All categories

2 years ago

How can a wave carry information?

2 answers:

7 0

A transmitter “encodes” or modulates messages by varying the amplitude or frequency of the wave – a bit like Morse code. At the other, a receiver tuned to the same wavelength picks up the signal and 'decodes' it back to the desired form

I think it’s A or D

Katen [24]2 years ago

4 0

Answer:

i think its A

Explanation:

You might be interested in

What type of wave is infrared light

GrogVix [38]

Answer: Heat waves? I’m not %100 sure

4 0

2 years ago

Read 2 more answers

A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W

Basile [38]

I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>

3 0

2 years ago

To heat the house, the boiler transfers 15 MJ of energy in 10 minutes.

Harrizon [31]

The formula is P = E/t, where P means power in watts, E means energy j , and t means time in seconds. This formula states that power is the consumption of energy per unit of time.

P = 15 M / 10*60

M = mega = 10⁶

15 *10⁶ / 600

= 25000 watt

8 0

2 years ago

If it requires 5.0 j of work to stretch a particular spring by 1.8 cm from its equilibrium length, how much more work will be re

ANEK [815]

15.277.. j. I did the problem using a proportion. an additional 3.7m to the current 1.8 cm=5.5cm.

Therefore, 5.0 j/1.8cm=x/5.5cm

5 0

3 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago