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nasty-shy [4]
3 years ago
9

State one way to decrease the moment of a given force about a given axis of rotation.​

Physics
2 answers:
wlad13 [49]3 years ago
8 0

Answer:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

LiRa [457]3 years ago
5 0

Answer:

hope it helps...

Explanation:

The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.

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Can animal cells be broken down further into a living unit
mezya [45]

Answer: No! Animal cells cannot be broken down further into living cells.

5 0
4 years ago
Read 2 more answers
How do you find the speed of an object given its mass and kinetic energy (what is the formula)?
madam [21]
   v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ; 
        
and solve for "v".

______________________________________________________
Explanation:
_____________________________________________________
The formula is:  KE = (½) * (m) * (v²) ;
_____________________________________
  
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note:  Velocity is similar to speed, in that velocity means "speed and direction";  however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ;  ↔  (½)*(m)* (v²) = (KE) ; 
___________________________________________________

→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
   
→ We divide each side of the equation by: "[(½)* (m)]" ; 
___________________________________________________
    
           →   [ (½)*(m)*(v²) ] /  [(½)* (m)]  = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
 to get: 
______________________________________________________
                           →   v²     =   (KE) / [(½)* (m)]
                     
                           →   v²     = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
                          →  √ (v²)  =  √ { 2*(KE) ] / m }
________________________________________________________

                          →     v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"]; 
       
and solve for "v".

______________________________________________________
8 0
3 years ago
A car drives past a pole at 40km/hr. Describe the motion from the point of view of a) the car, and b) the pole. Thanks in advanc
ki77a [65]
I was going to beg off until tomorrow, but this one is nothing like those others.
Why, at only 40km/hr, we can ignore any relativistic correction, and just go with Newton.

To put a finer point on it, let's give the car a direction.  Say it's driving North.

a).  From the point of view of the car, its driver, and passengers if any,
the pole moves past them, heading south, at 40 km/hour .

b).  From the point of view of the pole, and any bugs or birds that may be
sitting on it at the moment, the car and its contents whiz past them, heading
north, at 40 km/hour.

c).  A train, steaming North at 80 km/hour on a track that exactly parallels
the road, overtakes and passes the car at just about the same time as
the drama in (a) and (b) above is unfolding.

The rail motorman, fireman, and conductor all agree on what they have
seen. From their point of view, they see the car moving south at 40 km/hr,
and the pole moving south at 80 km/hr.

Now follow me here . . .

The car and the pole are both seen to be moving south.  BUT ... Since the
pole is moving south faster than the car is, it easily overtakes the car, and
passes it . . . going south.

That's what everybody on the train sees.

==============================================

Finally ... since you posed this question as having something to do with your
fixation on Relativity, there's one more question that needs to be considered
before we can put this whole thing away:

You glibly stated in the question that the car is driving along at 40 km/hour ...
AS IF we didn't need to know with respect to what, or in whose reference frame.
Now I ask you ... was that sloppy or what ? ! ? 

Of course, I came along later and did the same thing with the train, but I am
not here to make fun of myself !  Only of others.

The point is . . . the whole purpose of this question, obviously, is to get the student accustomed to the concept that speed has no meaning in and of itself, only relative to something else.  And if the given speed of the car ...40 km/hour ... was measured relative to anything else but the ground on which it drove, as we assumed it was, then all of the answers in (a) and (b) could have been different.

And now I believe that I have adequately milked this one for 50 points worth.


7 0
3 years ago
143°C = _____<br><br> 416 K<br> -130 K<br> 0 K<br> 143 K
dezoksy [38]
The answer is 0k because 143c equals nothing
5 0
2 years ago
Read 2 more answers
how do I solve how much time does it take for a satellite moving 15000 miles per hour travel to travel once around the earth 249
iogann1982 [59]

Answer:

The time taken by the satellite to orbit earth at its surface, t = 1.66 hr

Explanation:

Given data,

The velocity of the satellite, v = 15000 miles/hr

The distance of travel, d = 24901 miles,

It is equal to the circumference of earth,

So the time taken by the satellite to orbit earth at its surface,

                           v = d/t

                            t = d/v

Substituting the given values in the above equation,

                            t = 24901 / 15000

                              = 1.66 hr

Hence, the time taken by the satellite to orbit earth at its surface, t = 1.66 hr

3 0
3 years ago
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