What is the passenger's apparent weight at t=1.0s?

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight =  m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

Apparent Weight = 1035 N

For downward motion:

Apparent Weight =  m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

Apparent Weight = 435 N

reserve is the answer

the skeletons of most mammals reach a certain size and then stop growing. however, many animals, including some mammals, keep growing throughout their lives. kangaroos, for example, just keep growing and growing until they die. also, the longer an animal lives the greater the chance it might catch some deadly disease.