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Angelina_Jolie [31]
4 years ago
12

PLEASE HELP!

Physics
2 answers:
tino4ka555 [31]4 years ago
8 0

Answer:

option (c)

Explanation:

A small diameter open tube is called capillary tube.

When a capillary tube is immersed in a vessel containing mercury, then the mercury falls in the capillary tube and the shape is like outside of an umbrella.  

As the angle of contact for mercury and glass is obtuse, so it falls in the tube and the meniscus is convex.

maks197457 [2]4 years ago
4 0
The answer is A .
Due to air pressure, the mercury in the tube will be higher than that in the vessel.
Mercury in a container, the center of the liquid is always higher, so it is shaped like the outside of an umbrella.
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A car of mass 1600 kg can just be lifted what is the least force that the electronicmagnet must use to lift the car ?
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The car's mass is 1600 kg.

Its weight is (mass) x (gravity).  

On Earth, that's (1600 kg) x (9.8 m/s²)  =  15,680 Newtons.

At the moment, that's the only force acting on the car, directed downward and provided by gravity.

If you want to lift the car, then the net force has to be directed upward, and must either exactly cancel or exceed the force of gravity.

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Which terrestrial planet exhibits retrograde rotation?.
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Answer:

Planets that are farther from the sun than the earth (all but Mercury and Venus) will exhibit retrograde motion.

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(The planet will be on the side of the earth that is opposite that of the sun)

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8 0
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How is item A different from Item B?
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8 0
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Read 2 more answers
Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

  The speed of car A is  v_A  =  22 \ m/s

   The speed of car B is  v_B  = 29.0 \ m/s

     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

    The distance traveled by car B  =  The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is  

          d = v_B * t_A

Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

      d = v_A t_A  + \frac{1}{2}a_A t_A^2 - 300

Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

       v_B * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

     t_A  = 19.0 \ s

7 0
3 years ago
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