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Snezhnost [94]
3 years ago
12

A distant galaxy is determined to be 150 million light years distant and moving away from us; using the Hubble law determine its

velocity in terms
Physics
1 answer:
dlinn [17]3 years ago
4 0

Your question kind of petered out there towards the end and you didn't specify
the terms, so I'll pick my own.

The "Hubble Constant" hasn't yet been pinned down precisely, so let's pick a
round number that's in the neighborhood of the last 20 years of measurements:

             <em>70 km per second per megaparsec</em>.

We'll also need to know that 1 parsec = about 3.262 light years.

So the speed of your receding galaxy is

         (Distance in LY) x (1 megaparsec / 3,262,000 LY) x (70 km/sec-mpsc) =

              (150 million) x  (1 / 3,262,000) x (70 km/sec) =

                                 <em>3,219 km/sec  </em>in the direction away from us (rounded)

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A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming
Monica [59]

Answer:

The deceleration is  a =  - 76.27 m/s^2

Explanation:

From the question we are told that

   The height above  firefighter safety net is H  = 14 \ m

   The length by which the net is stretched is s =  1.8 \ m

   

From the law of energy conservation

    KE_T + PE_T =  KE_B + PE_B

 Where KE_T is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

   and  PE_T is the potential energy of the before jumping  which is mathematically represented at

          PE_T  = mg H

and  KE_B is the kinetic energy of the person just before landing on the safety net  which is mathematically represented at

        KE_B = \frac{1}{2} m v^2

and  PE_B is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

   So the above equation becomes

          mgH =  \frac{1}{2} m v^2

=>           v =  \sqrt{2 gH }

    substituting values

                v =  16.57 m/s

Applying the equation o motion

             v_f =  v  + 2 a s

Now the final velocity is zero because the person comes to rest

      So

         0 = 16.57 + 2 * a * 1.8

            a =  - \frac{16.57^2 }{2 * 1.8}

            a =  - 76.27 m/s^2

         

         

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2 years ago
The sum of two component vectors is referred to as the
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Describe using examples how objects can be at rest and in motion simultaneously
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An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N
algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

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2 years ago
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