If she is moving in a straight line at a constant speed,
then her acceleration is zero.
<span>The 16.2cm by 22.9cm paper has an area = 370.98 square cm = 0.037098 square meters. And the atmospheric pressure at sea level is 101,325 Newtons per square meter. So if multiply the quantities: 101,325 N/m^2 and 0.037098 m^2 we get an answer of 3758.95 N.
* If we respect the fact that 22.9cm and 16.2 cm only have three significant digits then our answer would be 3760N.</span>
consider the motion of the mass parallel to the incline
v₀ = initial velocity at the bottom of incline = 0 m/s
v = final velocity at the top of incline = 8.00 m/s
a = acceleration
d = displacement = L = length of incline = 15 m
using the equation
v² = v²₀ + 2 a d
8² = 0² + 2 a (15)
64 = 30 a
a = 64/30
a = 2.13 m/s²
F = applied force
from the force diagram, perpendicular to incline , force equation is given as
N = mg Cos30
μ = Coefficient of friction = 0.426
frictional force acting on the mass is given as
f = μ N
f = μ mg Cos30
parallel to incline , force equation is given as
F - f - mg Sin30 = ma
F - μ mg Cos30 - mg Sin30 = ma
inserting the values
F - (0.426 x 40 x 9.8) Cos30 - (40 x 9.8) Sin30 = 40 (2.13)
F = 425.82 N
Answer:
Dy = - 0.0789 [m]
Explanation:
The vertical component of the vector can be determined with the sine of the angle.
Dy = 0.250*sin(18.4)
Dy = 0.0789 [m]
As the y-component is pointing downwards the component is negative.
Dy = - 0.0789 [m]