Physics

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R
Raj Pandey

Contributor-Level 9

H = I 2 R t = 2 2 * R * 1 5

3 0 0 = 6 0 R

R = 5 Ω

Now for 3A, time = 10 sec

H ' = l 2 R t = 3 2 * 5 * 1 0 = 4 5 0 J

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2 months ago

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A
alok kumar singh

Contributor-Level 10

m ω 2 . 2 d 3 = G . m . 2 m d 2 ω = 3 G m d 3

T = 2 π ω = 2 π d 3 3 G m

 

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R
Raj Pandey

Contributor-Level 9

ceq = 2 4 * 8 3 2 = 6 μ F

 

 

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A
alok kumar singh

Contributor-Level 10

α = Δ l C Δ l E = 3 . 5 4 = 7 8

β = α 1 α = 7 / 8 1 7 / 8 = 7

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alok kumar singh

Contributor-Level 10

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + 1 2 r .

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Raj Pandey

Contributor-Level 9

H = λ 4

2 0 = λ 4

λ = 8 0 c m

Now first overtone for open organ pipe

L 2 = 4 λ 4 = λ = 8 0 c m

 

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A
alok kumar singh

Contributor-Level 10

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant.

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R
Raj Pandey

Contributor-Level 9

Q 1 = 2 0 ° C w a t e r 1 0 0 ° s t e a m

Q 1 = m c Δ T + m L

= m [ c Δ T + L ]

= 3 1 0 0 0 = 3 1 * 1 0 3 c a l

Q 2 = 2 9 3 3 7 3 * 3 1 * 1 0 3

 

 

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alok kumar singh

Contributor-Level 10

For part AM, slope of v – t graph is constant but negative. For part MB, slope of v – t graph is constant but positive.

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