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stiks02 [169]
3 years ago
7

In which situation is chemical energy being converted to another form of energy?

Physics
1 answer:
otez555 [7]3 years ago
5 0

Answer:

A burning candle. (chemical energy into energy of heat and light, i.e. thermal and wave)

Explanation:

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Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds th
Oksi-84 [34.3K]

Answer:

The  frequencies are  f_n  =  n (0.875 )

Explanation:

From the question we are told that

   The speed of the wave is  v  =  0.700 \  m/s

   The  length of vibrating  clothesline is  L  =  40.0 \  cm = 0.4 \ m

Generally the fundamental frequency is  mathematically represented as

        f =  \frac{v}{2 L  }

=>     f =  \frac{ 0.700 }{2 *  0.4   }

=>     f =  0.875 \  Hz

Now  this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of  the fundamental frequency

So  

   The  frequencies are mathematically represented as

       f_n  =  n  * f

=>     f_n  =  n (0.875 )

Where  n  =  1, 2, 3 ....

       

3 0
3 years ago
You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.
valentinak56 [21]

Answer:

F=6\times 10^{-7}\ N

Explanation:

Given:

  • quantity of point charge, q=-4\times 10^{-9}\ C
  • radial distance from the linear charge, r=0.09\ m
  • linear charge density, \lambda=3\times 10^{-9}\ C.m^{-1}

<u>We know that the electric field by the linear charge  is given as:</u>

E=\frac{\lambda}{2\pi.\epsilon_0.r}

E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}

E=150\ N.C^{-1}

<u>Now the force on the given charge can be given as:</u>

F=E.q

F=150\times 4\times 10^{-9}

F=6\times 10^{-7}\ N

3 0
3 years ago
0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc
weqwewe [10]

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

4 0
3 years ago
PLS HELP ME AS QUICK AS POSSIBLE,
Levart [38]

PLS HELP ME AS QUICK AS POSSIBLE,

THANKS :)) I'm a bit confused

Can you answer 1 and 2, then confirm 3 :))))

4 0
2 years ago
Read 2 more answers
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
3 years ago
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