Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where 
is the initial velocity.
(a).
When the projectile hits the 50m mark, 
; therefore,
solving for 
we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed 
of
Answer:
The circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.
Explanation:
Since the magnetic field, B points in my direction and the current, I is moving in a clockwise direction, the current is always perpendicular to the magnetic field and will thus experience a constant force, F = BILsinФ where Ф is the angle between B and L.
Since the magnetic field is in my direction, it is perpendicular to the plane of the circular loop and thus perpendicular to L where L = length of circular loop. Thus Ф = 90° and F = BILsin90° = BIL
According to Fleming's left-hand rule, the fore finger representing the magnetic field, the middle finger represent in the current and the thumb representing the direction of force on the circular loop.
At each point on the circular loop, the force is always directed towards the center of the loop and thus tends to compress it.
<u>So, the circular loop experiences a constant force which is always directed towards the center of the loop and tends to compress it.</u>
Given Information:
Angular displacement = θ = 51 cm = 0.51 m
Radius = 1.8 cm = 0.018 m
Initial angular velocity = ω₁ = 0 m/s
Angular acceleration = α = 10 rad/s
²
Required Information:
Final angular velocity = ω₂ = ?
Answer:
Final angular velocity = ω₂ = 21.6 rad/s
Explanation:
We know from the equations of kinematics,
ω₂² = ω₁² + 2αθ
Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.
ω₂² = (0)² + 2αθ
ω₂² = 2αθ
ω₂ = √(2αθ)
We know that the relation between angular displacement and arc length is given by
s = rθ
θ = s/r
θ = 0.51/0.018
θ = 23.33 radians
finally, final angular velocity is
ω₂ = √(2αθ)
ω₂ = √(2*10*23.33)
ω₂ = 21.6 rad/s
Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.
Answer:
The angular speed of the sphere at the bottom of the hill is 31.39 rad/s.
Explanation:
It is given that,
Weight of the sphere, W = 240 N
Radius of the sphere, r = 0.2 m
Angle with the horizontal, 
We need to find the angular speed of the sphere at the bottom of the hill if it starts from rest.
As per the law of conservation of energy, the total energy at the top is equal to the energy at the bottom.
Gravitational energy = translational energy + rotational energy
So,
I is the moment of inertia of the sphere, 
Also, 
h is the height of the ramp, 
On solving the above equation we get :
So, the angular speed of the sphere at the bottom of the hill is 31.39 rad/s. Hence, this is the required solution.
