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Elena L [17]
3 years ago
11

19. State any 3 applications of capillary action (3mk)​

Physics
1 answer:
zlopas [31]3 years ago
7 0

Explanation:

1. Movement of water, food and mineral salts in plants

2. Absorption of water by towels when wiping our bodies

3. It is used to absorb ink using a blotting paper or tissue

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Which objects would sink in honey, which has a density of 1.4 g/cm³? Check all that apply.
bazaltina [42]

Answer:

any object that has density more than 1.4

Explanation:

The object that has density more than 1.4 is denser than the honey

6 0
3 years ago
Read 2 more answers
Please help I need this fast
STALIN [3.7K]

Answer:

0 m/sec

Explanation:

b/c they were at rest and initial means at rest ,at rest means 0 HOPE THIS HELPS

8 0
2 years ago
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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
a certain projetor uses a concave mirror for projecting an object's image on a screen .it produces on image that is 4 times bigg
IrinaK [193]

Answer:

f = 1 m

Explanation:

The magnification of the lens is given by the formula:

M = \frac{q}{p}

where,

M = Magnification = 4

q = image distance = 5 m

p = object distance = ?

Therefore,

4 = \frac{5\ m}{p}\\\\p = \frac{5\ m}{4}\\\\p = 1.25\ m

Now using thin lens formula:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\\\\\frac{1}{f} = \frac{1}{1.25\ m}+\frac{1}{5\ m}\\\\\frac{1}{f} = 1\ m^{-1}\\\\

<u>f = 1 m</u>

6 0
2 years ago
Two equipotential surfaces surround a +3.10 x 10-8-c point charge. how far is the 290-v surface from the 41.0-v surface?
MrMuchimi
 T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r 

where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance. 
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect, 

Point 1: 
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r

Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m 
The distance between the two points then is equal to 7.07 m.


</span>
8 0
2 years ago
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