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Yakvenalex [24]
2 years ago
10

In our simple system, what type of energy was used?

Physics
1 answer:
seropon [69]2 years ago
6 0

Type of energy was used is kinetic energy.

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The kinetic energy of a sliding block came from the:
patriot [66]

Answer:

Correct sentence: gravitational potential energy of the mass on the hook.

Explanation:

The mechanical energy of a body or a physical system is the sum of its kinetic energy and potential energy. It is a scalar magnitude related to the movement of bodies and to forces of mechanical origin, such as gravitational force and elastic force, whose main exponent is Hooke's Law. Both are conservative forces. The mechanical energy associated with the movement of a body is kinetic energy, which depends on its mass and speed. On the other hand, the mechanical energy of potential origin or potential energy, has its origin in the conservative forces, comes from the work done by them and depends on their mass and position. The principle of conservation of energy relates both energies and expresses that the sum of both energies, the potential energy and the kinetic energy of a body or a physical system, remains constant. This sum is known as the mechanical energy of the body or physical system.

Therefore, the kinetic energy of the block comes from the transformation in this of the gravitational potential energy of the suspended mass as it loses height with respect to the earth, keeping the mechanical energy of the system constant.

3 0
3 years ago
When a wave has λ=3 m and f=15 Hz, what is the speed of the wave?
Tamiku [17]

Wave speed = (wavelength) x (frequency)

Wave speed = (3 m) x (15 Hz)

<em>Wave speed = 45 m/s</em>

5 0
2 years ago
What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?
Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

#SPJ4

5 0
11 months ago
What if m1 is initially moving at 3.4 m/s while m2 is initially at rest? (a) find the maximum spring compression in this case?
Lisa [10]
<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J max spring compression where both velocities are the same: conserve momentum: 1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J The remaining energy went into the spring: U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x² x = 0.0076 m ↠(a)</span>
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
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