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alina1380 [7]
3 years ago
10

What does the diameter in meters need to be of a round parachute that the designer has determined needs to have an area of 500 s

quare meters?
Physics
1 answer:
Marrrta [24]3 years ago
4 0
Area of a circle is
A= pi r^2
so
500m^2 = 3.14 r2
500/pi = r ^2
152.1549...=r^2
square root both sides
r=12.61566...
d=2r
d=25.2
to 3 sig fig
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9. [03.03]
Evgen [1.6K]

Answer:

Circuit one will have more current than circuit two

Explanation:

I am assuming that you have to see which circuit has the greater current in this case. Well, this is the perfect example of Ohm's Law, which states the following -

V = IR,

where V = voltage / potential difference, I = current, and R = resistance

If one circuit has twice the voltage and half the resistance of the second circuit, as voltage is directly proportional to the resistance -

2V = I( 1 / 2R ),

4V = IR,

I = 4V / R

Whereas in the second circuit -

V = IR,

I = V / R

As you can note, voltage is directly proportional to the current ( I ) as well as the resistance. The only difference between the two formulas I = 4V / R, and I = V / R is the difference in the voltage. With the voltage being 4 times greater in the first circuit, and current is 4 times greater in the first circuit as well.

<u><em>Hence, circuit one will have more current than circuit two</em></u>

5 0
3 years ago
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I would say the stone is being used as a fulcrum.

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A finite line of charge with linear charge density λ = 3.35 × 10-6 C/m, and length L = 0.808 m is located along the x axis (from
Andrews [41]

Answer: magnitude = 169.66N/C

direction = 8.6859°

Explanation:

Given from the question, we have that;

the Length of line of charge L = 0.808 m

Linear charge density λ = 3.35 × 10⁻⁶ C/m

charge q = -7.32 × 10⁻⁷ C

Coulombs force constant K = 1/(4π ε0) = 8.99 × 10⁹ N·m²/C².

NB. The picture uploaded gives a diagrammatic description of the problem.

From Pythagoras theorem we have,

tan Θ = 3.75 / (10.7-1.56)  

Θ = 22.3076 °

recall that the Electric field at point P due to the finite wire is;

È = Kλ (L / b(L+b)) Î .............. (1)

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹ × 3.35 × 10⁻⁶ (0.808/ 10.7(10.7 – 0.808))

È = 0.229905 × 10³ N/C Î

recall also that the Electric field at point -P due to -q is;

È  = (8.99 × 10⁹ × 7.32 × 10⁻⁷) / ((3.75)² + (10.75-1.56)²) = 0.6742 × 10² N/C

where E = -E₁cosθÎ  + E₁sinθĴ

E = - 0.62446 ×10²Î   + 0.2562 ×10²Ĵ

The Resultant Electric charge Er is given as;

Er = 1.6771 ×10²Î + 0.2562 ×10²

Er =  [√(1.6771)² + (0.2562)² ] × 10² = 169.66 N/C

∴ Magnitude = 169.66 N/C

Having gotten the magnitude, let us find the direction;

⇒ Direction = tan Ф = 0.25621/1.6771 = 8.6859°

Direction = 8.6859°

cheers i hope this helps

7 0
3 years ago
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