Answer:
2577 K
Explanation:
Power radiated , P = σεAT⁴ where σ = Stefan-Boltzmann constant = 5.6704 × 10⁻⁸ W/m²K⁴, ε = emissivity of bulb filament = 0.8, A = surface area of bulb = 30 mm² = 30 × 10⁻⁶ m² and T = operating temperature of filament.
So, T = ⁴√(P/σεA)
Since P = 60 W, we substitute the vales of the variables into T. So,
T = ⁴√(P/σεA)
= ⁴√(60 W/(5.6704 × 10⁻⁸ W/m²K⁴ × 0.8 × 30 × 10⁻⁶ m²)
= ⁴√(60 W/(136.0896 × 10⁻¹⁴ W/K⁴)
= ⁴√(60 W/(13608.96 × 10⁻¹⁶ W/K⁴)
= ⁴√(0.00441 × 10¹⁶K⁴)
= 0.2577 × 10⁴ K
= 2577 K
Answer:
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
Explanation:
We can answer this question by using Kepler's second law of planetary motion, which states that:
"A line connecting the center of the Sun with the center of each planet sweeps out equal areas in equal intervals of time"
This means that when a planet is further away from the Sun, it will move slower (because the line is longer, so it must move slower), while when the planet is closer to the Sun, it will move faster (because the line is shorter, so it must move faster).
In the text of this problem, it is written that the planet moves at 31 km/s when is close to the star and 35 km/s when it is farthest: this is in disagreement with what we said above, therefore the correct option is
E. Kepler's second law says the planet must move fastest when it is closest, not when it is farthest away.
The formula for calcium oxide is CaO.
Answer:
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
, r is radius
Let I is the displacement current. It is given by :
Here, 
is rate of increasing potential difference
So
So, the value of displacement current is 
.
Answer:
Distance = 13.9 meters
Explanation:
Given the following data;
Maximum speed = 150 km/hr to meters per seconds = 150 * 1000/3600 = 41.67 m/s
Decelerating speed = 3m/s
To find the distance travelled with this speed;
Distance = maximum speed/decelerating speed
Distance = 41.67/3
Distance = 13.9 meters
Therefore, the bus would travel a distance of 13.9 meters before stopping.
