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Vaselesa [24]
3 years ago
14

Which has the greatest frictional force among this surfaces

Physics
1 answer:
fomenos3 years ago
7 0

Answer:

D.sandy floor

Explanation:

hope this help you

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If the equipotential surfaces due to some charge distribution are vertical planes, what can you say about the electric field dir
Elina [12.6K]

Answer:

The correct option is

(e)either (c) or (d) could be correct.

Explanation:

The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.

Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since

Work = Force × distance = 0 where distance ≠ 0.

The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.

Therefore it is either horizontally to the left, or horizontally to the right.

7 0
3 years ago
When a football is kicked, the action and reaction forces do NOT cancel each other out because
AfilCa [17]

Answer:

C

Explanation:

The kicker exerts more force of the football which is why the football moves when its kicked.

6 0
3 years ago
Pls help pls due today
BaLLatris [955]
The mutualism I believe. So sorry if I’m wrong
7 0
2 years ago
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A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}, f = focal length of the mirror

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

5 0
3 years ago
What could result from regular participation in sports and exercise
shusha [124]
Common health issues that can be positively affected, prevented or controlled by exercise.
5 0
3 years ago
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