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Julli [10]
3 years ago
15

A Cephied is a star that belongs to a class of yellow supergiant pulsating stars; these stars vary periodically in brightness, a

nd the relationship between their periods and ________ is useful in deriving distances to them
Physics
1 answer:
mihalych1998 [28]3 years ago
4 0

Luminosities

Thanks to this relationship between period and luminosity, a Cepheid provide a practical and accurate method to evaluate their absolute magnitude. Once this is known, it is possible to know the distance of the Cepheid, calculating the difference with respect to the apparent magnitude.

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WILL GIVE BRAINLIEST TO THE CORRECT ANSWER
LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

F = 1000*3\\F=3000[N]

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

W=m*g\\W=1000*9.81\\W=9810 [N]

4 0
2 years ago
The bending of waves as they pass between media is called .?
Rufina [12.5K]

Refraction is the change in direction of waves that occurs when waves travel from one medium to another. Refraction is always accompanied by a wavelength and speed change. Diffraction is the bending of waves around obstacles and openings.

6 0
2 years ago
Read 2 more answers
A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0
2 years ago
How do gamma rays use electromagnetic waves to function....PLS HELP
BabaBlast [244]

Answer:

Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions.

Explanation:

Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.

Hope this helps!

Brain-LIst?

4 0
3 years ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
2 years ago
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