F=m*a
F=65 kg *9.8 m/s^2
F=637 N (Newtons) — this is the weight
Answer:
Option D: 1.5in in front of the target
Explanation:
The object distance is 
.
Because the surface is flat, the radius of curvature is infinity .
The incident index is 
and the transmitted index is 
.
The single interface equation is 
Substituting the quantities given in the problem,
The image distance is then 
Therefore, the coin falls 
in front of the target
Answer:
Explanation:
Frictional force acting on the child = μ mg cosθ
, μ is coefficient of kinetic friction , m is mass of child θ is inclination
work done by frictional force
μ mg cosθ x d , d is displacement on inclined plane
work done = .13 x 276 x cos34 x 5.9
= 175.5 J
This work will be converted into heat energy.
b ) Initial energy of child = mgh + 1/2 m v ² , h is height , v is initial velocity
= 276 x 5.9 sin34 + 1/2 x 276 / 9.8 x .518² [ mass m = 276 / g ]
= 910.59 + 3.77
= 914.36 J
loss of energy due to friction = 175.5
Net energy at the bottom
= 738.86 J
If v be the velocity at the bottom
1/2 m v² = 738 .86
.5 x (276 / 9.8) x v² = 738.86
v² = 52.47
v = 7.24 m /s .
Answer:
Market participation allows individuals to specialize and, with trade, ultimately consume more.
Answer:
Tt tT
tT tt
Explanation:
the rest I pretty sure don't understand
