r = radius of the circle traveled by the particle = 76 cm = 0.76 m
T = time period of revolution for the particle = 4.5 s
w = angular velocity of the particle
angular velocity of the particle is given as
w = 2π/T
inserting the values
w = 2 (3.14)/4.5
w = 1.4 rad/s
a = centripetal acceleration of the particle in the circle
centripetal acceleration is given as
a = r w²
inserting the values
a = (0.76) (1.4)²
a = 1.5 m/s²
Answer:
36km
Explanation:
Im pretty sure displacment is the start and finish in a straight line
Answer:
a) a = 4.9 m / s², N = 16.97 N and b) F = 9.8 N
Explanation:
a) For this exercise we will use Newton's second law, we write a reference system with the x axis parallel to the plane, see attached, in this system the only force we have to break down is weight, let's use trigonometry
sin 30 = Wx / W
cos 30 = Wy / W
Wx = W sin30
Wy = W cos 30
Let's write the equations on each axis
X axis
Wx = ma
Y Axis
N- Wy = 0
N = Wy = mg cos 30
N = 2.0 9.8 cos 30
N = 16.97 N
We calculate the acceleration
a = Wx / m
a = mg sin 30 / m
a = g sin 30
a =9.8 sin 30
a = 4.9 m / s²
b) For the block to move with constant speed, the acceleration must be zero, so the force applied must be equal to the weight component
F -Wx = 0
F = Wx
F = m g sin 30
F = 2.0 9.8 sin 30
F = 9.8 N
