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Lyrx [107]
3 years ago
9

An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin

der. In this problem we consider an imaginary cylinder with radius r around the axis AB where r

Physics
1 answer:
RUDIKE [14]3 years ago
4 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
nirvana33 [79]

Answer:

1.327363 m/s

0.00090243 m

Explanation:

u = Initial velocity

v = Final velocity

m = Mass of flea

Energy

E=\frac{1}{2}m(v^2-u^2)\\\Rightarrow 3.7\times 10^{-4}=2.1\times 10^{-4}(v^2-0)\\\Rightarrow v=\sqrt{\frac{3.7\times 10^{-4}}{2.1\times 10^{-4}}}\\\Rightarrow v=1.32736\ m/s

The velocity of the flea when leaving the ground is 1.327363 m/s

W=F\times s\\\Rightarrow s=\frac{W}{F}\\\Rightarrow s=\frac{3.7\times 10^{-4}}{0.41}\\\Rightarrow s=0.00090243\ m

The flea will travel 0.00090243 m upward

8 0
3 years ago
A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The
Slav-nsk [51]

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

8 0
3 years ago
A child who is swimming toward shore at 0.78 m/s sees shark and picks up his speed
Gnoma [55]

Answer:

0.085m/s²

Explanation:

Use v²=v0²+2a(d)

solve for a

v²-v0²/2d=a

Plug in givens

1.89²-0.78²/2*17.5=a

Plug into calculator

a=0.085m/s²

3 0
3 years ago
When the force at the ends of a bar pushes rather than pulls, the bar experiences?
Mariulka [41]

Answer:

I don't know sorry hopefully I can help with something else tho

4 0
3 years ago
The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the
taurus [48]

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.

The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>

3 0
3 years ago
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