Find the frequency and wavelength of the wave below, assuming it has a speed of 30 m/s

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Answer:

Series

Cells in Series connection.In series, cells are joined end to end so that the same current flows through each cell. In case if the cells are connected in series the emf of the battery connected to the sum of the emf of the individual cell,If E is the overall emf of the battery combined with n number cells and E1, E2,......Em is the EMFs of individual cell.

Then   

E= E1+E2+...............+Em.

Answer:

d

Explanation:

Answer:

The answer is "512 J".

Explanation:

bullet mass [tex]m_1 = 10 g= 10^{-2} \ kg\\\\[/tex]

initial speed [tex]u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\[/tex]

block mass [tex]m_2 = 4\ Kg[/tex]

initial speed [tex]v_2 =-4.2 \frac{m}{s}[/tex]

final speed [tex]v_2= 0[/tex]

Let [tex]v_1[/tex] will be the bullet speed after collision:

throughout the consevation the linear moemuntum

[tex]\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\[/tex]

                [tex]= 320 \frac{m}{s}[/tex]

The kinetic energy of the bullet in its emerges from the block

[tex]k=\frac{1}{2} m_1 v_1^2[/tex]

   [tex]=\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J[/tex]

Answer:

True

Explanation:

300/3=100km/hr

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Answer:

True

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The question is incomplete. The complete question is :

A plate of uniform areal density [tex]$\rho = 2 \ kg/m^2$[/tex] is bounded by the four curves:

[tex]$y = -x^2+4x-5m$[/tex]

[tex]$y = x^2+4x+6m$[/tex]

[tex]$x=1 \ m$[/tex]

[tex]$x=2 \ m$[/tex]

where x and y are in meters. Point [tex]$P$[/tex] has coordinates [tex]$P_x=1 \ m$[/tex] and [tex]$P_y=-2 \ m$[/tex]. What is the moment of inertia [tex]$I_P$[/tex] of the plate about the point [tex]$P$[/tex] ?

Solution :

Given :

[tex]$y = -x^2+4x-5$[/tex]

[tex]$y = x^2+4x+6$[/tex]

[tex]$x=1 $[/tex]

[tex]$x=2 $[/tex]

and [tex]$\rho = 2 \ kg/m^2$[/tex] , [tex]$P_x=1 \ $[/tex] , [tex]$P_y=-2 \ $[/tex].

So,

[tex]$dI = dmr^2$[/tex]

[tex]$dI = \rho \ dA \ r^2$[/tex]  ,           [tex]$r=\sqrt{(x-1)^2+(y+2)^2}$[/tex]

[tex]$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$[/tex]

[tex]$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$[/tex]

[tex]$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \ dy \ dx$[/tex]

[tex]$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$[/tex]

[tex]$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$[/tex]

[tex]$I=\frac{32027}{21} \times 2$[/tex]

  [tex]$= 3050.19 \ kg \ m^2$[/tex]

So the moment of inertia is  [tex]$3050.19 \ kg \ m^2$[/tex].

Answer:

3.33m

Explanation:

λ = C/f

Where,

λ (Lambda) = Wavelength in meters

c = velocity=1.5m/s

f = Frequency=0.45

λ = 1.5/0.45

=3.33m.

Answer:

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Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V