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fomenos
3 years ago
6

A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate

Physics
1 answer:
ra1l [238]3 years ago
6 0

Answer:

this is the answer......

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Which object represents a negatively charged particle? which object represents a positively charged molecule? which object repre
kari74 [83]

The answers to your questions are as written below:

  • The objects that represents a negatively charged particle is : Image B
  • The object that represents a positively charged molecule is : Image A
  • The object that represents an uncharged molecule is : Image C
  • The object the will not move when in an electric fied is : Image C

<h3>Different types of charges molecules</h3>

A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.

A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.

Hence we can concude that the answers to your questions are as listed above.

Learn more about electric charges :brainly.com/question/857179

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attached below is the missing image

8 0
2 years ago
Drag the tiles to the correct boxes to complete the pairs.
Dimas [21]
Brown dwarf is the first box
White dwarf is the second box
Black dwarf is the third box
Red giant is the fourth box
And
Black hole is the last box
5 0
2 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
2aS=v_f^2 - v_i^2
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
v_f =0
So we can rewrite the equation as
2(-9.81 m/s^2) h=-v_i^2
from which we can isolate h
h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
8 0
3 years ago
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