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Setler [38]
2 years ago
10

Can Vehicles us bike lanes when they are empty?

Physics
1 answer:
Allushta [10]2 years ago
7 0
No it is unsafe. Even if they are empty they cannot
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What is the magnitude (in N/C) and direction of an electric field that exerts a 3.50 ✕ 10−5 N upward force on a −1.55 µC charge?
IRISSAK [1]

Answer:

The magnitude of electric field is  22.58 N/C

Solution:

Given:

Force exerted in upward direction, \vec{F_{up}} = 3.50\times 10^{- 5} N

Charge, Q = - 1.55\micro C = - 1.55\times 10^{- 6} C

Now, we know by Coulomb's law,

F_{e} = \frac{1}{4\pi\epsilon_{o}\frac{Qq}{R^{2}}

Also,

Electric field, E = \frac{1}{4\pi\epsilon_{o}\frac{q}{R^{2}}

Thus from these two relations, we can deduce:

F = QE

Therefore, in the question:

\vec{E} = \frac{\vec F_{up}}{Q}

\vec{E} = \frac{3.50\times 10^{- 5}}{- 1.55\times 10^{- 6}}

\vec{E} = - 22.58 N/C

Here, the negative side is indicative of the Electric field acting in the opposite direction, i.e., downward direction.

The magnitude of the electric field is:

|\vec{E}| = 22.58\ N/C

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2 years ago
URGENT!!! PLEASE HELP!!!!<br> In what kind of ecosystem can the population of a Leopard be found?
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3 years ago
A skateboarder shoots off a ramp with a velocity of 5.1 m/s, directed at an angle of 55° above the horizontal. The end of the ra
ExtremeBDS [4]

Answer

given,

initial velocity of skateboard = 5.1 m/s

angle above the horizontal = 55°

height of the ramp = 1 m

a) maximum height of projectile

  H = \dfrac{u^2sin^2 \theta}{2g}

  H = \dfrac{5.1^2\times sin^2 55^0}{2\times 9.81}

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the maximum height of the skateboard above the ground

         = 1 + 0.889

         = 1.889 m

b) time to reach the height

   t = \dfrac{u\ sin\theta}{g}

   t = \dfrac{5.1\ sin55^0}{9.8}

          t = 0.426 s

horizontal distance = u cos θ × t

                                = 5.1 × cos 55° × 0.426

horizontal distance = 1.25 m

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3 years ago
Which statement best explains that sound waves are pressure waves?
kotykmax [81]
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What is the acceleration of a 10 kg mass pushed by a 5 N force?
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g-\ gravitational \ acceleration \\ g= \frac{G}{m} = \frac{5N}{10kg}=  \\ g=0,5 \frac{N}{kg}
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