Answer:
Part a)

Part b)

Part c)

Explanation:
As we know that acceleration is rate of change in velocity of the object
So here we know that


Part a)
differentiate x and y two times with respect to time to find the acceleration






Now the acceleration of the object is given as

at t= 1.1 s we have

now the net force of the object is given as



now magnitude of the force will be

Part b)
Direction of the force is given as



Part c)
For velocity of the particle we have




now at t = 1.1 s

now the direction of the velocity is given as



Answer:
the answer would be microwelds.
Answer:
The force is pull or push acting on the body which tends to change its state of rest or of motion is called force.
There are two types of force:
1.Contact force
2. Non-Contact
<h3><u>Answer;</u></h3>
Mechanical advantage
<h3><u>Explanation;</u></h3>
- Mechanical advantage is the ratio of force output from a machine divided by the force input into the machine.
- Mechanical advantage measures the machine's force-magnifying effect. It is an advantage gained by using simple machines to accomplish work with less effort.
- The formula is; M.A = output force/ Input force