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vfiekz [6]
2 years ago
9

After a great many contacts with the charged ball, how is any charge on the rod arranged (when the charged ball is far away)?The

re is positive charge on end B and negative charge on end A.There is negative charge spread evenly on both ends.There is negative charge on end A with end B remaining neutral.There is positive charge on end A with end B remaining neutral.The conducting rod is not grounded, so a negative charge accumulates on one end, but a charge cannot remain at a single end in a conductor, so it will spread to both ends:There is negative charge spread evenly on both ends
Physics
1 answer:
lozanna [386]2 years ago
8 0

Answer:

The correct option is that both the ends will remain neutral.

Explanation:

As the rod is grounded all of the prolonged charge will be converted to ground so the overall charge on the rod in absence of the charged ball will be equal to zero.Such that the both ends will not bear any charge.

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A 2.0-m-tall man is 5.0 m from the converging lens of a camera. His image appears on a detector that is 50 mm behind the lens. H
ladessa [460]

Answer:

20 cm

Explanation:

We can solve the problem by using the magnification equation:

M=\frac{h_i}{h_o}=-\frac{q}{p}

where

h_i is the size of the image

h_o = 2.0 m is the height of the real object (the man)

q=50 mm =0.050 m is the distance of the image from the lens

p = 5.0 m is the distance of the object (the man) from the lens

Solving the formula for h_i, we find

h_i = -\frac{q}{p}h_o=-\frac{0.050 m}{5.0 m}(2.0 m)=-0.02 m = -20 cm

And the negative sign means the image is inverted.

6 0
2 years ago
The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. If a rock weighs 75.0 N on earth, compu
IgorLugansk [536]

Answer:

g = 0.905 gE

W  = 67.9 N

Explanation:

given data

mass of Venus mv =  81.5% = 0.815

radius Rv = 94.9% = 0.949

weighs W = 75.0 N

solution

we apply here acceleration due to gravity at earth surface that is

g = \frac{Gm}{R^2}   = 9.80 m/s²  ............1

so

g = \frac{G(0.815)}{0.949R^2}  

g = 0.905 gE

and

W = m gv

W = 0.905 m gE

W  = 0.905 × 75

W  = 67.9 N

7 0
2 years ago
A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it
Ad libitum [116K]

Answer:

Δx=(v+v0/2)t

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the initial velocity v_0v  

0

​  

v, start subscript, 0, end subscript of the roller coaster.

7 0
2 years ago
Which of these shows unbalanced forces at work on an object? A. an ice skater turning as he skates around an ice rink B. a bicyc
rosijanka [135]
I’m pretty sure the answer is a
4 0
2 years ago
Read 2 more answers
The sound level at 1.0 m from a certain talking person talking is 60 dB. You are surrounded by five such people, all 1.0 m from
Hunter-Best [27]

Answer:

66.98 db

Explanation:

We know that

L_T=L_S+10log(n)

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n= number of sources

L_S= signal level from signal source.

L_T=60+10 log(5)

= 66.98 db

7 0
2 years ago
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