Could someone try helping me solve those 2 questions or write me a formula with equaion what to do , thanks for every bit of help
A 600kg elevator accelerates upward from rest for 10 seconds through 40.0meters. Find the cable tension
a) 5130 N
b) 5400 N
c) 6360 N
d) 6630 N
A car traveling 9 m/s slams on its brakes, coming to a stop in 2 seconds. Find how far the car traveling coming to a halt
a) 88.2 m
b) 44.1 m
c) 27.0 m
d) 9.0 m
Tension on the elevator cable= mg+ma
vf^2=vo^2+2*Acceleration*distance.
find distance when acceleration is -9m/s/3s= -3m/s^3
Sure, I can help you solve these questions.
1) To find the cable tension of the elevator, we can use the equation F = ma, where F is the force, m is the mass, and a is the acceleration. We can plug in the given values and solve for F.
Given:
Mass of the elevator (m) = 600 kg
Acceleration (a) = (change in velocity) / (change in time) = (final velocity - initial velocity) / time = (0 m/s - 0 m/s) / 10 s = 0 m/s^2 (since the elevator starts from rest)
Using the equation F = ma and plugging in the values:
F = 600 kg * 0 m/s^2 = 0 N
Therefore, the cable tension is 0 N. None of the given options (a, b, c, or d) are correct. Please double-check the question or the given options.
2) To find the distance the car travels while coming to a halt, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration.
Given:
Initial velocity (u) = 9 m/s
Time (t) = 2 s
Acceleration (a) = (change in velocity) / (change in time) = (final velocity - initial velocity) / time = (0 m/s - 9 m/s) / 2 s = -4.5 m/s^2 (since the car decelerates)
Using the equation s = ut + (1/2)at^2 and plugging in the values:
s = 9 m/s * 2 s + (1/2) * (-4.5 m/s^2) * (2 s)^2
s = 18 m + (-4.5 m/s^2) * 4 s^2
s = 18 m - 9 m
s = 9 m
Therefore, the car travels a distance of 9 meters while coming to a halt. The correct answer is option d) 9.0 m.