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3 years ago

Identify the properties of transverse and longitudinal waves.

Physics

2 answers:

Drupady [299]3 years ago

7 0

Answer:

1. C

2. A

3. C

4. C

5. C

Explanation:

i did the question on edge 2020 :>

Alona [7]3 years ago

5 0

Amplitude: both longitudinal and transverse (option c)

Compression: longitudinal (option a)

Frequency: both longitudinal and transverse (option c)

Period: both longitudinal and transverse (option c)

Wavelength: both longitudinal and transverse (option c)

hope this helps

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A bullet with a mass m b = 11.5 g is fired into a block of wood at velocity v b = 249 m/s. The block is attached to a spring tha

Ostrovityanka [42]

Answer:

0.358Kg

Explanation:

The potential energy in the spring at full compression = the initial kinetic energy of the bullet/block system

0.5Ke^2 = 0.5Mv^2

0.5(205)(0.35)^2 = 12.56 J = 0.5(M + 0.0115)v^2

Using conservation of momentum between the bullet and the block

0.0115(265) = (M + 0.0115)v

3.0475 = (M + 0.0115)v

v = 3.0475/(M + 0.0115)

plugging into Energy equation

12.56 = 0.5(M + 0.0115)(3.0475)^2/(M + 0.0115)^2

12.56 = 0.5 × 3.0475^2 / ( M + 0.0115 )

12.56 = 0.5 × 9.2872/ M + 0.0115

12.56 = 4.6436/ M + 0.0115

12.56 ( M + 0.0115 ) = 4.6436

12.56M + 0.1444 = 4.6436

12.56M = 4.6436 - 0.1444

12.56 M = 4.4992

M = 4.4992÷12.56

M = 0.358 Kg

4 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

The wetter the surface, the friction

allochka39001 [22]

Answer: Water can either increase or decrease the friction between surfaces.

7 0

3 years ago

If the pressure exerted by the liquid of density 500kg/m3 is 20,000Pa. What is height of liquid column.

gregori [183]

The height of the liquid column is 4.08 metres.

4 0

3 years ago

A5.0 kg TNT explosive, initially at rest, explodes into two pieces. One of the pieces weighing 2.0 kg flies off to

nordsb [41]

Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

It explodes into two pieces.

One of the pieces weighing 2.0 kg flies off to the left at 36 m/s. Let left be negative and right be positive.

The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

$5\times 0=2\times (-36)+3\times v\\\\-72=-3v\\\\v=24\ m/s$

So, the final speed of the remaining piece is 24 m/s and it is in the right direction.

4 0

3 years ago