Answer:
the work done by the 30N force is 4156.92 J.
For this problem, they don´t ask you to determine the work of the total force applied in the block. They only want the work done for the force of 30N, with an angle of 30º respectively of the displacement and a traveled distance of 160m. So:
W=F·s·cos(α)=30N·160m·cos(30º)=4156.92J
Answer:
Velocity is the displacement (change In position of an an object it is similar to distance ) divided by the time taken. SI unit m/s (metre per second)
These are the Kepler's laws of planetary motion.
This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.
The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.
The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.
Answer:
148(m/s)
Explanation:
V_final = V_current + (acceleration) x (time)
= 4 + 80 x 1.8 = 148 (m/s)
Answer:
Explanation:
Given data:
weigh (head+arms + head) w_1 = 438 N
centre of gravity y_1= 1.28 m
weigh (upper leg) w_2 = 144 N
Center of gravity y_2 = 0.760 m
weigh ( lower leg + feet) = 87 N
centre of gravity = y_3 = 0.250 m
location of center of gravity 
