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Fittoniya [83]
4 years ago
8

Can saturn be seen at night without a telescope?

Physics
2 answers:
Gnoma [55]4 years ago
5 0

yes it can be observed at night during the right time.

irina1246 [14]4 years ago
4 0

yes but its harder without

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En un rio una Onda viaja con una velocidad de propagación de 50 m/s con una longitud de Onda de 40 metros. Hallar la frecuencia
sattari [20]

Answer:

Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

  • Velocidad = 50 m/s
  • Longitud de onda = 40 metros

Para encontrar la frecuencia de la onda;

Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

Haciendo de la frecuencia el tema de la fórmula, tenemos;

Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

Sustituyendo en la fórmula, tenemos;

Frequencia = \frac {50}{40}

<em>Frequencia = 1.25 Hz</em>

7 0
3 years ago
Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.
andreev551 [17]

Answer:

the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = m_A

then, the mass of object B = m_B = \frac{m_A}{4}

work done on object A = -18 J

work done on object B = -18 J

let v_i be the initial speed

let v_f be the final speed

For object A;

K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2  = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ -  v_i^2 )\ =- 18\\\\v_f^2 \ -  v_i^2  = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ -  v_i^2  = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\

v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\

Thus, the final speed of object A changed by a factor of  \frac{1}{\sqrt{3} } = 0.58

To obtain the change in the final speed of object B, apply the following equations.

K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B =  \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\

\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i

Thus, the final speed of object B changed by a factor of \sqrt{\frac{5}{3} } = 1.29

3 0
3 years ago
What magnet maintains its magnetic proaperties even in the absence of an external magnetic field? a) Ferromagnet b Paramagnet c)
insens350 [35]

Answer:

(a) Ferromagnet

Explanation:

Ferromagnetism is defined as the property by which certain magnets form the permanent magnets.

It is tone of the strong magnetism and it is common phenomenon of magnet in the everyday life of magnetism.

Permanent magnets are made up of ferromagnetic material, in this if the magnetic field is applied then this material is magnetized but do not losses its magnetic property after removal of external magnetic field.

5 0
4 years ago
A certain AM radio wave has a frequency of 1.12 x 100 Hz. Given that radio waves travel at
riadik2000 [5.3K]

Answer: 267 m

Explanation:

2.99x10^8 m/s

———————-

1.12 x 10^6 Hz

3 0
3 years ago
What is the equation for momentum
Marina CMI [18]

Answer:

p=mv

Explanation:

i looked it up.

5 0
3 years ago
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