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serious [3.7K]
3 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]3 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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4 0
2 years ago
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How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?
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The amount of heat needed to raise the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
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Cs is its specific heat capacity
\Delta T is the increase in temperature

For oxygen, the specific heat capacity is approximately 
C_s = 0.92 J/(g K)
The variation of temperature for the sample in our problem is 
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4 0
3 years ago
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hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac
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A parallel-plate capacitor is constructed from two aluminum foils of 1 square centimeter area each placedon both sides of a rubb
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3 years ago
A 0.106-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find
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Answer:

The time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

Explanation:

Given :

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Where \epsilon _{o} = 8.85 \times 10^{-12}

   C = \frac{8.85 \times 10^{-12 } \times 36 \times 10^{-4}  }{4 \times 10^{-3} }

   C = 7.9 \times 10^{-12} F

But   C = \frac{Q}{V}

Where Q = It

  C = \frac{It}{V}

  V = \frac{It}{C}

Now differentiate above equation wrt. time,

  \frac{dV}{dt} = \frac{I}{C}

       = \frac{0.106}{7.9 \times 10^{-12} }

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Therefore, the time rate of change of flux is 1.34 \times 10^{10} \frac{V}{s}

8 0
3 years ago
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