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Leni [432]
3 years ago
6

Can someone help asap plz don’t understand how to do this

Physics
1 answer:
shtirl [24]3 years ago
3 0
Homie I don’t know either‍♀️Drop out of schl ig ;-;
You might be interested in
If Frequency F, velocity v, and density D are considered fundamental units, the dimensional formula for momentum will be :
gizmo_the_mogwai [7]

Let's see

Momentum be P

\\ \rm\Rrightarrow P=[Frequency]^a[velocity]^b[Density]^c

\\ \rm\Rrightarrow [P]=[F]^a[v]^b[D]^c

\\ \rm\Rrightarrow [M^1L^1T^{-1}]=[T^{-1}]^a[L^1T^{-1}]^b[M^1L^{-3}]^c

\\ \rm\Rrightarrow MLT^{-1}=T^{-a}L^bT^{-b}M^cL^{-3c}

\\ \rm\Rrightarrow MLT^{-1}=T^{-a-b}L^{b-3c}M^c

On comaparing

  • c=1

So

  • b-3c=1
  • b-3=1
  • b=1+3
  • b=4

and

  • -a-b=-1
  • -a-4=-1
  • -a=-1+4=3
  • a=-3

So the unit is

  • DV⁴/F³
5 0
2 years ago
Read 2 more answers
Pics pls you know what I mean
V125BC [204]

boiiii I wont show you nun

4 0
3 years ago
As the mass of an object increases, the weight of the object will ______?
Sergio039 [100]

Answer:

also increase

Explanation:

If the mass increases so will the weight.

4 0
2 years ago
Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
2 years ago
The planet Saturn has a mass that is 95 times Earth's mass and a radius that is 9.4 times Earth's radius. What is the accelerati
ziro4ka [17]

Answer:

10.55111 m/s²

Explanation:

M = Mass of Saturn = 95\times 5.972\times 10^{24}\ kg

r = Radius of Saturn = 9.4\times 6.371\times 10^6\ m

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

Acceleration due to gravity is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 95\times 5.972\times 10^{24}}{(9.4\times 6.371\times 10^6)^2}\\\Rightarrow g=10.55111\ m/s^2

The acceleration due to gravity on Saturn is 10.55111 m/s²

4 0
3 years ago
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