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m_a_m_a [10]
3 years ago
5

What is the wavelength of this wave? 1 2 3 4

Physics
2 answers:
Katen [24]3 years ago
4 0
Its 2. i jus took test and it was right

Mama L [17]3 years ago
3 0
2 would be your answer
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1. बालू के एक कण की त्रिज्या 1.6 x 10-4मी है। इस कण की त्रिज्या
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3 years ago
A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separa
Elanso [62]

Answer:

Explanation:

plate separation = 2.3 x 10⁻³ m

capacity C₁ = ε A / d

= ε A / 2.3 x 10⁻³

C₂ = ε A / 1.15 x 10⁻³

\frac{C_2}{C_1} = \frac{2.3}{1.15}

a ) when charge remains constant

energy = \frac{q^2}{2C}

q is charge and C is capacity

energy stored initially E₁= \frac{q^2}{2C_1}

energy stored finally E₂ = \frac{q^2}{2C_2}

\frac{E_1}{E_2} = \frac{C_2}{C_1} = \frac{2.3}{1.15}

E_2 = \frac{1.15}{2.3 } \times E_1

= \frac{1.15}{2.3 } \times 8.38

= 4.19 J

b )

In this case potential diff remains constant

energy of capacitor = 1/2 C V²

energy is proportional to capacity as V is constant .

\frac{E_2}{E_1} = \frac{C_2}{C_1}

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8 0
3 years ago
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C

Charge 2, q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

F=k\dfrac{q_1q_2}{r^2}

r=\sqrt{\dfrac{kq_1q_2}{F}}

r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}

r = 0.8009 meters

or

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

4 0
3 years ago
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