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Lina20 [59]
3 years ago
13

A photographer in a helicopter ascending vertically at a constant rate of 11.5 m/s accidentally drops a camera out the window wh

en the helicopter is 70.0 m above the ground.
A) How long will the camera take to reach the ground?
B) What will its speed be when it hits?
Physics
1 answer:
Degger [83]3 years ago
6 0
The problem is a free fall problem as the camera drops. The vertical velocity is zero. It is the action of gravity that takes place here. To calculate the time it takes to reach the ground, we do as follows:

t = √2y/g = √(2)(70)/9.81 = 3.78 s

The velocity as it impacts the ground would be,

v = √2gy = 37.04 m/s

Hope this helps.
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The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

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Suppose we want to calculate the moment of inertia of a 56.5 kg skater, relative to a vertical axis through their center of mass
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a. 0.342 kg-m² b. 2.0728 kg-m²

Explanation:

a. Since the skater is assumed to be a cylinder, the moment of inertia of a cylinder is I = 1/2MR² where M = mass of cylinder and r = radius of cylinder. Now, here, M = 56.5 kg and r = 0.11 m

I = 1/2MR²

= 1/2 × 56.5 kg × (0.11 m)²

= 0.342 kgm²

So the moment of inertia of the skater is

b. Let the moment of inertia of each arm be I'. So the moment of inertia of each arm relative to the axis through the center of mass is (since they are long rods)

I' = 1/12ml² + mh² where m = mass of arm = 0.05M, l = length of arm = 0.875 m and h = distance of center of mass of the arm from the center of mass of the cylindrical body = R/2 + l/2 = (R + l)/2 = (0.11 m + 0.875 m)/2 = 0.985 m/2 = 0.4925 m

I' = 1/12 × 0.05 × 56.5 kg × (0.875 m)² + 0.05 × 56.5 kg × (0.4925 m)²

= 0.1802 kg-m² + 0.6852 kg-m²

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The total moment of inertia from both arms is thus I'' = 2I' = 1.7308 kg-m².

So, the moment of inertia of the skater with the arms extended is thus I₀ = I + I'' = 0.342 kg-m² + 1.7308 kg-m² = 2.0728 kg-m²

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