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Nataly [62]
3 years ago
8

What is one disadvantage of sending information over long distances

Physics
2 answers:
xxTIMURxx [149]3 years ago
5 0

Answer:

The answer is A. The signals become weaker the farther the receiver is from the source.  Because and this is common sense signals can't travel for ever.

Explanation:

A P E X

P L E A S E    L I K E

shusha [124]3 years ago
4 0

Answer:

A is the correct answer.

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Which statements describe the motion of the object represented by the graph? Check all that apply.
Sav [38]

Answer:a, d, e

Explanation:

6 0
3 years ago
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If a metal has a mass of 78.2 g and a volume of 11.4 mL, what is its density?
hram777 [196]

Answer:

6.86g/mL

Explanation:

4 0
3 years ago
Gold, which has a mass of 19.32 g for each cubic centimeter of volume, is the most ductile metal and can be pressed into a thin
Dennis_Churaev [7]

Answer:

area = 5733.33  cm²

length = 5.47 ×10^{7} cm

Explanation:

Given data

density = 19.32 g/cm³

mass = 33.16 g

thickness = 3.000 µm = 3 ×10^{-4} cm

radius r = 1.000 µm = 1 ×10^{-4} cm

to find out

area of the leaf and  length of the fiber

solution

we know volume formula that is

volume = mass / density

volume = 33.16 /  19.32

volume = 1.72 cm³

we know that volume = thickness × area

so area

area = volume / thickness

area = 1.72 / 3 ×10^{-4}

area = 5733.33  cm²

and

we know volume = πr²L

so L = volume /  πr²

length = 1.72 / π(1×10^{-4})²

length = 5.47 ×10^{7} cm

3 0
3 years ago
a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc
Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W

So, 0.25 W power is delivered to the bulb.

5 0
3 years ago
A person hits a tennis ball with a mass of 0.058 kg against a wall.
horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.058\times (11-(-11))}{2.1}

F = 0.607 N

(b) Area of wall, A=3\ m^2

Let P is the average pressure on that area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{0.607\ N}{3\ m^2}

P = 0.202 Pa

Hence, this is the required solution.

8 0
3 years ago
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