Answer:
(A⃗ ×B⃗ )⋅C⃗ = 69.868
Explanation:
We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment
Answer:
A jet plane flying straight and at level at constant speed
Explanation:
The<em> inertial frame </em>of reference is a frame of reference in which all <em>Newton law is valid</em> ie Newton second law of motion and therefore newton first law of motion holds good. <em>The frame of reference does not accelerate.</em>
All the object that is in the frame of reference are at rest or moving with constant rectilinear motion with constant velocity unless acted upon by any force.
Answer:
a.52.9 km/h
b.90 km
Explanation:
We are given that






Time spend on eating lunch and buying ga=15 min.
a.Total time=30+12+45+15=102 minute=
1 hour=60 minutes
Distance=



Total distance=
Average speed=
Using the formula
Average speed=
b.Total distance between the initial and final city lies along the route=90 km
AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)