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
Answer:112.82 m/s
Explanation:
Given
range of arrow=68 m
as the arrow travels it acquire a vertical velocity 
-------1
Range is given by
R=ut
where u=initial velocity
substitute the value of t in eqn 1
--------2
and 
substitute it in 2
u=112.82 m/s
Answer:
The distance traveled by the ball is 8.5 m
Explanation:
Initial height of the ball, h₁ = 1.5 m above the ground
final height of the ball, h₂ = 5m
Upward distance = distance traveled by the ball from a height of 1.5m to 5m = 5m - 1.5m = 3.5 m
Downward distance = distance traveled by the ball from 5m height to the ground =5m - 0 = 5m
Total distance traveled = upward distance + downward distance
Total distance traveled = 3.5 m + 5m = 8.5 m
Therefore, the distance traveled by the ball is 8.5 m
Answer:
3.0 x10^-3 J
Explanation:
The potential energy of a spring is given by PE = (0.5)k*x^2
Where
K: Spring Constant = 60 N/m
x: displacement of the spring from its equilibrium position = 1cm = 0.01m
Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J
Answer: c
Explanation:
C Air is a compound of two or more components that keep their own identifying properties, while water is composed of mixtures that combine to form a compound.
