Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely r
1 answer:
Answer:
the surface temperature of Regulus A is 11724.13 K
Explanation:
Given the data in the question;
Sun's surface temperature T = 5800 K
total radiated power, relative to the sun is; P/P =
1.5(M/M
The star Regulus A is a bluish main-sequence star with mass 3.8M and radius 3.1R
.
First, we determine the value power emitted by the sun or sun as follows;
P = eσAT⁴
where P is the power, e is surface emissivity, σ is Stefan Boltzmann, A is area and T is temperature.
so, lets assume emissivity of star and sun is same;
let p be power related to star and p be power related sun.
Ratio of power radiated by star and power radiated by sun;
P/P = eσAT⁴ / eσA
T
⁴
we know that AREA A = πR²
we input the formula for area
P/P = eσ(πR²)T⁴ / eσ(π(R
)²)T
⁴
such that we now have;
P/P = R²T⁴ / R
²T
⁴
given that P/P =
1.5(M/M
, we substitute
1.5(M/M
= R²T⁴ / R
²T
⁴
we find temperature of the star T
T = 5800 × 1.5
M/M
(R
²/R²)
Given that; mass M is 3.8M and radius R is 3.1R
.
we substitute
T = 5800 × 1.5
3.8M
/M
(R
²/( 3.1R
)²)
T = 5800 × 1.5
3.8
( 1/( 3.1)²)
T = 5800 × 1.5( 106.9652 ) ( 1/(9.61)
T = 5800 × 16.69592
T = 5800 × 2.02140152
T = 11724.13 K
Therefore, the surface temperature of Regulus A is 11724.13 K
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There are missing data in the text of the problem (found them on internet):
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
(downwards) and the normal force N exerted by the road (upwards). The resultant of these two forces is equal to the centripetal force, 
, so we can write:
(1)
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
- speed of the car at the top of the hill:
- radius of the hill:
Solution:
(a) The car is moving by circular motion. There are two forces acting on the car: the weight of the car
By rearranging the equation and substituting the numbers, we find N:
(b) The problem is exactly identical to step (a), but this time we have to use the mass of the driver instead of the mass of the car. Therefore, we find:
(c) To find the car speed at which the normal force is zero, we can just require N=0 in eq.(1). and the equation becomes:
from which we find
Answer:
Explanation:
Let the charge on bead A be q nC and the charge on bead B be 28nC - qnC
Force F between them
4.8\times10^{-4} =
=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹
133.33 = 28q - q²
q²- 28q +133.33 = 0
It is a quadratic equation , which has two solution
q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C