Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Answer:
The final velocity of the object is,
= 27 m/s
Explanation:
Given,
The acceleration of the object, a = 1000 m/s²
The initial displacement of the object,
= 0 m
The final displacement of the object,
= 0.75 m
The initial velocity of the object will be,
= o m/s
The final velocity of the object,
= ?
The average velocity of the object,
v = (
-
)/ t
= 0.75 / t
The acceleration is given by the relation
a = v / t
1000 m/s² = 0.75 / t²
t² = 7.5 x 10⁻⁴
t = 0.027 s
Using the I equation of motion,
= u + at
Substituting the values
= 0 + 1000 x 0.027
= 27 m/s
Hence, the final velocity of the object is,
= 27 m/s
Answer:
The correct answer is B
Explanation:
To calculate the acceleration we must use Newton's second law
F = m a
a = F / m
To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface
P = I / c absorbent surface
P = F / A
F / A = I / c
F = I A / c
The area of area of a circle is
A = π r²
We replace
F = I π r² / c
Let's calculate
F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸
F = 8.375 10⁻²³ N
Density is
ρ = m / V
m = ρ V
m = ρ (4/3 π r³)
m = 4500 (4/3 π (1 10⁻⁶)³)
m = 1,885 10⁻¹⁴ kg
Let's calculate the acceleration
a = 8.375 10⁻²³ / 1.885 10⁻¹⁴
a = 4.44 10⁻⁹ m/s² absorbent surface
The correct answer is B
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