The angular speed of the earth, the linear speed of an object on the surface, and the acceleration of the object will be 7.288 × 10⁻⁵ m/sec,446.36 m/sec, and 31.27 m/s² respectively.
The rate of velocity change concerning time is known as acceleration.
Unit conversion;
1 hour = 3600 sec
Given data;
Velocity, v= m/s
Time elapsed, t = 23 hours 56 minutes and 4 seconds
The radius of the earth is, R= 6.37×103 km.
The total taken in the second is;
T=23 hr × 3600 sec + 56 min × 60 sec + 4 sec
T= 86164 sec
The angular speed of the earth;
[tex]\rm \omega_e = \frac{2 \pi}{T} \\\\ \omega_e =\frac{ 2 \times 3.14 }{86164 \ sec} \\\\ \omega_e =7.28 8 \times 10^{-5 } \ rad /sec[/tex]
The linear speed of an object on the surface of the radius vector from the center of the earth is;
[tex]\rm v = r \times \omega \\\\ v= 6123 \ km \times 7.29 \times 10 ^{-5} \\\\ v = 446.36 \ m/sec[/tex]
The acceleration of the object on the surface of the earth is;
[tex]\rm a = \frac{v^2}{r} \\\\ a=\frac{4446.32^2}{6.37 \times 10^3} \\\\ a= 31.27 \ m/s^2[/tex]
Hence,the angular speed of the earth, the linear speed of an object on the surface, and the acceleration of the object will be 7.288 × 10⁻⁵ m/sec,446.36 m/sec, and 31.27 m/s²
To learn more about acceleration, refer to the link https://brainly.com/question/2437624
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