What do safeguarding devices do to protect the worker? (Select 3)
A) Detects operator contact with the machines point of operation

B) Prevents operator contact with the machines point of operation

C) Stops hazardous action if a workers body part comes to close

D) Acts as a barrier between the worker and the machine.

(A) Detects operator contact with the machines point of operation

(C) Stops hazardous action if a workers body part comes too close

(D) Acts as a barrier between the worker and the machine.

Explanation:

(A) Detects operator contact with the machines point of operation: This is a necessary point for the Safeguarding device because when it will detect it can make sure that whatever the person is working on, is safe or not. Therefore the safeguarding device should be detecting operator's contact but not interrupting unnecessarily.

(C) Stops hazardous action if a workers body part comes too close: Safeguarding device must protect the worker body part from that comes too close to any harmful or hazardous machines, like a safeguarding device should prevent the operator from going too close to the crusher machine as it can lead to injuries.

(D) Acts as a barrier between the worker and the machine: The safeguarding device should be acting as a barrier between the operator and the machine as it will prevent any injuries to the operator. The device should not be preventing the worker to work but should act a protective shield between the machines and the operator.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied