Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
I think that the answer is friction
As per the question Bob drops the bag full with feathers from the top of the building.
The mass of the bag(m)= 1.0 lb
Let the air resistance is neglected.As the bag is under free fall ,hence the only force that acts on the bag is the force of gravity which is in vertical downward direction.
Here the acceleration produced on bag due to the free fall will be nothing else except the acceleration due to gravity i.e g =9.8 m/s^2
Here we are asked to calculate the distance travelled by the bag at the instant 1.5 s
Hence time t= 1.5 s
From equation of kinematics we know that -
S=ut + 0.5at^2 [ here S is the distance travelled]
For motion under free fall initial velocity (u)=0.
Hence S= 0×1.5+{0.5×(-9.8)×(1.5)^2}
⇒ -S =0-11.025 m
⇒ S= 11.025 m
=11 m
Here the negative sign is taken only due to the vertical downward motion of the body .we may take is positive depending on our frame of reference .
Hence the correct option is B.
Answer:
Explanation:
Given that:
the initial angular velocity 
angular acceleration
= 4.44 rad/s²
Using the formula:

Making t the subject of the formula:

where;

∴

t = 0.345 s
b)
Using the formula:

here;
= angular displacement
∴



Recall that:
2π rad = 1 revolution
Then;
0.264 rad = (x) revolution

x = 0.042 revolutions
c)
Here; force = 270 N
radius = 1.20 m
The torque = F * r

However;
From the moment of inertia;

given that;
I = 84.4 kg.m²

For re-tardation; 
Using the equation



t = 0.398s
The required time it takes= 0.398s