Explanation:
PRIMERO HACES EL RECUENTO DEL TIEMPO Y LO CONVIERTES EN
SEGUNDOS Y ENTONCES
<em>t</em> = 227 s
= 227 S - 38 s = 189 s
= 38 s
LUEGO USANDO LA ECUACIÓN DE GALILEO GALILEI SSUPONIENDO
QUE EL MOVIL VIAJA A VELOCIDAD CONSTANTE
<em>v</em> = 3.50 m/189 s = 0.0185 m/s
PARA LA DISTANCIA NTRE B Y C
= 0.0185 m/S( 38 s) = 0.703 m
LA HORA EN QUE EL MOVIL PASA POR A ES
11:43:15 - 38 s - 189 s = 11:39:29
Answer:
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,
Explanation:
When a satellite is in orbit the most important force is the docking of gravity with the Earth
F = m a
where the acceleration is centripetal and F is the force of universal attraction
centripetal acceleration is
a = v² / r
F = m v² / r
In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.
The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.
Answer:
weight on earth is mg
which is 5*9.8
49 Newton
weight on moon is 1/6 th of weight on earth
1/6*49
8.166 Newton..
Answer:
a) 2nd case rate of rotation gives the greater speed for the ball
b) 1534.98 m/s^2
c) 1515.04 m/s^2
Explanation:
(a) v = ωR
when R = 0.60, ω = 8.05×2π
v = 0.60×8.05×2π = 30.34 m/s
Now in 2nd case
when R = 0.90, ω = 6.53×2π
v = 0.90×6.53×2π = 36.92 m/s
6.35 rev/s gives greater speed for the ball.
(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2
(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2
According to Newton's first law of motion, it is the natural tendency of all moving objects to continue in motion in the same direction that they are moving ... unless some form of unbalanced force acts upon the object to deviate its motion from its straight-line path.
Hope this helped, have a great day!