At the center of a 40.1 m diameter circular (frictionless) ice rink, a 72.8 kg skater travelling north at 1.14 m/s collides with and holds onto a 61.2 kg skater who had been heading west at 4.74 m/s. How long will it take them to glide to the edge of the rink? Ans: 8.90 s b) Where will they reach it? Give your answer as an angle north of west. (Attached below is what I attempted for b), but apparently it is wrong)
At the center of a 40.1 m diameter circular (frictionless) ice rink, a 72.8 kg skater travelling north at 1.14 m/s collides with and holds onto a 61.2 kg skater who had been heading west at 4.74 m/s. How long will it take them to glide to the edge of the rink? Ans: 8.90 s b) Where will they reach it? Give your answer as an angle north of west. (Attached below is what I attempted for b), but apparently it is wrong)
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
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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(7a) At the center of a 40.1 m diameter circular (frictionless) ice rink, a 72.8 kg skater travelling north at 1.14 m/s collides with and holds onto a 61.2 kg skater who had been heading west at 4.74 m/s. How long will it take them to glide to the edge of the rink? Ans: 8.90 s
b) Where will they reach it? Give your answer as an angle north of west. (Attached below is what I attempted for b), but apparently it is wrong)
![4:16 PM Tue Oct 31
QPHYSICS 1A03 Notebook
5
<
Cost:
V
vor vcaso
using
T
2.16m15
Î
tano
In west
b) tano = 0.619
-2.16
direction
In north
direction
-0.2865
0 16.0⁰
:
Home Insert Draw
!
Patter
P before collision =
before collision
m₂v₂i = (m₂ + m₂) v coso
2
(61.2) (4.74): 134 1080
vioso: -2.16 mls
after collision
View
vsing=
P after collision
m₂ V₂i
= (m₁+m₂) using
:
(72.8) (1.14) : (72.8 +61.2) SING
V sine = 82.992
134
0.619m 15
+
Uf = √ (2.16)² + (0.619) ²
vi: Vf: 2.25 m²s
+-
X÷
4x= vit +!
20.05
2.25t
t = 8.91 S
8
0
स्वहर
69%
K
71](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F999df56d-3a08-4973-bf69-06d978976eb4%2Fa2b42d0f-fee2-41e0-9aa9-751f4b7b3eae%2F1qt6inna_processed.png&w=3840&q=75)
Transcribed Image Text:4:16 PM Tue Oct 31
QPHYSICS 1A03 Notebook
5
<
Cost:
V
vor vcaso
using
T
2.16m15
Î
tano
In west
b) tano = 0.619
-2.16
direction
In north
direction
-0.2865
0 16.0⁰
:
Home Insert Draw
!
Patter
P before collision =
before collision
m₂v₂i = (m₂ + m₂) v coso
2
(61.2) (4.74): 134 1080
vioso: -2.16 mls
after collision
View
vsing=
P after collision
m₂ V₂i
= (m₁+m₂) using
:
(72.8) (1.14) : (72.8 +61.2) SING
V sine = 82.992
134
0.619m 15
+
Uf = √ (2.16)² + (0.619) ²
vi: Vf: 2.25 m²s
+-
X÷
4x= vit +!
20.05
2.25t
t = 8.91 S
8
0
स्वहर
69%
K
71
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