A 1.00-kilogram sphere M, from point P, is lifted to a height h. The sphere is released and passes through the lowest point in ts swing at a speed of 10.0 meters per second. Dlogloct friction.]
A 1.00-kilogram sphere M, from point P, is lifted to a height h. The sphere is released and passes through the lowest point in ts swing at a speed of 10.0 meters per second. Dlogloct friction.]
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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please help with questions 94,95 and 96
![Base your answers to questions 94 through 97 on
the information and diagram below.
A 1.00-kilogram sphere M, suspended by a string
from point P, is lifted to a height h. The sphere is
released and passes through the lowest point in
its swing at a speed of 10.0 meters per second.
[Neglect friction.]
M.
10.0 m
1.00 kg
94. Calculate the height from which the sphere was
released.
95. Calculate the magnitude of the centripetal force
on the sphere as it passes through the lowest
point in its swing.
96. The magnitude of the centripetal force on the
sphere could be halved as it passes through the
lowest point in its swing by doubling the
(1) weight of the sphere, only
(2) length of the string, only
(3) height h and the weight of the sphere
(4) the length of the string and height h](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc69cdd85-cca7-4d84-b022-8594381da460%2Faa69e68f-573c-4cb9-b2ab-4af5f9909734%2Flgdjie_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Base your answers to questions 94 through 97 on
the information and diagram below.
A 1.00-kilogram sphere M, suspended by a string
from point P, is lifted to a height h. The sphere is
released and passes through the lowest point in
its swing at a speed of 10.0 meters per second.
[Neglect friction.]
M.
10.0 m
1.00 kg
94. Calculate the height from which the sphere was
released.
95. Calculate the magnitude of the centripetal force
on the sphere as it passes through the lowest
point in its swing.
96. The magnitude of the centripetal force on the
sphere could be halved as it passes through the
lowest point in its swing by doubling the
(1) weight of the sphere, only
(2) length of the string, only
(3) height h and the weight of the sphere
(4) the length of the string and height h
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