Answer:
a. 125 kJ
Explanation:
Her total energy is the same as the potential energy she had at the top of the hill:
PE = mgh
= (52 kg)(9.8 m/s^2)(245 m) = 124,852 J
≈ 125 kJ . . . . matches choice A
_____
After skiing down 112 m, some of her initial energy is converted to kinetic energy, and some remains as potential energy. We assume the ski slope is essentially frictionless, and air resistance is negligible.
Answer:
1)
Force on bar magnet = 0
Torque on bar magnet = 0
2)
Force on bar magnet = 0
Torque on bar magnet = 0.177 Nm
3)
Force on bar magnet = 0
Torque on bar magnet = 0.25 Nm
Explanation:
Part 1)
net force on bar magnet in uniform magnetic field is always zero
Torque on bar magnet is given as

when bar magnet is inclined along z axis along magnetic field
then we will have

Part 2)
net force on bar magnet in uniform magnetic field is always zero
Torque on bar magnet is given as

when bar magnet is pointing 45 degree with z axis then we will have



Part 3)
net force on bar magnet in uniform magnetic field is always zero
Torque on bar magnet is given as

when bar magnet is pointing 90 degree with z axis then we will have



Answer:
A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
, this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
we have a fluctuation of the intensity emitted by the stars. Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
Explanation:
The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.
λ T = 2,898 10⁻³
therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm
The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.
The burst with A = 4300K has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.
Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.
When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
To find the answer, we have to know about the rules followed by drawing ray-diagram.
<h3>What are the rules obeyed by light rays?</h3>
- If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
- The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
- The refracted ray follows the same path if the incident light passes through the center of the curve.
Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.
Learn more about refraction by a lens here:
brainly.com/question/13095658
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