10.4 N
Given
m = 1.10 kg
θ = 15.0°
g = 9.81 m/s2
Solution
Fnet, y = ΣF y = Fn − Fg, y = 0
Fn = Fg, y = Fg
cosθ = mgcosθ
Fn = (1.10 kg)(9.81 m/s2
)(cos15.0°) = 10.4 N
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:

where q is the charge and

is the potential difference.
In our problem, the work done is W=39 J while the potential difference of the battery is

, so we can find the charge transferred by the battery:
Answer:
Explanation:
28 / 70 = 0.3857142... = 0.39 hr
280 / 100 = 2.8 hrs.
(100 - 0) / 10 = 10 m/s²
(60 - 20) / 4 = 10 m/s²
A. True. You can use displacement to determine the volume of solids and liquids.
W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.
735 N = m* 9.8
735/9.8 = m
75 = m
Mass , m = 75 kg. B.