The radius of Venus (from the center to just above the atmosphere) is 6050 km (6050✕103 m), and its mass is 4.9✕1024 kg. An obje
ct is launched straight up from just above the atmosphere of Venus. (a) What initial speed is needed so that when the object is far from Venus its final speed is 8000 m/s? vinitial = m/s (b) What initial speed is needed so that when the object is far from Venus its final speed is 0 m/s? (This is called the "escape speed.") vescape = m/s
1 answer:
Answer:
(a) The initial speed required is 13116 m/s
(b) The escape speed is 10394 m/s
This problem involves the application of newtons laws of gravitation. The forces in action here are conservative and as a result mechanical energy is conserved.
The full calculation can be found in the attachment below.
Explanation:
In both parts (a) and (b) the energy conservation equation were used. Assumption was made that when the object is very far from the planet the distance from the planet's center approaches infinity and the gravitational potential energy approaches zero.
The calculation can be found below.
You might be interested in
<h2>5.3 km</h2>
Explanation:
This question involves continuous displacement in various directions. When it becomes difficult to imagine, vector analysis becomes handy.
Let us denote each of the individual displacements by a vector. Consider the unit vectors as the unit vectors in the direction of East and North respectively.
By simple calculations, we can derive the unit vectors in the directions North,
South of West and
North of West respectively.
So Total displacement vector = Sum of individual displacement vectors.
Displacement vector =
Magnitude of Displacement =
∴ Total displacement =
Answer:
b. AG, work function=4.74eV
Explanation:
Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:
(wavelength of lowest-energy ultraviolet light)
So, the lowest energy of ultraviolet light can be found by using the formula
where
h is the Planck constant
c is the speed of light
Substituting,
And keeping in mind that
This energy converted into electronvolts is
The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:
b. AG, work function=4.74eV
Because for all the other metals, visible light will be enough to extract photoelectrons.