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IrinaK [193]
3 years ago
5

Why is it easier to swim in sea than in a pond or river?​

Physics
1 answer:
butalik [34]3 years ago
6 0

Answer:

Since sea water contains alot of salt the density of the sea water is higher as to the river water . So it makes it easier to swim in sea water

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A cat walks along a plank with mass M= 6.00 kg. The plank is supported by two sawhorses. The center of mass of the plank is a di
dimulka [17.4K]

Answer:

d₂ = 1.466 m

Explanation:

In this case we must use the rotational equilibrium equations

        Στ = 0

        τ = F r

we must set a reference system, we use with origin at the easel B and an axis parallel to the plank , we will use that the counterclockwise ratio is positive

      + W d₁ - w_cat d₂ = 0

      d₂ = W / w d₁

      d₂ = M /m d₁

      d₂ = 5.00 /2.9    0.850

      d₂ = 1.466 m

6 0
4 years ago
What is the primary outgoing radiation put off by the sun?
ira [324]

I really don’t know but I think it’s D

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3 years ago
If a sealed syringe is plunged into cold water, in which direction will the syringe piston slide? *
Tamiku [17]

Answer:

a

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3 years ago
This particle would have to have at least how many protons in order to have a positive charge?
jasenka [17]

Answer: You don't do work.

Explanation: If you don't do the work you don't have to ask this question. With work comes labor. With labor comes change. With change comes resistance and people do not like change. IT causes anger, so if you stop doing work and stick to the important supply like food, water, and shelter you don't have to go through the trouble of this and we can live in a peaceful environment for this world to be better than it is. People do to much, it's a natural fact but you can change the entire picture. If you read all of this thank you for accepting the fact of life.

6 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m

4 0
3 years ago
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