Answer:
F = Gm1m2/r^2 where G = 6.67x10^-11, m1 =1300, m2 = 7800, r = 0.23m
F = 6.67x10^-11 *1300*7800/(0.23)^2 = 0.0127852N
Explanation:
<span>The answer to your question is choice: D</span>
if we are walking on a perfectly smooth ground which has no friction our force would simply cancel out the force reverted by the ground and we would fall.
We need it to help push out feet off the ground
Hope those helps :)
Answer:
The speed of the particle is 2.86 m/s
Explanation:
Given;
radius of the circular path, r = 2.0 m
tangential acceleration, = 4.4 m/s²
total magnitude of the acceleration, a = 6.0 m/s²
Total acceleration is the vector sum of tangential acceleration and radial acceleration
where;
is the radial acceleration
The radial acceleration relates to speed of particle in the following equations;
where;
v is the speed of the particle
Therefore, the speed of the particle is 2.86 m/s
Answer:
7.85 m/s^2
Explanation:
linear or tangential acceleration= dv/dt
⇒
=0.83 m/s^2
radial acceleration is given by =
⇒
= 7.81 m/s^2
total acceleration
putting values we get
= 7.85 m/s^2