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3 years ago

Which of these are terrestrial planets? Select all that apply.

2 answers:

5 0

The answers to the question above would be letters a, b, c, and d. The terrestrial planets are those that are close to the sun, namely, Mercury, Venus, Earth, and Mars. The others are known as the gas giants since they have a thick atmosphere and are considerably cold.

lawyer [7]3 years ago

4 0

Answer:

<h2>a. Mars </h2><h2>b. Earth </h2><h2>c. Venus</h2>

Explanation:

Terrestrial planets are defined as planets which are formed by silicate. An important characteristic is that these planets have a solid surface. On the other hand, there are gaseous planets, which are bigger and more separated of the Sun.

In our Solar system, there are 4 terrestrial planets, which are Mercury, Venus, Earth, and Mars. Notice that they are the inner planets of the system.

So, in this case, the right chice is a. Mars, b. Earth and c. Venus. All three are terrestrial planets, that is, they have a rock solid surface.

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A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the radio wave in this problem,

$\lambda = 0.3 m$

$c=300,000,000 m/s = 3\cdot 10^8 m/s$

Therefore, the frequency is:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz$

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4 0

2 years ago

Which two formulas are used to calculate potential and kinetic energy

goblinko [34]

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

6 0

2 years ago

Read 2 more answers

An electron that has a velocity with x component 2.6 × 106 m/s and y component 3.2 × 106 m/s moves through a uniform magnetic fi

elena55 [62]

Answer:

a) $\vec F_{B} = 8.766\times 10^{-14}\,T\,k$ , b) $\vec F_{B} = -8.766\times 10^{-14}\,T\,k$

Explanation:

a) The magnetic force experimented by a particle has the following vectorial form:

$\vec F_{B} = q\cdot \vec v \times \vec B$

The charge of the electron is equal to $-1.602\times 10^{-19}\,C$ . Then, cross product can be solved by using determinants:

$\vec F_{B} = \begin{vmatrix}i&j&k\\-4.165\times 10^{-13}\, C\cdot \frac{m}{s} &-5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$

The magnetic force is:

$\vec F_{B} = 8.766\times 10^{-14}\,T\,k$

b) The charge of the proton is equal to $1.602\times 10^{-19}\,C$ . Then, cross product has the following determinant:

$\vec F_{B} = \begin{vmatrix}i&j&k\\4.165\times 10^{-13}\, C\cdot \frac{m}{s} &5.126\times 10^{-13}\,C\cdot \frac{m}{s} &0\,C\cdot \frac{m}{s} \\0.041\,T&-0.16\,T&0\,T\end{vmatrix}$