Answer:
The value of g in Kepler-12b is 
Explanation:
To determine the value of g at Kepler-12b it is necessary to combine the equation of the weight and the equation for the Universal law of gravity:
(1)
Where m is the mass and g is the value of the gravity
(2)
Equation (1) and equation (2) will be equal, since the weight is a force acting on the object as a consequence of gravity:
(3)
Then g will be isolated from equation 3:
(4)
The radius of Jupiter has a value of 69911000 meters, so its diameter can be determined by:
(5)
Where d and R are the diameter and radius of Jupiter
Procedure for finding the radius of Kepler-12b:
For the case of Kepler-12b it has a diameter that is 1.7 times that of Jupiter
By means of equation 5 the radius of Kepler-12b can be known:
Procedure for finding the mass of Kepler-12b:
The mass of Jupiter has a value of 
kilograms
For the case of Kepler-12b it has a mass that is 0.43 times that of jupiter
The value of g in Kepler-12b can be found by replacing its radius and mass in equation 4:
Hence, in Kepler-12b the gravity has a value of
Answer:
a. 4.733 × 10⁻¹⁹ J = 2.954 eV b i. yes ii. 0.054 eV = 8.651 × 10⁻²¹ J
Explanation:
a. Find the energy of the incident photon.
The energy of the incident photon E = hc/λ where h = Planck's constant = 6.626 × 10⁻³⁴ Js, c = speed of light = 3 × 10⁸ m/s and λ = wavelength of light = 420 nm = 420 × 10⁻⁹ m
Substituting the values of the variables into the equation, we have
E = hc/λ
= 6.626 × 10⁻³⁴ Js × 3 × 10⁸ m/s ÷ 420 × 10⁻⁹ m
= 19.878 × 10⁻²⁶ Jm ÷ 420 × 10⁻⁹ m
= 0.04733 × 10⁻¹⁷ J
= 4.733 × 10⁻¹⁹ J
Since 1 eV = 1.602 × 10⁻¹⁹ J,
4.733 × 10⁻¹⁹ J = 4.733 × 10⁻¹⁹ J × 1 eV/1.602 × 10⁻¹⁹ J = 2.954 eV
b. i. Is this energy enough for an electron to leave the atom
Since E = 2.954 eV is greater than the work function Ф = 2.9 eV, an electron would leave the atom. So, the answer is yes.
ii. What is its maximum energy?
The maximum energy E' = E - Ф = 2.954 - 2.9
= 0.054 eV
= 0.054 × 1 eV
= 0.054 × 1.602 × 10⁻¹⁹ J
= 0.08651 × 10⁻¹⁹ J
= 8.651 × 10⁻²¹ J
On the dog's return trip (between <em>t</em> = 10 and <em>t</em> = 12.5 seconds), the slope of the position function is steeper than during the first 5 seconds, which means the dog ran home faster. The only option that captures this is D.
You can check to make sure that the dog indeed runs twice as fast on the return trip. The slope of the position function during the first 5 seconds is
(change in position) / (change in time) = (5 - 0) / (5 - 0) = 5/5 = 1
while during the return trip, it is
(0 - 5) / (12.5 - 10) = -5/2.5 = -2
Ignoring the sign (which only indicates the direction in which the dog was running), we see that the dog's speed on the return trip was indeed twice as high as during the first 5 s.
Answer:
C , E , A , D , B
Explanation:
We evaluate the accelerations for each case, using the formula: a = (vf - vi) / t
A) a = (10.3 - 0.5 ) / 1 = 9.8 m/s^2 --> magnitude: 9.8 m/s^2
B) a = (0 - 20) / 1 = - 20 m/s^2 --> magnitude : 20 m/s^2
C) a = (0.02 - 0.004) / 1 = 0.016 m/s^2 --> magnitude : 0.016 m/s^2
D) a = (4.3 - 0) / 0.4 = 10.75 m/s^2 --> magnitude : 10.75 m/s^2
E) a = (1 - 2) / 8.3 = - 0.12 m/s^2 --> magnitude: 0.12 m/s^2
Then, comparing magnitudes from least to greatest:
C , E , A , D , B
