For a rigid body in rotational motion, what can be stated about the angular velocity of all of the particles? The linear velocit
1 answer:
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There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
where
is the coefficient of friction, while N is the normal force. So we have:
since we know that F=300 N and
, we can find N, the magnitude of the normal force:
b) The problem is identical to that of the first part; however, this time the coefficienct of friction is
due to the presence of the oil. Therefore, we have:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
where
since we know that F=300 N and
b) The problem is identical to that of the first part; however, this time the coefficienct of friction is
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
