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sesenic [268]
3 years ago
9

Describe succinctly the relationship between how far a galaxy is from us (its distance), versus how fast it is moving.

Physics
1 answer:
Eva8 [605]3 years ago
7 0

Answer:

Distance is directly proportional to the velocity

Explanation:

In 1929, Edwin Hubble's wrote an article that talked about relationship between the distance and recession speed/velocity of galaxies which led to what is known as the Hubble Law. This law states that galaxies are moving away from the earth at velocities proportional to their distances.

Thus is written as;

v = H_o•d

Where;

v is velocity

d is distance

H_o is Hubble's constant rate of cosmic expansion.

He came to this conclusion by generating a graph known as Hubble's classic graph which was a graph of observed velocity vs distance for nearby galaxies.

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An object has a position given by r = [2.0 m + (5.00 m/s)t] i^ + [3.0m−(2.00 m/s2)t2] j^, where all quantities are in SI units.
makkiz [27]

Answer:

Acceleration of the object is 4\ m/s^2.

Explanation:

It is given that, the position of the object is given by :

r=[2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j

Velocity of the object, v=\dfrac{dr}{dt}

Acceleration of the object is given by :

a=\dfrac{d^2r}{dt^2}

a=\dfrac{d^2}{dt^2}([2\ m+(5\ m/s)t]i+[3\ m-(2\ m/s^2)t^2]j)

Using the property of differentiation, we get :

a=\dfrac{d^2r}{dt^2}=-4\ m/s^2

So, the magnitude of the acceleration of the object at time t = 2.00 s is 4\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
If you divide a speed in miles per minute by 60, you get the same speed
Mariulka [41]
Yes so x multiply by the hour but u add 59 for each hour to get the exath speed per minute
6 0
3 years ago
What do you call the procedure that helps you determine the volume of an irregularly shaped object, while using a graduated cyli
Lena [83]
Water displacement. You fill a graduated cylinder with an amount of water, place the object inside the graduated cylinder, and then measure the new water level. The change in volume of the water is the volume of the object, assuming the object was completely submerged. 
8 0
3 years ago
f the magnitude of the acceleration of a propeller blade's tip exceeds a certain value amaxamax, the blade tip will fracture. If
Luden [163]

Answer:

The angular velocity is   w= \sqrt[4]{\frac{a_{max}^2}{r^2}  - \alpha ^2}      

Explanation:

Generally the acceleration experienced by the propeller blade's is broken down into

          The Radial acceleration which is mathematically represented as

                              a_r = \frac{v^2}{r}  = w^2r

And the Tangential  acceleration which is mathematically represented as

                                a_r = \alpha r

  The net acceleration is evaluated as

                      a = \sqrt{a_r^2 + a_t^2}

       

Now since angular speed varies directly with angular acceleration so when acceleration is maximum the angular velocity is maximum also and this point if the propeller blade's tip exceeds it the blade would fracture

                 

So at maximum angular acceleration we a have

             a_{max} = \sqrt{a_r^2 + a_t^2}

                     a_{max}^2 = a_r^2 + a_t^2

                    a_{max}^2 = (w^2r)^2 + (\alpha r)^2

                 a_{max}^2 =  r^2 w^4 + r^2 \alpha ^2

                  a_{max}^2 = r^2 (w^4 + \alpha^2 )

                w^4 +\alpha ^2 = \frac{a_{max}^2}{r^2}

                         w^4 = \frac{a_{max}^2}{r^2}  - \alpha ^2

                         w= \sqrt[4]{\frac{a_{max}^2}{r^2}  - \alpha ^2}        

                     

3 0
3 years ago
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

  • ∑F is net force
  • m is mass of the third child
  • a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

  • the wagon
  • the children outside the wagon

<h3>Free body diagram</h3>

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

<h3>Acceleration of the  child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

7 0
2 years ago
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