Question

Земля делает 1 оборот вокруг своей оси за 24 часа.вычислить угловую и линейную скорости вращения точки на поверхностисземли,кот.находится на экваторе. радиус =6400 км

Answer 1

N=1
t=86400 c
R=6.4*10^5 м
w-?
V-?        w=360/86400=0.0042 1/c  V=wR    
V=0.0042*6.4*10^5=2.688*10^3 м/с

Answer 2

Calculating Angular and Linear Speed of a Point on the Equator

To calculate the angular and linear speeds of a point on the Earth's equator, we can use the following formulas:

1. Angular Speed (ω): - The angular speed of an object in circular motion is given by the formula: \[ \omega = \frac{2\pi}{T} \] where ω is the angular speed in radians per second, and T is the period of rotation in seconds.

2. Linear Speed (v): - The linear speed of a point on the Earth's surface can be calculated using the formula: \[ v = r \times \omega \] where v is the linear speed in meters per second, r is the radius of the Earth, and ω is the angular speed in radians per second.

Given: - Radius of the Earth (r) = 6400 km = 6400000 meters - Period of rotation (T) = 24 hours = 24 * 60 * 60 seconds

Calculations

1. Angular Speed (ω): \[ \omega = \frac{2\pi}{T} \] \[ \omega = \frac{2 \times 3.1416}{24 \times 60 \times 60} \] \[ \omega ≈ 7.27 \times 10^{-5} \, \text{radians per second} \]

2. Linear Speed (v): \[ v = r \times \omega \] \[ v = 6400000 \times 7.27 \times 10^{-5} \] \[ v ≈ 463.4 \, \text{meters per second} \]

Results

- Angular Speed (ω): Approximately 7.27 × 10^-5 radians per second - Linear Speed (v): Approximately 463.4 meters per second