Chlorine has the smallest atomic radius since the atomic radius decreases as you travel to the right and up
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Answer:
a = 2.275 10⁻⁴ m
Explanation:
This is a diffraction problem that is described by the equation
a sin θ = m λ
The first dark minimum occurs for m = 1
a = λ / sin θ
The angle can be found by trigonometry,
tan θ = y / x
θ = tan⁻¹ y / x
Let's reduce the magnitudes to the SI system
y = 8.24 mm = 8.24 10⁻³ m
λ = 625 nm = 625 10⁻⁹ m
θ = tan⁻¹ 8.24 10⁻³ / 3.00
θ = 0.002747 rad
We calculate
a = 625 10⁻⁹ / sin 0.002747
a = 2.275 10⁻⁴ m
B) no light pass through
Explanation:
An opaque material is a material that does not allow light to pass through. Opaque materials are the most common materials, they do not allow light to pass through.
- All opaque materials will cast shadows along their path when there is a screen.
- Transparent materials allows light to pass through for example glasses.
- Translucent materials allows light to go through partially.
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Transmission of light brainly.com/question/8032392
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