Form concentric circles around the wire
Answer:
(a)0.0002778
(b)
Explanation:
(a) The minute hand has a period of 60 minutes ( or 60 * 60 = 3600 seconds) for 1 circle. Its frequency per second would be
1 / 3600 = 0.0002778
(b) The hour hand has a period of 24 hours ( or 24*60 * 60 = 86400
seconds) for 1 circle. Its frequency per second would be

Answer:B
Explanation:Grote reber was the first scientist to map the milky way galaxy using radio waves.
Answer:
1.73 m/s²
3.0 cm
Explanation:
Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.
Sum of forces in the y direction:
∑F = ma
T cos 10° − mg = 0
T cos 10° = mg
T = mg / cos 10°
Sum of forces in the x direction:
∑F = ma
T sin 10° = ma
mg tan 10° = ma
g tan 10° = a
a = 1.73 m/s²
Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
D − mg = 0
D = mg
½ ρₐ v² C A = ρᵢ V g
½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g
3 ρₐ v² C = 8 ρᵢ r g
r = 3 ρₐ v² C / (8 ρᵢ g)
r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))
r = 0.030 m
r = 3.0 cm
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:
brainly.com/question/14548124