Answer: a) 361.23m
b) 47.38m
Explanation:
Friction is opposite in direction to an applied force
Find the attached file for the solution
(b) The distance traveled with the friction of rain-free conditions is 47.38 m.
<em>D. Less luminous than those on the right []</em>
Answer:
r = 5.08 m
Explanation:
The electric force of attraction or repulsion is given by :
We need to find how far above the electron would the proton have to be if you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above it.
So, the force from the proton is balanced by the mass of the electron.
r is distance
So, proton have to be at a distance of 5.08 meters above the electron.
in the service producing sector
Answer:
6957.04N
Explanation:
Using
vf2=vi2+2ad
But vf = 0 .
So convert 50km/hr to m/s, and you need to convert 61 cmto m
(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s
61cm * (1m/100cm) = .61m
So n
0 = (13.9m/s)^2 + 2a(.61m)
a = 158.11m/s^2
So
using F = ma
F = 44kg(158.11m/s^2) = 6957.04N