Y Part C What is the velocity of the blocks 4.50 s after the blocks have started moving? Assume that the rope joining the two blocks in long enough so that, at this time, block A is on the frictionless surface while block 8 is still on the rough surface (Figure 2) Express your answer numerically in meters per second to three significant figures. View Available Hint(s) IVE ΑΣΦ 12= 20.923 vec 1 CE? m/s
Y Part C What is the velocity of the blocks 4.50 s after the blocks have started moving? Assume that the rope joining the two blocks in long enough so that, at this time, block A is on the frictionless surface while block 8 is still on the rough surface (Figure 2) Express your answer numerically in meters per second to three significant figures. View Available Hint(s) IVE ΑΣΦ 12= 20.923 vec 1 CE? m/s
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
Solve Part C
![Part A
Two blocks, each of mass m = 120 kg, are connected by a massless rope and start sliding down a slope of incline 0 = 39.0° att= 0.000s. The slope's top portion is a rough surface whose coefficient of kinetic
friction is 0.200. At a distance d = 2.40 m from block A's initial position the slope becomes frictionless. (Figure 1)What is the velocity of the blocks when block A reaches this frictional transition point? Assume
that the blocks' width is negligible
Express your answer numerically in meters per second to four significant figures.
►View Available Hint(s)
v=4.724 m/s
All attempts used; correct answer displayed
Part B
Previous Answers
How long does it take block A to reach the transition point?
Express your answer numerically in seconds to four significant figures.
View Available Hint(s)
t₁ = 1.016 s
✓
Part C
Previous Answers
Correct
Correct answer is shown Your answer 1.02 s was either rounded differently or used a different number of significant figures than required for this part.
Important: If you use this answer in later parts, use the full unrounded value in your calculations.
You can verify your result by computing this time using the kinematic equations that describe the motion of a particle with constant acceleration t₁ = √√/2d/a a t₁=v/a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00030fb4-2bf0-41c7-8151-28e1f2b4f688%2Fbfa4d137-a20e-45da-a340-ddd66e36637a%2Fq7nnm4o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Part A
Two blocks, each of mass m = 120 kg, are connected by a massless rope and start sliding down a slope of incline 0 = 39.0° att= 0.000s. The slope's top portion is a rough surface whose coefficient of kinetic
friction is 0.200. At a distance d = 2.40 m from block A's initial position the slope becomes frictionless. (Figure 1)What is the velocity of the blocks when block A reaches this frictional transition point? Assume
that the blocks' width is negligible
Express your answer numerically in meters per second to four significant figures.
►View Available Hint(s)
v=4.724 m/s
All attempts used; correct answer displayed
Part B
Previous Answers
How long does it take block A to reach the transition point?
Express your answer numerically in seconds to four significant figures.
View Available Hint(s)
t₁ = 1.016 s
✓
Part C
Previous Answers
Correct
Correct answer is shown Your answer 1.02 s was either rounded differently or used a different number of significant figures than required for this part.
Important: If you use this answer in later parts, use the full unrounded value in your calculations.
You can verify your result by computing this time using the kinematic equations that describe the motion of a particle with constant acceleration t₁ = √√/2d/a a t₁=v/a
![How long does it take block A to reach the transition point?
Express your answer numerically in seconds to four significant figures.
View Available Hint(s)
t₁ = 1.016 s
Previous Answers
✓Correct
Correct answer is shown. Your answer 1.02 s was either rounded differently or used a different number of significant figures than required for this part.
Important: If you use this answer in later parts, use the full unrounded value in your calculations.
You can verify your result by computing this time using the kinematic equations that describe the motion of a particle with constant acceleration: ₁
Part C
What is the velocity of the blocks 4.50 s after the blocks have started moving? Assume that the rope joining the two blocks in long enough so that, at this time, block A is on the frictionless surface while block B is still
on the rough surface. (Figure 2)
Express your answer numerically in meters per second to three significant figures.
UIM
▸ View Available Hint(s)
ΠΠ ΑΣΦ
₂= 20.923
11 vec 3
Submit Previous Answers
CE ?
X Incorrect; Try Again; 5 attempts remaining
√2d/a or t₁ = v/a
m/s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F00030fb4-2bf0-41c7-8151-28e1f2b4f688%2Fbfa4d137-a20e-45da-a340-ddd66e36637a%2Fohub3ep_processed.jpeg&w=3840&q=75)
Transcribed Image Text:How long does it take block A to reach the transition point?
Express your answer numerically in seconds to four significant figures.
View Available Hint(s)
t₁ = 1.016 s
Previous Answers
✓Correct
Correct answer is shown. Your answer 1.02 s was either rounded differently or used a different number of significant figures than required for this part.
Important: If you use this answer in later parts, use the full unrounded value in your calculations.
You can verify your result by computing this time using the kinematic equations that describe the motion of a particle with constant acceleration: ₁
Part C
What is the velocity of the blocks 4.50 s after the blocks have started moving? Assume that the rope joining the two blocks in long enough so that, at this time, block A is on the frictionless surface while block B is still
on the rough surface. (Figure 2)
Express your answer numerically in meters per second to three significant figures.
UIM
▸ View Available Hint(s)
ΠΠ ΑΣΦ
₂= 20.923
11 vec 3
Submit Previous Answers
CE ?
X Incorrect; Try Again; 5 attempts remaining
√2d/a or t₁ = v/a
m/s
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