The resistance in this circuit is 39.8 ohms.
Explanation:
Any circuit having resistor, battery and ammeter connected in series will obey the ohm's law in basic case. So according to the Ohm's law, the current flowing in the circuit through the ammeter will be equal to the voltage shown in the voltmeter or battery and resistor is the proportionality constant. So with this law
V = IR
So, Resistance R = V/I
As the voltage is given as 23.90 V and the current is given as 0.6 A, then resistance is
R = 23.90/0.6 = 39.8 ohms.
So, the resistance in this circuit is 39.8 ohms.
Answer:
the value of the final pressure is 0.168 atm
Explanation:
Given the data in the question;
Let p₁ be initial pressure, v₁ be initial volume.
After expansion, p₂ is final pressure and v₂ is final volume.
So using the following equations;
p₁v₁ = nRT
p₂v₂ = nRT
hence, p₁v₁ = p₂v₂
we find p₂
p₂ = p₁v₁ / v₂
given that; initial volume v₁ = 0.175 m³, Initial pressure p₁ = 0.350 atm,
final volume v₂ = 0.365 m³
we substitute
p₂ = ( 0.350 atm × 0.175 m³ ) / 0.365 m³
p₂ = 0.06125 atm-m³ / 0.365 m³
p₂ = 0.168 atm
Therefore, the value of the final pressure is 0.168 atm
Answer:
282 m
Explanation:
Given:
v₀ = 20.1 m/s
v = 33.2 m/s
t = 10.6 s
Find: Δx
Δx = ½ (v + v₀) t
Δx = ½ (33.2 m/s + 20.1 m/s) (10.6 s)
Δx ≈ 282 m
A geologic event causes changes to the physical makeup of a particular place and occurs slowly.
Geological events are what causes numerous changes and phenomena on the Earth's surface. Examples of these events include cliff erosion, volcanic eruption, or sedimentation at a mouth of a river.
Geological processes are extremely slow. However, because of the immense lengths of time involved, huge physical changes do occur - mountains are created and destroyed, continents form, break up and move over the surface of the Earth, coastlines change and rivers and glaciers erode huge valleys.
Geological events are both classified as internal and external. This means that these events occur both in the Earth's surface and interior.
So, If the silica cyliner of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
To estimate the operating temperature of the radiant wall heater, we need to use the equation for power radiated by the radiant wall heater.
<h3>Power radiated by the radiant wall heater</h3>
The power radiated by the radiant wall heater is given by P = εσAT⁴ where
- ε = emissivity = 1 (since we are not given),
- σ = Stefan-Boltzmann constant = 6 × 10⁻⁸ W/m²-K⁴,
- A = surface area of cylindrical wall heater = 2πrh where
- r = radius of wall heater = 6 mm = 6 × 10⁻³ m and
- h = length of heater = 0.6 m, and
- T = temperature of heater
Since P = εσAT⁴
P = εσ(2πrh)T⁴
Making T subject of the formula, we have
<h3>Temperature of heater</h3>
T = ⁴√[P/εσ(2πrh)]
Since P = 1.5 kW = 1.5 × 10³ W
Substituting the values of the variables into the equation, we have
T = ⁴√[P/εσ(2πrh)]
T = ⁴√[1.5 × 10³ W/(1 × 6 × 10⁻⁸ W/m²-K⁴ × 2π × 6 × 10⁻³ m × 0.6 m)]
T = ⁴√[1.5 × 10³ W/(43.2π × 10⁻¹¹ W/K⁴)]
T = ⁴√[1.5 × 10³ W/135.72 × 10⁻¹¹ W/K⁴)]
T = ⁴√[0.01105 × 10¹⁴ K⁴)]
T = ⁴√[1.105 × 10¹² K⁴)]
T = 1.0253 × 10³ K
T = 1025.3 K
So, If the silica cylinder of the radiant wall heater is rated at 1.5 kw its temperature when operating is 1025.3 K
Learn more about temperature of radiant wall heater here:
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