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2 years ago

True or False: Modulated waves can be analog or digital.

Physics

1 answer:

Firdavs [7]2 years ago

6 0

Answer:

True

Explanation:

Modulation waves(schemes) can be analog or digital. An analog modulation scheme has an input wave that varies continuously like a sine wave.

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PLEASE HELP! LAST DAY TO TURN IN AND IM SO BEHIND!!!

kolbaska11 [484]

<h3>Question 1</h3>

Answer

option C) velocity

Explanation

acceleration = Δv ÷ Δt

<h3>Question 2</h3>

Answer

option C) m/s²

Explanation

Δv ÷ Δt

= m/s ÷ s

= m/s x 1/s

= m/s²

<h3>Question 3</h3>

Answer

option B) velocity has both direction and speed.

That is why velocity can be negative but speed can not and velocity is rate of change of displacement where as speed is rate of change of distance.

7 0

3 years ago

Read 2 more answers

Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i

Tanzania [10]

The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
$v= \frac{d}{t}$ ⇒ $d=v*t$

It is given:
$v = \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s} = 3.2 m/s$
t = 10 s
d = ?

So:
$d= v*t=3.2m/s*10s = 32m$

3 0

2 years ago

When you measure something in meters cubed, you are measuring ____.

Shalnov [3]

I think your answer is volume

7 0

2 years ago

Read 2 more answers

Waves on a pond are an example of which kind of wave?

KATRIN_1 [288]

<span>
</span><span>Waves on a pond are an example of which kind of wave?

</span>B. surface waves

6 0

3 years ago

When cars are equipped with flexible bumpers, they will bounce off each other during low-speed collisions, thus causing less dam

Len [333]

Answer:

Explanation:

Given that,

Mass of the heavier car m_1 = 1750 kg

Mass of the lighter car m_2 = 1350 kg

The speed of the lighter car just after collision can be represented as follows

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\frac{(1850)(1.4)+(1450)(-1.10)-(1850)(0.250)}{1450} \\\\=\frac{2590+(-1595)-(462.5)}{1450} \\\\=\frac{2590-1595-462.5}{1450} \\\\=\frac{532.5}{1450}\\\\=0.367m/s$

b) the change in the combined kinetic energy of the two-car system during this collision

$\Delta K.E=(\frac{1}{2} m_1v_1^2+\frac{1}{2} m_2v_2^2)-(\frac{1}{2} m_1u_1^2+\frac{1}{2} m_2u_2^2)\\\\=\frac{1}{2} (m_1(v_1^2-u_1^2)+m_2(v_2^2-u_2^2))$

substitute the value in the equation above

$=\frac{1}{2} (1850((0.250)^2-(1.4)^2)+(1450((0.3670)^2-(-1.10)^2)\\\\=\frac{1}{2}(11850(0.0625-1.96)+(1450(0.1347)-(1.21))\\\\= \frac{1}{2}(11850(-1.8975))+(1450(-1.0753))\\\\=\frac{1}{2} (-3510.375+(-1559.185)\\\\=\frac{1}{2} (-5069.56)\\\\=-2534.78J$

Hence, the change in combine kinetic energy is -2534.78J

8 0

3 years ago