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:
Explanation:
Let T be the tension
For linear motion of hoop downwards
mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .
For rotational motion of hoop
Torque by tension
T x R , R is radius of hoop.
Angular acceleration be α,
Linear acceleration a = α R
So TR = I α
= I a / R
a = TR² / I
Putting this value in earlier relation
mg -T = m TR² / I
mg = T ( 1 + m R² / I )
T = mg / ( 1 + m R² / I )
mg / ( 1 + R² / k² )
Tension is less than mg or weight because denominator of the expression is more than 1.
Answer:0.3
Explanation:
Given
velocity of car=15 m/s
truck brought to halt in a distance of 38 m
We know
Final velocity (v)=0
(deceleration)
Therefore minimum coefficient of friction \mu will be
here we have to calculate the net positive charge present on he surface of the conducting sphere.
as the sphere is a conducting one the,the charge will be stored on its surface.
though the charge is present at the surface,still it will behave just like the total charge is concentrated at the centre of the sphere.
the electric field at outside of the sphere is E=
as E= 
so V=
here V=27 volt,radius of sphere is 0.800 m and the point at which the potential is considered is at 1.20 m from th centre of the sphere.hence the point is at the outside of the sphere .
hence we have 27=
q=27×
coulomb
=3.6×
coulomb [ans]
