What Are the type's of Tidal turbines?
2 answers:
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Answer: = 16.49N
Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,
= horizontal component of the weight = mgsinФ
= vertical component of weight = mgcosФ
Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.
by using newton's law of motion, we have that
mgsinФ - = ma
where m = mass of object=5kg
a = acceleration= unknown
Ф = angle of inclination = 37°
g = acceleration due to gravity = 9.8
= frictional force = unknown
we need to first get the acceleration before the frictional force which is gotten by using the equation below
where v = final velocity = 2m/s
u = initial velocity = 0m/s (because the object started from rest)
a= unknown
S= distance covered = length of plane = 5m
we slot in a into the equation below to get frictional force
mgsinФ - = ma
3 * 9.8 * sin 37 - = 3* 0.4
17.9633 - = 1.2
= 17.9633 - 1.2
= 16.49N
Answer:
a) 20 rad/s
b) 6 m/s
Explanation:
b) Force acting on the wheel is 200 N
mass of the wheel is 100 kg
From Newton's second law of motion, F = m × a
Where F is the net force acting on the body
m is mass of the body
a is the acceleration of the body
By substituting the values we get, a = 2 m/s²
As acceleration is constant, we can use the below formula for calculating the final velocity of the object
v = u + a × t
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
u = 0 (∵ it starts from rest)
By substituting the values we get
v = 0 + 2 × 3 = 6 m/s
∴ Speed of center of mass after 3 seconds = 6 m/s
a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel
∴ Radius of the wheel = 300 mm
Torque acting on the wheel about axis of rotation = 300 mm × 200 N =
60 N·m
Torque = (Moment of inertia) × (angular acceleration)
Assuming that the mass of spokes of the wheel to be negligible,
Moment of inertia of the wheel about axis of rotation = 100 × 300² × = 9 kg·m²
Then,
60 = 9 × (angular acceleration)
∴ angular acceleration ≈ 6·67 rad/s²
As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed
=
+ α × t
Where
is the final angular velocity
is the initial angular velocity
α is the angular acceleration
t is the time taken
is 0 (∵ initially it starts from rest)
By substituting the values we get
= 6·67 × 3 = 20 rad/s
∴ Angular velocity of the wheel after three seconds = 20 rad/s