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iragen [17]
3 years ago
10

What could you do to change the volume of a gas?

Physics
1 answer:
Lorico [155]3 years ago
8 0
The ONLY way to change the volume of a sample of gas is to transfer it to a container with different volume.
Simply changing its temperature or pressure in the same jar won't do it. Any amount of gas always fills whatever container you keep it in.
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Point P and point charge Q are separated by a distance R. The electric field at point P has magnitude E. How could the magnitude
Angelina_Jolie [31]
<span>We know , E = kQ/r^2 where q = charge and r is separation between point and point charge. Now, At P, E= kQ/r^2 Since, Q can't be changed, we can do that by varying r 2E = 2kq/r^2 2E = kq/ (r/ sqrt2)^2 Hence, if we bring Q closer such that distance between P and Q becomes r/ sqrt 2, E will get doubled.</span>
5 0
3 years ago
D the element in blue square has a full outer shell, due to it’s location on the periodic table, so it will not react with other
Juliette [100K]
The correct response that will be used to describe this particular element would be the third option, since all of the other options are incorrect and apply to different elements in their groups. The element is a metal and will react with a non metal.






5 0
3 years ago
A rock is thrown down from the top of a cliff with a velocity of 3.61 m/s (down). The cliff is 28.4 m above the ground. Determin
Katarina [22]
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4 0
2 years ago
Suppose that you hear a clap of thunder 16 s after seeing the associated lightning strike. How far are you from the lightning st
riadik2000 [5.3K]

Answer:

d=5.376km

Explanation:

Since <em>light is so fast</em> we can assume no time passes between the lightning strikes and we observe it. We want to know then how far away did the strike occur for the sound to take 16s to reach our ears. Since the definition of velocity tells us that v=d/t, we can write d=vt=(336m/s)(16s)=5376m=5.376km

4 0
3 years ago
Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and
lawyer [7]

1 kilometre=1000 metre

      1 hour = 3600 second

       1\ km/hr=\frac{1000}{3600} m/s

       1\ km/hr=\frac{5}{18} m/s

The initial velocity of car A is 35.0 km/hr i.e

                                         35.0\ km/hr=35*\frac{5}{18} m/s

                                                                   = 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as  25\ km/hr^2

                                            =\ 25*\frac{1000}{3600*3600} m/s^2

                                            =0.00192901234 m/s^2

The time taken by car A = 15 min.

From equation of kinematics we know that-

                                 v= u+at      [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A,  v = 9.72 m/s +[0.00192901234×15×60]m/s

                                   =11.456111106 m/s

The acceleration of B is given as    15\ km/hr^2

                                    =0.00115740740740 m/s^2

The time taken by car B =20 min

The final velocity of B is -

                             v= u+at

                               = u-at    [Here a is negative due to deceleration]

                               =12.5 m/s +[0.0011574074074×20×60]

                               =13.8888888.....

                               =13.9

The acceleration of C is given as    40\ km/hr^2          

                                                            =\ 0.003086419753 m/s^2

The time taken by car C =30 min

The final velocity of C is-

                                v = u+at

                                   =8.89 m/s+[0.003086419753×30×60] m/s

                                   =14.4455555555..m/s

                                   =14.45 m/s

The car C is decelerating.The deceleration is given as-  60\ km/hr^2

                                                                      =0.0046296296296m/s^2

The time taken by car D= 45 min.

The final velocity of the car D is-

                     v =u+at

                        =30.56 -[0.00462962962962×45×60]m/s

                        =18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

                 


3 0
3 years ago
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