We need to directly measure the spectral type in order to determine the surface temperature of a star.
<h3>How do you find the properties of a star?</h3>
Astronomers can determine the temperature of a star by looking at its color and spectrum. The apparent brightness of a star describes how luminous it looks to us. The brightness of a star tells us how bright it really is. The luminance can be determined using both the perceived brightness and the distance.
A star's luminosity, or the total amount of energy it emits each second, is determined by two factors: The stellar photosphere's "Effective Temperature," T. the star's total surface area, which is influenced by its radius, R.
Because it controls how much fuel a star has and how quickly it burns it, a star's mass is its most fundamental characteristic. The majority of a star's life is spent burning hydrogen into helium in its core, which generates energy. The star needs to achieve a balance between gravity and outward pressure in order to continue to be "alive."
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Answer:yes
Explanation:The constan acceleration means that it wont stop moving but if you kick it a different direction then it will change direction
Answer:

Explanation:
Given that,
Initial angular velocity, 
Acceleration of the wheel, 
Rotation, 
Let t is the time. Using second equation of kinematics can be calculated using time.

Let
is the final angular velocity and a is the radial component of acceleration.

Radial component of acceleration,

So, the required acceleration on the edge of the wheel is
.
Answer:

Explanation:
It is given that, a proton moves at constant velocity, through a region in which there is an electric field and a magnetic field such that,
The electric field is, E = 800 V/m
Magnetic field, B = 0.25 T
We know that the net force in the region of magnetic and electric field is given by Lorentz forces. But here, the proton moves with constant velocity. So, the net force acting on it is 0.
i.e.

Hence, this is the required solution.
Answer:
0.661 s, 5.29 m
Explanation:
In the y direction:
Δy = 2.14 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.661 s
In the x direction:
v₀ = 8 m/s
a = 0 m/s²
t = 0.661 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²
Δx = 5.29 m
Round as needed.