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inysia [295]
3 years ago
13

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

Physics
2 answers:
____ [38]3 years ago
4 0
One way the action force of a canoe could be increased is by peddling with an oar while sitting in the canoe. The point here is that the oar pushes against the current of the water, causing the canoe to move further.
Gemiola [76]3 years ago
4 0
One way the action force of a canoe could be increased is by peddling with an oar while sitting in the canoe
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A scientist discovered an elementary particle paired with another particle. It has a positive charge equal to two–thirds of the
julia-pushkina [17]
W boson has +1e or - 1e charge, Z boson has 0 charge.

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Photons have 0 charge.

Only quarks have a charge of +2/3e or -1/3e of an electron charge. 

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6 0
3 years ago
What happens to the volume of a gas when pressure is applied? A. increases B. decreases C. Stays the same D. evaporates
BaLLatris [955]

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Explanation:

8 0
3 years ago
In water what happens to a base?
Ratling [72]
Your answer would be B. It forms hydrogen ions (H+)
4 0
3 years ago
Read 2 more answers
*PHYSICS HELP*
sveta [45]
My calculator is about 1cm thick, 7cm wide, and 13cm long.

Its volume is (length) (width) (thick) = (13 x 7 x 1) = 91 cm³ .

The question wants me to assume that the density of my calculator
is about  the same as the density of water.  That doesn't seem right
to me.  I could check it easily.  All I have to do is put my calculator
into water, watch to see if sinks or floats, and how enthusiastically. 
I won't do that.  I'll accept the assumption.

If its density is actually 1 g/cm³, then its mass is about 91 grams.

The choices of answers confused me at first, until I realized that
the choices are actually 1g, 10² g, 10⁴ g, and 10⁶ g.

My result of 91 grams is about 100 grams ... about 10² grams.

Your results could be different.
3 0
3 years ago
Find the direction of the sum of<br> these two vectors:<br> 12.0 m<br> 45.0°<br> 9.66 m. A<br> m A,
Rina8888 [55]

Answer:

8.57

Explanation:

4 0
3 years ago
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