Answer:
Explanation:
<u>Horizontal Launch</u>
When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
vx=v
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
The horizontal component of the velocity is always the same:
The vertical component at t=5.5 s is:
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:
The angular displacement is given as the form:
In the equlibrium we have to 
and in the given position we have to
Derived the expression we will have the equivalent to angular velocity
Replacing,
Finally
Therefore the maximum angular displacement is 9.848°
Answer:
Explanation:
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