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serious [3.7K]
4 years ago
11

Dump Tower is 96 stories tall. A small, 1.2-kg object is dropped over the side of the roof of the tower and accelerates toward t

he ground. The mass of the planet housing Dump Tower is 2.7 times the mass of Ganymede and the radius of the body is 1.7 times the radius of Makemake. You will track the object for its entire fall. Each story of this tower is 3.05 meters tall.
How many seconds will it take the object to reach the ground and what will be its impact speed with the ground?

Physics
1 answer:
laiz [17]4 years ago
8 0

Answer:

6.0 s

98 m/s

Explanation:

The radius of the planet is much bigger than the height of the tower, so we will assume the acceleration is constant.  Neglect air resistance.

Acceleration due to gravity on this planet is:

a = GM / r²

a = (6.67×10⁻¹¹ m³/kg/s²) (2.7 × 1.48×10²³ kg) / (1.7 × 750,000 m)²

a = 16.4 m/s²

The height of the tower is:

Δy = 96 × 3.05 m

Δy = 293 m

Given v₀ = 0 m/s, find t and v.

Δy = v₀ t + ½ at²

(293 m) = (0 m/s) t + ½ (16.4 m/s²) t²

t = 6.0 s

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (16.4 m/s²) (293 m)

v = 98 m/s

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Answer:

0.247 μC

Explanation:

As both sphere will be at the same level at wquilibrium, the direction of the electric force will be on the x axis. As you can see in the picture below, the x component of the tension of the string of any of the spheres should be equal to the electric force of repulsion. And its y component will be equal to the weight of one sphere. We can use trigonometry to find the components of the tensions:

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In equilibrium, the distance between the spheres will be equal to 2 times the length of the string times sin(50):

r = 2*L*sin(50) = 2 * 0.1m * sin(50) 0.1532 m

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F_x = T_x - F_e = 0\\T_x = F_e = k*\frac{q^2}{r^2}

q = \sqrt{T_x *\frac{r^2}{k}} = \sqrt{0.0234 N * \frac{(0.1532m)^2}{9*10^9 N*m^2/C^2} } = 2.4704 * 10^-7 C

O 0.247 μC

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