Answer:
Moment of the force is 20 N-m.
Explanation:
Given:
Force exerted by the person is, 
Distance of application of force from the point about which moment is needed is, 
Now, we know that, moment of a force 'F' about a point at a perpendicular distance of 'd' from the same point is given as the product of the force and the perpendicular distance.
Therefore, the moment of the force about the end of the claw hammer is given as:
Hence, the moment of the force exerted by the person about the end of the claw hammer is 20 N-m.
Because they are different they all show different traits.
The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m
s = ut + 1 / 2 at²
s = Distance
u = Initial velocity
t = Time
a = Acceleration
Vertically,
s = 15.4 m
u = 0
a = 9.8 m / s²
15.4 = 0 + ( 1 / 2 * 9.8 * t² )
t² = 3.14
t = 1.77 s
Horizontally,
u = 3.01 m / s
a = 0 ( Since there is no external force )
s = ( 3.01 * 1.77 ) + 0
s = 5.34 m
Therefore, the distance travelled by the ball before hitting the ground is 5.34 m
To know more about distance travelled
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Collision domain is a portion in the network where there is a possibility of formation of packets. This occurs when two or more devices are able to send a packet to a single switch or port on the network that is shared, on the same time. It was noted that this collision domain reduces the efficiency of the network.
For this item, the first packet is the whole switch with the three devices. Next one would be first of the three devices that is connected to the other port. Similarly, the third one would be the second of the three devices that is also connected to the switch. Therefore, the answer is 3.
