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:
Step 7- Communicate. Present/share your results. Replicate.
Step 1- Question.
Step 2-Research.
Step 3-Hypothesis.
Step 4-Experiment.
Step 5-Observations.
Step 6-Results/Conclusion
Explanation:
Answer:
The net magnetic field ta the center of square is
.
Explanation:
Current, I = 12 A , side ,a = 10 cm = 0.1 m
Let the magnetic field due to the one side is B.
The magnetic field is given by
Net magnetic field at the center of the square is
B' = 4 B
(a) The velocity (in m/s) of the rock after 1 second is 11.28 m/s.
(b) The velocity of the rock after 2 seconds is 7.56 m/s.
(c) The time for the block to hit the surface is 4.03.
(d) The velocity of the block at the maximum height is 0.
<h3>
Velocity of the rock</h3>
The velocity of the rock is determined as shown below;
Height of the rock after 1 second; H(t) = 15(1) - 1.86(1)² = 13.14 m
v² = u² - 2gh
where;
- g is acceleration due to gravity in mars = 3.72 m/s²
v² = (15)² - 2(3.72)(13.14)
v² = 127.23
v = √127.23
v = 11.28 m/s
<h3>Velocity of the rock when t = 2 second</h3>
v = dh/dt
v = 15 - 3.72t
v(2) = 15 - 3.72(2)
v(2) = 7.56 m/s
<h3>Time for the rock to reach maximum height</h3>
dh/dt = 0
15 - 3.72t = 0
t = 4.03 s
<h3>Velocity of the rock when it hits the surface</h3>
v = u - gt
v = 15 - 3.72(4.03)
v = 0
Learn more about velocity at maximum height here: brainly.com/question/14638187
Answer:
592.92 x 10³ Pa
Explanation:
Mole of ammonia required = 10 g / 17 =0 .588 moles
We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .
From the relation
PV = nRT
P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )
P = 592.92 x 10³ Pa
