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salantis [7]
3 years ago
5

PLEASE HELP

Physics
1 answer:
gulaghasi [49]3 years ago
7 0

Answer:

The answer should be Playing Area

Explanation:

I HOPE THIS HELPS.....GOOD LUCK!!!

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A sail boat moves north for a distance of 10 km when blown by a wind 30° east of south with the force of 5.00×10^4 N. how much w
erma4kov [3.2K]
<h3><u>Answer;</u></h3>

a) 5.00 x 10^8 J

<h3><u>Explanation;</u></h3>

The work done to move the sailboat is calculated through the equation;

W = F x d

where F is force and d is the distance.

Substituting the known values from the given above,

                             W = (5.00 x 10⁴ N)(10 km)(1000 m/ 1km)

                                 = 5.00 x 10⁸ J

Thus, the work done is <u>5.00 x 10⁸Joules</u>

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3 years ago
State how heat loss by radiation is minimized in a thermos flask​
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Thermos bottles are equipped with the parts that can prevent the hot water from cooling down by the three ways: Supporting the inner container by a few heat-insulating supporters to minimize the heat loss through heat conduction, using a vacuum space between the outer and the inner vessels to eliminate the heat loss by the air convection, and giving a high reflectivity to the inner surface of the outer vessel and the whole surface of the inner container to reduce the heat loss due to the radiation cooling.
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2 years ago
A wave is traveling through a medium ad it travels it displaces particles of matter in the same direction as the wave is traveli
Tamiku [17]
I think it's longitudinal wave because the particles move parallel to the direction that the wave is traveling. 
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4 years ago
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A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr
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Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = \frac{9\times 10^9\times3\times10^{-9}}{d^2}

Field due to charge = 4 X 10⁻⁹ C

E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}

These two fields will be equal and opposite to make net field zero

\frac{9\times 10^9\times3\times10^{-9}}{d^2} = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

\frac{3}{d^2} =\frac{4}{(2-d)^2}

\frac{2-d}{d} =\frac{2}{1.732}

d = 0.928

5 0
3 years ago
What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at
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let Coefficients of Friction of Rubber on asphalt (dry) =0.7

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When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

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