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raketka [301]
2 years ago
9

A capacitor is constructed of two large, identical, parallel metal plates separated by a small distance d

Physics
1 answer:
yawa3891 [41]2 years ago
4 0
I think its true i dont kno for sure
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What type of circuit is illustrated?
Reil [10]

Answer:

I beleive its B

Explanation:

If not then A but I'm positive its B

6 0
2 years ago
On arrival at a vehicle collision, you observe a small fire in the engine compartment. A bystander is attempting to smother the
mina [271]

Answer:

<em>Aim at the base of the fire and use short bursts until the fire is out.</em>

<em></em>

Explanation:

Fire extinguishers use CO2 (Carbondioxide) as the extinguishing agent. This is because CO2 is denser than air, and does not support combustion.

Aiming at the base of the fire causes the CO2 to fall on the base of the fire, where the source of the fire is, trapping it, and preventing it from further reacting with air in a combustion reaction. Also, the short burst creates a strong wind that forces the flame to blow out.

3 0
3 years ago
What element is formed in the following nuclear reaction
MissTica

Answer:

Oxygen or more precisely, the O-15 isotope.

4 0
2 years ago
A veritical brass rod of circular section is loaded by placing a 10 kg wt on top of it .if it's length is 1 m. it's radius of cr
Inga [223]

Answer:

4.37 * 10^-4 J

Explanation:

Energy stored :

mgΔl / 2

m = mass = 10kg ; g = 9.8m/s² ; r = cross sectional Radius = 1cm = 1 * 10-2 m

Δl = mgl / πr²Y

Y = Youngs modulus = Y=3.5 ×10^10 ; l = Length = 1m

Δl = (10 * 9.8 * 1) / π * (1 * 10^-2)²* 3.5 ×10^10

Δl = 98 / 3.5 * π * 10^6

Δl = 0.00000891267

Energy stored :

mgΔl / 2

(10 * 9.8 * 0.00000891267) / 2

= 0.00043672083 J

4.37 * 10^-4 J

3 0
2 years ago
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7
Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I=\frac{P_{avg}}{A}

I=\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}

I=7.296\times 10^{-6} W/m^2

Amplitude of electric field at receiver end

E_{max}=\sqrt{2I\mu _0c}

Amplitude of induced emf

=E_{max}d

=\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75

=17.591\times 10^{-2}=0.1759 v

7 0
3 years ago
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