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Montano1993 [528]
2 years ago
13

In Burglar alarm LDR acts as a/an a. off switch b. on switch c. AND gate d. OR gate

Physics
1 answer:
Gelneren [198K]2 years ago
4 0

In Burglar alarm, LDR acts an AND gate.

Answer: C

Explanation

The LDR is light dependent resistor. The principle used in the working of LDR is that the resistance is inversely proportional to the intensity of light falling on the diode.

In burglar alarm, LDR diode is combined with an IC 555.

Normally an LED source is made to be incident on the LDR diode with same intensity such that the resistance will be maintained constant.

As the LDR is connected with IC, the voltage will be high when light is falling on the diode.

The IC will give only two output states that is high and low. This confirms that LDR in burglar alarm act as AND gate.

As the thief enters and crosses the LED light, the intensity of the light falling on the diode will decrease leading to decrease in the voltage which will cause the alarm to beep.

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Two speakers face each other, and they each emit a sound of wavelength λ. One speaker is 180∘ out of phase with respect to the o
mel-nik [20]

Answer:

a. 3/4λ

d. 1/4λ

Explanation:

When the wavelength of the sound waves is λ and the two waves are having same frequency the waves are said to be out of phase if their phase difference is in the multiples of \frac{\lambda}{2} or 180°.

When the two waves are out of phase then their opposite maxima coincide at the same time resulting in the minimum amplitude of the resulting wave throughout.

  • As we observe from the schematic that the a wave has sinusoidal pattern of variation and we get a maxima after each \frac{\lambda}{4} of the distance.
  • Here we have two speakers out of phase therefore on shifting one of the speakers by the odd multiples of \frac{\lambda}{2} we have the maxima or the extreme amplitudes.

So, we must place the microphone at  3/4λ and 1/4λ to pickup the loudest sound.

4 0
2 years ago
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
2 years ago
A plane traveling north at 100.0 km/h through the air gets caught in a 40.0 km/h crosswind blowing west. This turbulence caused
mojhsa [17]

Answer:

Explanation:

If I assume that the wind did not cause the plane to chage its velocity.

The plane will have a velocity of vp =  (0*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (0*i + 100*j) + (0*i - 20*j) = (0*i + 80*j) km/h relative to the ground.

If I assume that the wind move the plane:

The plane will have a velocity of vp =  (-40*i + 100*j) km/h relative to ground

The cart has a velocity of vc = (0*i - 20*j) km/h relative to the plane

vc' = vc + vp

vc' = (-40*i + 100*j) + (0*i - 20*j) = (-40*i + 80*j) km/h relative to the ground.

In reality the wind would move the plane a little, not to the full speed of the wind, somewhere between these two values, but without more data it cannot be calculated.

6 0
2 years ago
A rectangular beam 10 cm wide, is subjected to a maximum shear force of 50000 N, the corresponding maximum shear stress being 3
nika2105 [10]

Answer:

Option B is the correct answer.

Explanation:

Shear stress is the ratio of shear force to area.

We have

       Shear stress = 3 N/mm² = 3 x 10⁶ N/m²

       Area = Area of rectangle = 10 x 10⁻² x d = 0.1d

       Shear force = 50000 N

Substituting

        \texttt{Shear stress}=\frac{\texttt{Shear force}}{\texttt{Area}}\\\\3\times 10^6=\frac{50000}{0.1d}\\\\d=0.1667m=16.67cm

Width of beam = 16.67 cm

Option B is the correct answer.

6 0
2 years ago
A uniform sphere with mass M and radius R is rotating with angular speed ω1 about a frictionless axle along a diameter of the sp
liq [111]

Answer:

W_2=\sqrt{\frac{3}{5} }W_1

Explanation:

For the first ball, the moment of inertia and the kinetic energy is:

I_1 =\frac{2}{5}MR^2

K_1 = \frac{1}{2}IW_1^2

So, replacing, we get that:

K_1 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

At the same way, the moment of inertia and kinetic energy for second ball is:

I_2 =\frac{2}{3}MR^2

K_2 = \frac{1}{2}IW_2^2

So:

K_2 = \frac{1}{2}(\frac{2}{3}MR^2)W_2^2

Then, K_2 is equal to K_1, so:

K_2 = K_1

\frac{1}{2}(\frac{2}{3}MR^2)W_2^2 = \frac{1}{2}(\frac{2}{5}MR^2)W_1^2

\frac{1}{3}MR^2W_2^2 = \frac{1}{5}MR^2W_1^2

\frac{1}{3}W_2^2 = \frac{1}{5}W_1^2

Finally, solving for W_2, we get:

W_2=\sqrt{\frac{3}{5} }W_1

5 0
2 years ago
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