1.05 (b) The elevator continues upward at constant velocity for 8.2 s. (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upward "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. ZF, T-mg = 1 xxm.a (ii) Calculate the tension in the cable supporting the elevator. T= 18620 N S (iii) How high has the elevator moved during this time? Ay= 8.61 X m m (c) The elevator decelerates at a rate of 0.35 for 2.75 s. m S (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. ΣF₁=T-m.g= xxm.a (ii) Calculate the tension in the cable supporting the elevator. T= 17955 N (iii) ow high has the elevator moved during this time? Ay= 1.61
1.05 (b) The elevator continues upward at constant velocity for 8.2 s. (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upward "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. ZF, T-mg = 1 xxm.a (ii) Calculate the tension in the cable supporting the elevator. T= 18620 N S (iii) How high has the elevator moved during this time? Ay= 8.61 X m m (c) The elevator decelerates at a rate of 0.35 for 2.75 s. m S (i) Newton's Law in the y-direction can be written as: Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards "-1" if the acceleration is downwards, and pick "0" if there is no acceleration. ΣF₁=T-m.g= xxm.a (ii) Calculate the tension in the cable supporting the elevator. T= 17955 N (iii) ow high has the elevator moved during this time? Ay= 1.61
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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can someone help me answer these questions please?
![ΣF₁=T-m₂g = -1
Xxm,a
(ii) Calculate the tension in the cable supporting the elevator.
T= 19950
XVN
(iii) How high has the elevator moved during this time?
Ду= |0.7875
X√m
(iv) Calculate the velocity of the elevator after this time.
v(t = 1.5 s) = 1.05
=
(b) The elevator continues upward at constant velocity for 8.2 s.
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upward
"-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
ΣF, T-mg = 1
Xxm.a
(ii) Calculate the tension in the cable supporting the elevator.
T= 18620
√N
X
X √
m
S
(iii) How high has the elevator moved during this time?
Ay= 8.61
X√ m
What is
(c) The elevator decelerates at a rate of 0.35 for 2.75 s.
m
s²
2
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards
"-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
ΣF₁=T-m.g=
xxm.a
(ii) Calculate the tension in the cable supporting the elevator.
T= 17955
XVN
X
(iii) How high has the elevator moved during this time?
Ay= 1.61
m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92d13e7a-55ed-4295-b91a-cbfac319b8b3%2F8c20f381-1da3-4daf-862a-9b07a02a7f37%2Ffu9cqnp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:ΣF₁=T-m₂g = -1
Xxm,a
(ii) Calculate the tension in the cable supporting the elevator.
T= 19950
XVN
(iii) How high has the elevator moved during this time?
Ду= |0.7875
X√m
(iv) Calculate the velocity of the elevator after this time.
v(t = 1.5 s) = 1.05
=
(b) The elevator continues upward at constant velocity for 8.2 s.
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upward
"-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
ΣF, T-mg = 1
Xxm.a
(ii) Calculate the tension in the cable supporting the elevator.
T= 18620
√N
X
X √
m
S
(iii) How high has the elevator moved during this time?
Ay= 8.61
X√ m
What is
(c) The elevator decelerates at a rate of 0.35 for 2.75 s.
m
s²
2
(i) Newton's Law in the y-direction can be written as:
Instruction: If a is the magnitude of the acceleration, pick "1" if the acceleration is upwards
"-1" if the acceleration is downwards, and pick "0" if there is no acceleration.
ΣF₁=T-m.g=
xxm.a
(ii) Calculate the tension in the cable supporting the elevator.
T= 17955
XVN
X
(iii) How high has the elevator moved during this time?
Ay= 1.61
m
![1. An elevator shown below filled with passengers has a mass of 1700 kg. The elevator does motions (a)
through (c) in succession.
T
y
L.
X
I
meg
For each of the parts below draw a free body diagram of the elevator in your notebook for each of the
parts (a) to (c). Draw the acceleration and velocity vectors in the boxes.
m
E
(a) The elevator accelerates upward from rest at a rate of 0.75
2
S
a
V
for 1.5 s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F92d13e7a-55ed-4295-b91a-cbfac319b8b3%2F8c20f381-1da3-4daf-862a-9b07a02a7f37%2Fswyuhde_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. An elevator shown below filled with passengers has a mass of 1700 kg. The elevator does motions (a)
through (c) in succession.
T
y
L.
X
I
meg
For each of the parts below draw a free body diagram of the elevator in your notebook for each of the
parts (a) to (c). Draw the acceleration and velocity vectors in the boxes.
m
E
(a) The elevator accelerates upward from rest at a rate of 0.75
2
S
a
V
for 1.5 s.
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