Answer:
3.53 m/s towards the left
Explanation:
= Mass of first glider = 400.5 g
= Mass of Second glider = 500 g
= Initial Velocity of first object = 2.3 m/s
= Initial Velocity of second object = -2.95 m/s
As momentum and Energy is conserved
From the two equations we get
The velocity of first glider after collision is 3.53 m/s towards the left
Two runners are not allowed to occupy the same base. If two runners are touching the same base, the lead runner is entitled to the base. Most coaches will teach their defensive players to tag both runners when they are occupying the same base.
So it is false
Answer:
See the answer below
Explanation:
<em>The best thing one can do in this case would be to return the microscope's objective to low power and then </em><em>re-center the specimen </em><em>before switching back to high-dry power.</em>
Most of the time, <u>what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage</u>. Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.
After doing the above and the specimen still does not come into focus, then the instructor can be called upon.
25Wwwwwwwww has a greater resists
Answer:
Charge enter a 0.100 mm length of the axon is
Explanation:
Electric field E at a point due to a point charge is given by
where is the constant =
is the magnitude of point charge and is the distance from the point charge
Charges entering one meter of axon is
Charges entering 0.100 mm of axon is
substituting the value of in above equation, we get charge enter a 0.100 mm length of the axon is