Refer to the diagram shown.

Because the surface is frictionless, the resistive for, R, is zero.

Let m = the mass of the object.
Let a = acceleration due to the applied force.
Therefore
12.7 N = (m kg)*(a m/s²)
a = 12.7/m m/s²

The object travels 16.1 m in 2.5 s, starting from rest. Therefore
16.1 N = (1/2)*(12.7/m m/s²)*(2.5 s)² = 39.6875/m N
m = 16.1/39.6875 = 0.4057 kg

For freefall, let g = acceleration due to gravity.
The time to fall from 10.3 m is 2.88 s, therefore
10.3 m = (1/2)*(g m/s²)*(2.88 s)² = 4.1472g m
g = 10.3/4.1472 = 2.484 m/s²

Answer:
The gravitational acceleration on the planet is 2.5 m/s² (nearest tenth)

3 0

3 years ago

A low orbit satellite at 100 km altitude had a camera that resolved a car location to within 0.3 meter. Find the minimum diamete

NeX [460]

Answer:

Minimum diameter of the camera lens is 22.4 cm

The focal length of the camera's lens is 300cm

Explanation:

y = Resolve distance = 0.3 m

h = Height of satellite = 100 km

λ = Wavelength = 550 nm

Angular resolution

$tan\theta\approx \theta =\frac{y}{h}\\\Rightarrow \theta=\frac{0.3}{100\times 10^3}=3\times 10^{-6}$

From Rayleigh criteria

$sin\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{sin\theta}\\\Rightarrow D=1.22\frac{550\times 10^{-9}}{sin3\times 10^{-6}}=0.2236\ m=22.4\ cm$

Minimum diameter of the camera lens is 22.4 cm

Relation between resolvable feature, focal length and angular resolution

$d=f\Delta \theta\\\Rightarrow f=\frac{d}{\Delta \theta}\\\Rightarrow f=\frac{9\times 10^{-6}}{3\times 10^{-6}}=3\ m=300\ cm$

The focal length of the camera's lens is 300cm

3 0

2 years ago