If the sack weighs 210 newtons, then an upward force of 210 newtons
exactly cancels the downward force of gravity, and makes the net vertical
force on the bag zero.
ANY upward force that's greater than 210 newtons makes the net force
act upward on the bag, and causes it to accelerate upward.
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a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N
b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>
Answer:
a. The disk
b. Because it has the smallest rotational inertia
Explanation:
a. Which object do you expect to reach the bottom of the inclined plan first?
I would expect the disk to reach the bottom first.
b. Why?
This is because the disk has the smallest rotational inertia.
The rotational inertial of the hollow sphere, disk and ring are 2/3MR², 1/2MR² and MR² respectively.
Since the three objects are rolling from the same height, they have the same mechanical energy.
But, since the disk has the smallest rotational inertia, it would have the smallest rotational kinetic energy and largest translational kinetic energy. The disk's smaller rotational kinetic energy will cause to rotate less but translate more than the other objects and thus reach the bottom first.
The minimum size copper equipment grounding conductor that is required is ; #4 copper
<h3>Equipment grounding conductor </h3>
An Equipment grounding conductor is a conductive part of ground fault current path which connects the non-conducting metallic parts of the euipment together and also connects the system grounded conductor.
For a copper euipment grounding conductor required to serve a control center connected to a 300 amp over current device the minimum size of copper is #4 copper
Hence we can conclude that The minimum size copper equipment grounding conductor that is required is ; #4 copper.
Learn more about Equipment grounding conductor : brainly.com/question/14124204
Answer:
Right shoe
Explanation:
Let the mass and velocity of incoming puck be m and v respectively.
Momentum of the colliding puck will be mv
In case of first case , the momentum of puck becomes zero so change in momentum after collision with left shoe
= mv - 0 = mv
If time duration of collision be t
rate of change of momentum
= mv / t
This is the force exerted by puck on the left shoe .
Now let us consider collision with right shoe
momentum after collision with right shoe
- mv
change in momentum
= mv - ( - mv ) = 2mv
If time duration of collision be t
rate of change of momentum
= 2mv / t
This is the force exerted by puck on the right shoe .
Since the force on the right shoe is more , this shoe will have greater speed
after collision.
