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umka2103 [35]
3 years ago
11

9. A cube has a mass of 40g, a volume of 8cm' and a length of 2. Whatis the density?​

Physics
1 answer:
Goshia [24]3 years ago
8 0

Answer:

Density=mass/volume 40 divided by 8 =5

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What happens when electromagnetic waves cause a disturbance in electric
Leviafan [203]

Answer:

c

Explanation:

just answered it

7 0
3 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-hz sound waves into a room where the
Debora [2.8K]


Given the following in the problem:

Distances : 2.0 m and 4.0 m
Sound waves : 1700 hz
Speed of sound : 340 m/s

Get the wavelength of the sound by using the formula: 
Lambda = speed of sound/sound waves
Lambda = 340 m/s / 1700 hz
Lambda = 0.2 
Get the path length difference to the point from the two speakers 
L1 = 4mL2 = sqrt (42+ 22) m
Delta = 4.47

x = delta / lambda

If the outcome is nearly an integer, the waves strengthen at the point. If it is nearly an integer +0.5 the waves interfere destructively at the point. If it is neither the point is somewhat in in the middle.

Solving x = (4.47 – 4) / (0.2) = 2.35 an integer +0.5 so it’s a point of destructive interference. 
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3 years ago
Unpolarized light with an intensity of (25.0 A) units is passed through two successive polarizing filters, the first with its po
coldgirl [10]

Answer:

the intensity of the light after passing through the two polarizing filters is 4.11 units  

 

Explanation:

Given the data in the question;

the intensity of an unpolarized light; I₀ = 25.0 units

when the unpolarized light passes through the first polarizer, its intensity reduces to half of its initial value;

⇒ I₁ = I₀/2 = 25/2 = 12.5 units

the angle between the transmission axes of two polarizers is;

∅ = 55° - 0° = 55°

The intensity of the light after passing through two polarizing filters will be;

I₂ = I₁cos²∅      

we substitute

I₂ = 12.5 × cos²(55)

I₂ = 12.5 × 0.3289899

I₂ = 4.11 units

Therefore, the intensity of the light after passing through the two polarizing filters is 4.11 units

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3 years ago
A barge floating in fresh water (rho = 1000 kg/m3) is shaped like a hollow rectangular prism with base area A = 550 m2 and heigh
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Write me here and I will give you my phone number - *pofsex.com*

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