a 1.0 N-weight is hanging at rest as shown. In each case, write in the magnitude of the unknown force.

a) chain with a weight of 2.5 N has a 1.0 N hanging weight
b) chain with a weight of .5 N has a 1.0 N hanging weight
c) stretched elastic band with negligible weight has a 1.0 N hanging weight

an insect inside a bus flies from the back toward the front at 2 m/s the bus is moving in a straight line at 20 m/s what is the speed of the insect relative to earth's surface

THIS IS NO HELP ???

a) F(unknown)= 2.5N + 1.0N

b) F(unknown)= .5N +1.0N
c) F(unknown)= 0N + 1.0N

a) The unknown force is the difference between the weight of the chain (2.5 N) and the hanging weight (1.0 N). So, the magnitude of the unknown force is 2.5 N - 1.0 N = 1.5 N. But let's add a bit of humor to it... The unknown force says, "Just hangin' around, but with 1.5 N of magnitude!"

b) In this case, the weight of the chain (0.5 N) is smaller than the hanging weight (1.0 N). So, that means the unknown force is pulling up harder than the weight of the chain can handle. It's saying, "Hey there, Mr. Chain, I need more oomph! The magnitude of the unknown force is 1.0 N - 0.5 N = 0.5 N."

c) Ah, the stretched elastic band. It's like a superhero, way more flexible than a rigid chain. Since the elastic band has negligible weight, we can consider the entire 1.0 N weight as the unknown force pulling it down. So, the magnitude of the unknown force is simply 1.0 N. The elastic band stretches a bit and says, "Don't worry, I got this! With 1.0 N of magnitude, I can handle anything!"

To determine the magnitude of the unknown force in each case, we need to consider the equilibrium condition for the hanging weight. When an object is at rest, the sum of forces acting on it must be zero. In other words, the weight of the hanging object must be balanced by an equal and opposite force.

a) In this case, a chain with a weight of 2.5 N is attached to a 1.0 N hanging weight. To find the magnitude of the unknown force acting on the chain, we subtract the weight of the hanging object from the weight of the chain: 2.5 N - 1.0 N = 1.5 N. Therefore, the magnitude of the unknown force is 1.5 N.

b) Here, a chain with a weight of 0.5 N is attached to a 1.0 N hanging weight. Similar to the previous case, we subtract the hanging weight from the weight of the chain: 0.5 N - 1.0 N = -0.5 N. The negative sign indicates that the unknown force is acting in the opposite direction, pulling the chain upward. Therefore, the magnitude of the unknown force is 0.5 N.

c) In this scenario, a stretched elastic band with negligible weight is connected to a 1.0 N hanging weight. Since the elastic band is stretched, it exerts an upward force on the hanging weight to counteract its downward weight. The magnitude of the unknown force in this case would be equal to the weight of the hanging object, which is 1.0 N.

In summary:
a) The magnitude of the unknown force is 1.5 N.
b) The magnitude of the unknown force is 0.5 N, acting in the opposite direction.
c) The magnitude of the unknown force is 1.0 N.