frac{9}{5} C+32=F\\\frac{9}{5} C+32=C\\\frac{9}{5} C-\frac{5}{5} C =-32\\\frac{4}{5} C==32\\C= \frac{-32\,*\,5}{4} \\C=-40

This indicates that -40°C = -40°F

And this temperature in Kelvin is:

-40°C + 273.15 = 233.15 K

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inna [3103]

La force agissant pendant 9 s et la décélération pendant 12 - 9 = 3 s.

Distance totale parcourue = 990 m

vitesse initiale u = 0

Distance parcourue pendant l'accélération

s₁ = 1/2 a 9² où a est l'accélération

= 40.5 a

vitesse finale après 9 s

v = at = 9a

pendant la décélération

v² = u² - 5 x s₂

0 = (9a)² - 5 s₂

s₂ = 16.2 a²

Distance parcourue pendant la décélération = 16.2 a²

s₁ + s₂ = 990

40.5 a + 16.2 a² = 990

16.2 a² + 40.5 a - 990 = 0

a = 6.5

4 0
1 month ago
If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
Maru [3345]

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
1 month ago
A long cylindrical rod of diameter 200 mm with thermal conductivity of 0.5 W/m⋅K experiences uniform volumetric heat generation
Ostrovityanka [3204]

Answer:

a, 71.8° C, 51° C

b, 191.8° C

Explanation:

Given the data:

D(i) = 200 mm

D(o) = 400 mm

q' = 24000 W/m³

k(r) = 0.5 W/m.K

k(s) = 4 W/m.K

k(h) = 25 W/m².K

The heat generation formula can be articulated as follows:

q = πr²Lq'

q = π. 0.1². L. 24000

q = 754L W/m

Thermal conduction resistance, R(cond) = 0.0276/L

Thermal conduction resistance, R(conv) = 0.0318/L

Applying the energy balance equation,

Energy In = Energy Out

This equates to q, which is 754L

From the initial analysis, the temperature at the interface between the rod and sleeve is found to be 71.8° C

Additionally, the outer surface temperature records as 51° C

Furthermore, based on the second analysis, the calculated temperature at the center of the rod is determined to be 191.8° C

6 0
2 months ago
A rare and valuable antique chest is being moved into a truck using a 4.00 m long ramp. the kj weight of the chest plus packing
ValentinkaMS [3465]

We start by finding the angle of inclination with the sine function,

sin θ = 1 m / 4 m

θ = 14.48°

 

Next, we compute the work done by the movers using the following formula:

W = Fnet * d

 

We need to first determine Fnet. It is the weight force minus the frictional force.

Fnet = m g sinθ – μ m g cosθ

Fnet = 1,500 sin14.48 – 0.2 * 1,500 * cos14.48

Fnet = 84.526 N

 

The work done is therefore:

W = 84.526 N * 4 m

<span>W = 338.10 J</span>

7 0
1 month ago
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