Answer:
ω = 0.0347 rad/s²
a ≅ 1.07 m/s²
Explanation:
Given that:
mass of the model airplane = 0.741 kg
radius of the wire = 30.9 m
Force = 0.795 N
The torque produced by the net thrust about the center of the circle can be calculated as:
where;
F represent the magnitude of the thrust
r represent the radius of the wire
Since we have our parameters in set, the next thing to do is to replace it into the above formula;
So;
(b)
Find the angular acceleration of the airplane when it is in level flight rad/s²
where;
I = moment of inertia
ω = angular acceleration
The moment of inertia (I) can also be illustrated as:
I = ( 0.741) × (30.9)²
I = 0.741 × 954.81
I = 707.51 Kg.m²
Making angular acceleration the subject of the formula; we have;
ω = 
ω = 0.0347 rad/s²
(c)
Find the linear acceleration of the airplane tangent to its flight path.m/s²
the linear acceleration (a) can be given as:
a = ωr
a = 0.0347 × 30.9
a = 1.07223 m/s²
a ≅ 1.07 m/s²
Answer:
option A is the correct answer
Answer:
The time taken for the paint ball to hit the ground is 
The distance of the landing point from the tower is 
Explanation:
From the question we are told that
The height of the tower is 
The speed of the paintball in the horizontal direction is 
Generally from kinematic equation we have that
Here u is the initial velocity of the paintball in the vertical direction and the value is 0 m/s , this because the ball was fired horizontally
a is equivalent to 
t is the time taken for the paintball to hit the ground
So
=>
Generally the distance of its landing position from the tower is
=> 
=>
Answer:
r = 2.55 m
Explanation:
Given that,
First charge, q₁ = 32.2 μC
Second charge, q₂ = + 12.3 μC
The force between charges, F = 0.544 N
We need to find the distance between charges. The force between two charges is given by the formula as follows :
So, the charges are 2.55 m apart.
Answer: It is given that A body weighs 20gf in air and 18. 0gf in water. Hence, the answer X-3 = 7.
