Respuesta :
Answer:
A torque of 102.5375 Nm must be exerted by the fireman
Explanation:
Given:
The rate of water flow = 6.31 kg/s
The speed of nozzle = 12.5 m/s
Now, from the Newton's second law we have
The reaction force to water being redirected horizontally (F) = rate of change of water's momentum in the horizontal direction
thus we have,
F = 6.31 kg/s x 12.5m/s
or
F = 78.875 N
Now,
The torque (T) exerted by water force about the fireman's will be
T = (F x d)
or
T = 78.875 N x 1.30 m
T = 102.5375 Nm
hence,
A torque of 102.5375 Nm must be exerted by the fireman
The torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.
What is Newtons second law of motion?
Newtons second law of motion shows the relation between the force mass and acceleration of a body. It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
For the flow of water, the second law of motion can be given as,
[tex]F=Q\times v[/tex]
As he rate of water flow is 6.31 kg/s, and the nozzle speed is 12.5 m/s. Thus the force of this can be given as,
[tex]F=6.31\times12.5\\F=78.875\rm N[/tex]
The torque for the fireman exert on the hose is equal to the product of force applied and the distance traveled. Therefore the value of torque is,
[tex]\tau=78.875\times1.30\\\tau=102.5375\rm Nm[/tex]
Thus, the torque must the fireman exert on the hose toward a burning building is 102.5375 Nm.
Learn more about the Newtons second law of motion here;
https://brainly.com/question/25545050
