Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
Answer:
80% (Eighty percent)
Explanation:
The material has a refractive index (n) of 1.25
Speed of light in a vacuum (c) is 2.99792458 x 10⁸ m/s
We can find the speed of light in the material (v) using the relationship
n = c/v, similarly
v = c/n
therefore v = 2.99792458 x 10⁸ m/s ÷ (1.25) = 239 833 966 m/s
v = 239 833 966 m/s
Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as
(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%
Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)
Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is, 
Tangential speed of the skater is, 
Mass of the skater is, 
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:
Now, plug in the given values of the known quantities and solve for centripetal force, 
. This gives,
Therefore, the centripetal force acting on the skater is 48.32 N.
Answer:
Explanation:
mass of car, m = 1000 kg
initial velocity, u = 20 m/s
final velocity, v = 0 m/s
distance, s = 120 m
Let a be the acceleration of motion
use third equation of motion
v² = u² + 2 as
0 = 20 x 20 + 2 x a x 120
a = - 1.67 m/s²
Let F be the force
Force, F mass x acceleration
F = - 1000 x 1.67
F = - 1666.67 N
The direction of force is towards south and the magnitude of force is 1666.67 N.
