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3 years ago

How could the action force of a canoe moving through water be increased?.

1 answer:

4 0

There are several actions that can be used to increase the action force of a canoe that is moving through water. One can increase the rate of striking or strike harder than before with a bigger paddle. Using a lighter canoe can also be helpful for increasing the speed. I hope the answer comes to your help.

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Assume you are given an int variable named nElements and a two-dimensional array that has been created and assigned to a2d. Writ

Alex777 [14]

Answer:

Explained

Explanation:

public int dimension(int [][]a2d,int nElements)

{

int count = 0;

for(int i = 0;i < a2d.length ; i++)

{

count = count + a2d[i].length;

}

return count;

}

3 0

3 years ago

Which of the following is not an example of a molecule? A.H2O2 B.NCl3 C.F D.O3

lara [203]

I believe the answer is F. This is because, F stands for Fluorine, which is an element and consists of one atom. A molecule consists of TWO OR MORE atoms held together by chemical bonds. F consists of only one atom, which is Fluorine. So the answer is C. Hope helped.

7 0

3 years ago

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Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at

dedylja [7]

Answer:

$50.91 \mu C$

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for $q_{1}$ and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by $q_{1}$ on q. Then we can know the magnitude of the force exerted by $q_{2}$ about q, finally this will allow us to know the magnitude of $q_{2}$

$q_{1}$ exerts a force on q in +y direction, and $q_{2}$ exerts a force on q in -y direction.

$F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\$

The net force on q is:

$F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}$

Rewriting for $q_{2}$ :

$q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C$

8 0

3 years ago

How do you calculate elastic potential energy

Nimfa-mama [501]

U=1/2kx2

This image sums it up

5 0

3 years ago

How much would a 15.0 kg object weigh on Neptune?

yan [13]

The gravity on Neptune is 11.15 m/s²
the gravity on earth is 9.81 m/s²
divide the Neptune and earth gravity we get 1.13
which means object on neptune is 1.13 heavier than earth
yield, weigh of the object on neptune is 1.13×15=17.04kg

5 0

3 years ago

Read 2 more answers