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2 years ago

A boat is stationary at 12 meters away from a dock. The boat then begins to move toward the dock with

Physics

1 answer:

just olya [345]2 years ago

8 0

Answer: 2.19 seconds

Explanation:

<u>Given:</u>

Initial speed, u = 0 m/s

Acceleration of the boat, a = 5 m/s^2

Distance between the boat and dock, d = 12 m

Using the third kinematics equation to solve for time:

d = u*t + (1/2)*a*t^2

12 = 0*t + (1/2)*5*t^2

t = sqrt (12*2/5)

t = 2.19 seconds

Therefore, it will take the boat approximately 2.19 seconds to reach the dock

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Answer:

A. External

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External stimulus includes touch/pain, vision, smell, taste, and sound.

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2 years ago

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Answer:

The surface gravity is inversely proportional to the square of the radius of the planet

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2 years ago

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2 years ago

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3 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

Now,

The net Electric field is given by:

$\vec{E_{net}} = \vec{E} - \vec{E'}$

$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

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3 years ago