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Vika [28.1K]
2 years ago
10

People who drink coffee are more likely to develop cancer

Physics
2 answers:
krok68 [10]2 years ago
8 0

Answer:

The answer is False

Rashid [163]2 years ago
8 0
False

Hope this helps
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A small metal sphere has a mass of 0.19 gg and a charge of -23.0 nCnC. It is 10.0 cmcm directly above an identical sphere that h
RoseWind [281]

Answer:

  a = -7.29 m / s²

Explanation:

For this exercise we must use Newton's second law,

          F -W = m a

Force is electrical force

         F = k q₁ q₂ / r²

         k q₁ q₂ / r² -mg = m a

indicate that the charge of the two spheres is equal

         q₁ = q₂ = q

         a = (k q² / r² - m g) / m

         a = k q² / m r² - g

Let's reduce the magnitudes to the SI system

        m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg

        q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C

        r = 10.0 cm (1m / 100cm) = 0.1000 m

let's calculate

        a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8

        a = -7.29 m / s²

The negative sign indicates that the direction of this acceleration is downward

6 0
2 years ago
The early workers in spectroscopy (Fraunhofer with the solar spectrum, Bunsen and Kirchhoff with laboratory spectra) discovered
Anestetic [448]

The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.

<h3><u>Explanation: </u></h3>

A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.

Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen.  "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.

The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.

8 0
3 years ago
Si pudieras viajar a la Luna:
Nostrana [21]

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).

La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

4 0
3 years ago
Match the correct term with the statement that is true about it
Sholpan [36]
Do you have an image?
5 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
2 years ago
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