Temperature determines the hotness or coldness. The surface temperature of Canopus is 7350 K. The surface temperature of Vega is 9602 K. Because the surface temperature of Vega is greater than Canopus, Vega is hotter.
Brightness is determined by the star's magnitude. The apparent magnitude of Canopus is (-0.72) and that of Vega is (0.03). we know that, lesser the magnitude, brighter is the star. Hence, Canopus is brighter.
Vega is hotter but Canopus is brighter. This is because, brightness depends on the star's distance and size of the star as well other than the factor temperature.
Answer:
The magnitude of the angular acceleration ∝ = ![\frac{rxF}{2.8[tex]r^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BrxF%7D%7B2.8%5Btex%5Dr%5E%7B2%7D)
}[/tex]
Explanation:
The angular acceleration ∝ is equal to the torque (radius multiplied by force) divided by the mass times the square of the radius. The magnitude of angular acceleration ∝ will have the equation above but we have to replace the mass in the equation by 2.8kg as stated.
Answer:
(a) Since net charge remains same,after immersion Q is same
(b) I. 14.56pF ii. 3.05V
(c) ΔU = 5.204nJ
Explanation:
a)
C = kεA/d
k=1 for air
ε is 8.85x10-12F/m
A = .0025m2
d = .125m
C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF
Q = CV = .177pF * 244V = 43.188pC
Since net charge remains same,after immersion Q is same
b)
C = kεA/d, for distilled water k is approx. 80
Cwater = Cair x k
=0.177pF x 80 = 14.16pF
Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V
c) Change in energy: ΔU = Uwater - Uair
Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ
Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ
ΔU = 5.204nJ
Answer:
An element is a pure substance that cannot be separated into simpler substances by chemical or physical means. There are about 117 elements.
Explanation:
Answer:
a) 9.99 s
b) 538 m
c) 20.5 s
d) 1160 m
Explanation:
Given:
x₀ = 0 m
y₀ = 49.0 m
v₀ = 113 m/s
θ = 60.0°
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find t.
vᵧ = aᵧ t + v₀ᵧ
(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)
t ≈ 9.99 s
b) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.
vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)
(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)
y ≈ 538 m
c) When the projectile lands, y = 0 m. Find t.
y = y₀ + v₀ᵧ t + ½ aᵧ t²
(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²
You'll need to solve using quadratic formula:
t ≈ -0.489, 20.5
Since negative time doesn't apply here, t ≈ 20.5 s.
d) When the projectile lands, y = 0 m. Find x. (Use answer from part c).
x = x₀ + v₀ₓ t + ½ aₓ t²
x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²
x ≈ 1160 m
