Answer:
Tension in the cable is T = 16653.32 N
Explanation:
Give data:
Cross section Area A = 1.3 m^2
Drag coefficient CD = 1.2
Velocity V = 4.3 m/s
Angle made by cable with horizontal =30 degree
Density
Drag force FD is given as
Drag force = 14422.2 N acting opposite to the motion
As cable made angle of 30 degree with horizontal thus horizontal component is take into action to calculate drag force
TCos30 = F_D
T = 16653.32 N
Answer:
The impulse transferred to the nail is 0.01 kg*m/s.
Explanation:
The impulse (J) transferred to the nail can be found using the following equation:
Where:
: is the final momentum
: is the initial momentum
The initial momentum is given by:
Where 1 is for the hammer and 2 is for the nail.
Since the hammer is moving down (in the negative direction):
And because the nail is not moving:
Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:
Since the hammer and the nail are moving in the negative direction:
=
= -9.7 m/s
Finally, the impulse is:
Therefore, the impulse transferred to the nail is 0.01 kg*m/s.
I hope it helps you!
Answer:
A) attached file
B) attached file
C) attached file
D) Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Explanation:
Given that the lifts are analogous to batteries, and the runs are analogous to resistors.
So from all the figures. The resistors represent the runs while the lift represents the battery.
Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Please find the attached file for the sketch
