Answer:
1.23×10⁸ m
Explanation:
Acceleration due to gravity is:
a = GM / r²
where G is the universal gravitational constant,
M is the mass of the planet,
and r is the distance from the center of the planet to the object.
When the object is on the surface of the Earth, a = g and r = R.
g = GM / R²
When the object is at height i above the surface, a = 1/410 g and r = i + R.
1/410 g = GM / (i + R)²
Divide the first equation by the second:
g / (1/410 g) = (GM / R²) / (GM / (i + R)²)
410 = (i + R)² / R²
410 R² = (i + R)²
410 R² = i² + 2iR + R²
0 = i² + 2iR − 409R²
Solve with quadratic formula:
i = [ -2R ± √((2R)² − 4(1)(-409R²)) ] / 2(1)
i = [ -2R ± √(1640R²) ] / 2
i = (-2R ± 2R√410) / 2
i = -R ± R√410
i = (-1 ± √410) R
Since i > 0:
i = (-1 + √410) R
R = 6.37×10⁶ m:
i ≈ 1.23×10⁸ m
I think this is the answer:
<span>Matter can change its state because of the pressure and/or the temperature.</span>
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I believe the answer is A. shorter wires
Answer:
the time interval that an earth observer measures is 4 seconds
Explanation:
Given the data in the question;
speed of the spacecraft as it moves past the is 0.6 times the speed of light
we know that speed of light c = 3 × 10⁸ m/s
so speed of spacecraft v = 0.6 × c = 0.6c
time interval between ticks of the spacecraft clock Δt₀ = 3.2 seconds
Now, from time dilation;
t = Δt₀ / √( 1 - ( v² / c² ) )
t = Δt₀ / √( 1 - ( v/c )² )
we substitute
t = 3.2 / √( 1 - ( 0.6c / c )² )
t = 3.2 / √( 1 - ( 0.6 )² )
t = 3.2 / √( 1 - 0.36 )
t = 3.2 / √0.64
t = 3.2 / 0.8
t = 4 seconds
Therefore, the time interval that an earth observer measures is 4 seconds
