Answered

Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

It takes the elevator in a skyscraper 4.0 s to reach its cruising speed of 10 m/s. A 60 kg passenger gets aboard on the ground floor.
1. What is the passenger's apparent weight before the elevator starts moving?
2. What is the passenger's apparent weight whilethe elevator is speeding up?
3. What is the passenger's apparent weight afterthe elevator reaches its cruising speed?

Sagot :

Numenius

Answer:

1. 588 N

2. 738 N

3. 588 N

Explanation:

time, t = 4 s

initial velocity, u = 0

final velocity, v = 10 m/s

mass, m= 60 kg

1.

Weight of passenger before starts

W =m g = 60 x 9.8 = 588 N

2.

When the elevator is speeding up

v = u + a t

10 = 0 + a x 4

a = 2.5 m/s2

Now the weight is

W' = m (a + g) = 60 (9.8 + 2.5) = 738 N

3.

When he reaches the cruising speed, the weight is

W = 588 N

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.