<span>After an exoplanet has been identified using a given detection method, scientists attempt to identify the basic properties of the planet which can tell us what it might be made of, how hot it might be, whether or not it contains an atmosphere, how that atmosphere might behave, and finally, whether the planet may be suitable for life. It is often useful to first determine basic properties of the parent star (such as mass and distance from the Earth). This is then followed by the use of planetary detection methods to calculate planetary mass, radius, orbital radius, orbital period, and density. The density calculation will provide clues as to what the planet is made of and whether or not it contains a significant atmosphere.
Mass and Distance of Parent Star
The mass and distance of an exoplanet's parent star must often be calculated first, before certain measurements of the exoplanet can be made. For example, determining the star's distance is an important step in determining a star's mass (see below). Knowing the mass of a star then allows the mass of the planet to be measured, for example when using the Radial Velocity Method.</span>
A compound Machine is 2 machines that work together in order to make a task easier.
Answer: Option d.)
R_eq = (1/R_1 + 1/R_2 + 1/R_3)^-1
Explanation:
Since, there are three resistors connected in parallel, the reciprocal of the total resistance of the resistor combination (R_eq) is obtained by adding the reciprocal of each resistance.
i.e 1/R_eq = (1/R_1 + 1/R_2 + 1/R_3)
So, R_eq = (1/R_1 + 1/R_2 + 1/R_3)^-1
Thus, the total resistance (R_eq) is equal to the inverse of the sum of the reciprocal of each resistance.
Answer:
E = 1.64 x 10⁶ V/m
Explanation:
The electric field in the region between the plates can be given by the following formula:
where,
E = Electric Field = ?
ΔV = Poetential Difference across the plates = 23 KV = 23000 V
d = distance between plates = 1.4 cm = 0.014 m
Therefore, using these values in the equation, we get:
<u>E = 1.64 x 10⁶ V/m</u>
Answer:
1.87 A
Explanation:
τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s
d = diameter of copper wire = 2 mm = 2 x 10⁻³ m
Area of cross-section of copper wire is given as
A = (0.25) πd²
A = (0.25) (3.14) (2 x 10⁻³)²
A = 3.14 x 10⁻⁶ m²
E = magnitude of electric field = 0.01 V/m
e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
m = mass of electron = 9.1 x 10⁻³¹ kg
n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³
= magnitude of current
magnitude of current is given as
= 1.87 A
