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3 years ago

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Why do remote controls for TV’s use infrared waves to communicate?

1 answer:

Delvig [45]3 years ago

3 0

Answer: An IR remote (also called a transmitter) uses light to carry signals from the remote to the device so it can be controlled. It emits pulses of invisible infrared light that correspond to specific binary codes. These codes represent commands, such as power on, volume up, or channel down.

Explanation:

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An 84% efficient single pulley is used to lift a 230 kg piano 3.5 m. How much work must be input?

Sati [7]

Answer:

35%

Explanation:

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3 years ago

Arlecino [84]

i think the data is not complete but that's according to me

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3 years ago

Please help me it's URGENT (#10 btw and also how do I solve for time

Afina-wow [57]

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

5 0

3 years ago

A cloud of interstellar gas is rotating. Because the gravitational force pulls the gas particles together, the cloud shrinks, an

Fynjy0 [20]

Answer:

Greater than

Explanation:

Here, angular momentum is conserved.

$l_1\omega_1 =l_2\omega_2$

When the cloud shrinks under the right conditions, a star may be formed.

Thus, Diameter of clouds are much higher than a star.

Moment of inertia of cloud is greater than the star's inertial.

so, angular velocity of the star would be greater than angular velocity of the rotating gas.

3 0

4 years ago

Read 2 more answers

A thin taut string is fixed at both ends and stretched along the horizontal x-axis with its left end at x = 0. It is vibrating i

Fofino [41]

Answer:

(a) Wavelength is 0.436 m

(b) Length is 0.872 m

(c) 11.518 m/s

Solution:

As per the question:

The eqn of the displacement is given by:

$y(x, t) = (1.22 cm)sin[14.4 m^{- 1}x]cos[(166\ rad/s)t]$ (1)

n = 4

Now,

We know the standard eqn is given by:

$y = AsinKxcos\omega t$ (2)

Now, on comparing eqn (1) and (2):

A = 1.22 cm

K = $14.4 m^{- 1}$

$\omega = 166\ rad/s$

where

A = Amplitude

K = Propagation constant

$\omega$ = angular velocity

Now, to calculate the string's wavelength,

(a) $K = \frac{2\pi}{\lambda}$

where

K = propagation vector

$\lambda = \frac{2\pi}{K}$

$\lambda = \frac{2\pi}{14.4} = 0.436\ m$

(b) The length of the string is given by:

$l = \frac{n\lambda}{2}$

$l = \frac{4\times 0.436}{2} = 0.872\ m$

(c) Now, we first find the frequency of the wave:

$\omega = 2\pi f$

$f = \frac{\omega}{2\pi}$

$f = \frac{2\pi}{166} = 26.42\ Hz$

Now,

Speed of the wave is given by:

$v = f\lambda$

$v = 26.419\times 0.436 = 11.518\ m/s$

4 0

3 years ago