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3 years ago

14

A charge of 19 nC is uniformly distributed along a straight rod of length 15 m that is bent into a circular arc with a radius of

5.7 m. What is the magnitude of the electric field at the center of curvature of the arc?

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C

Explanation:

Please see the attachments below

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2 years ago

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3 years ago

Why is a flask is wider on the bottom than on the top? allows for a more precise measurement allows for better thermal equilibri

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3 years ago

A diver springs upward from a board that is 2.70 m above the water. At the instant she contacts the water her speed is 10.9 m/s

TiliK225 [7]

Answer:

vo=5.87m/s

Explanation:

Hello! In this problem we have a uniformly varied rectilinear movement.

Taking into account the data:

α =69.2

vf = 10m / s

h=2.7m

g=9.8m/s2

We know we want to know the speed on the y axis.

We calculate vfy

vfy = 10m / s * (sen69.2) = 9.35m / s

We can use the following equation.

$vf^{2} =vo^{2}+2*g*h\\$

We clear the vo (initial speed)

$vo=\sqrt{vf^{2}-2*g*h }$

$v0=\sqrt{(9.35m/s)^{2}-2*9.8m/s^{2} *2.7m}$

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3 years ago

Convert 70 mi/h to m/s. 1 mi = 1609 m.<br><br> Answer in units of m/s.<br><br> Plz help me now

melisa1 [442]

Answer:

70mi/h

1mile =1609 meters

1hour=3600 seconds

So,

70×1609/3600

112,630/3600

31.286 m/s

Hope this helps you

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3 years ago