Someone help me please i need to finish this
2 answers:
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Answer:
τ =9.41 * 10⁻⁴ N*m
Explanation:
Kinematics of the CD
The CD rotates with constant angular acceleration and its angular acceleration is calculated as follows:
Formula (1)
Where:
α : angular acceleration. (rad/s²)
ωf: final angular velocity (rad/s)
ωi : initial angular velocity (rad/s)
t = time interval (s)
Data
ωf= 20 rad/s
ωi =0
t = 0.65 s.
Calculating of the angular acceleration of the CD
We replace data in the formula (1)
α = 30.77 rad/s²
Newton's second law in rotation:
F = ma has the equivalent for rotation:
τ = I * α Formula (2)
where:
τ : It is the net torque applied to the body. (N*m)
I : it is the moment of inertia of the body with respect to the axis of rotation (kg*m²)
α : It is angular acceleration. (rad/s²)
Calculating of the moment of inertia of the CD
The moment of inertia of a disk with respect to an axis perpendicular to the plane and passing through its center is calculated by the following formula:
I = (1/2) M*R² Formula (3)
Data
M= 17 g = 17/1000 kg = 0.017 kg : CD mass
R= 6.0 cm= 6/100 m = 0.06 m : CD radius
We replace data in the formula (3) :
I = (1/2) ( 0.017 kg)*(0.06 m)² = 3.06 * 10⁻⁵ kg*m²
Calculating of the net torque acting on the CD
Data
α = 30.77 rad/s²
I = 3.06 * 10⁻⁵ kg*m²
We replace data in the formula (2) :
τ = I * α
τ =( 3.06 * 10⁻⁵ kg*m²) * (30.77 rad/s²)
τ =9.41 * 10⁻⁴ N*m
Answer:
Explanation:
The equations of kinematics will be used to solve this question:
At its maximum height, the projectile has zero velocity in the y-direction. But its velocity in the x-direction is unaffected.
First, let's apply the above equations to the x-direction.
There is no acceleration in the x-direction. So, its velocity in the x-direction is constant during the motion.
Therefore, the velocity vector of the projectile is
The speed of the projectile is the same.
The acceleration vector is constant during the motion and equal to the gravitational acceleration, which is -9.8 downwards.