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Ainat [17]
3 years ago
14

The Drake equation calculates the number of technically advanced civilizations in the galaxy as a product of seven factors: R* f

p ne fl fi fc L. In the sample calculation in the text, these values are taken as 1 × 1 × 0.1 × 1 × 1 × 1 × 100 = 10. If we were to redo the calculation, increasing each increasable factor by a multiple of ten, what would the result be?
Physics
1 answer:
juin [17]3 years ago
3 0
The Drake equation calculates the civilization parameter as
R*fp*ne*fl*fi*fc*L = 1 x 1 x 0.1 x 1 x 1 x 1 x 100 = 10.

If each increasable factor is multipled by 10, the result will be
(1*10) x (1*10) x (0.1*10) x (1*10) x (1*10) x (1*10) x (100*10)
= 10 x 10⁷
= 10⁸

Answer: 10⁸



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In Hans Oersted’s experiment, why did a compass needle move when an electric current flowed through a nearby wire?
Zinaida [17]

Answer:

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Explanation: The magenetic field Would not move a place without a magnetic field.

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A heat engine has three quarters the thermal efficiency of a carnot engine operating between temperatures of 65°C and 435°C if
const2013 [10]

Answer:

The value is  P_e  =  31275.2 \  W

Explanation:

From the question we are told that

   The efficiency of the carnot engine is  \eta

    The efficiency of a heat engine is k =  \frac{3}{4}  *  \eta

    The operating temperatures of the carnot engine is  T_1  =  65 ^oC =338 \ K  to  T_2 = 435  ^oC = 708 \  K

    The rate at which the heat engine absorbs energy is  P =  44.0 kW  = 44.0 *10^{3} \  W

Generally the efficiency of the carnot engine is mathematically represented as

          \eta =  [ 1 - \frac{T_1 }{T_2}  ]

=>       \eta =  [ \frac{T_2 - T_1}{T_2} ]

=>       \eta = 0.3856

Generally the efficiency of the heat engine is

           k  =  \frac{3}{4} * 0.3856

=>        k  = 0.2892

Generally the efficiency of the heat engine is also mathematically represented as

          k  =  \frac{W}{P}

Here W is the work done which is mathematically represented as

        W =  P - P_e

Here P_e  is the heat exhausted

So

       k  =  \frac{P - P_e}{P}

=>    0.2892   =  \frac{44*10^{3} - P_e}{44*10^{3}}

=>   P_e  =  31275.2 \  W

8 0
3 years ago
A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Fin
Ira Lisetskai [31]

Answer:

a)     d = (6.00 t i ^ + 0.500 t²) m , b)   v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d)  v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

       v = x / t

        x = v t

we substitute

        x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

         y = v₀ t + ½ a t²

         y = ½ at t²

we substitute

        y = ½ 1.00 t²

        y = 0.500 t²

in vector position is

       d = x i ^ + y j ^

       d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

       vₓ = v₀

       vₓ = 6.00 m / s

y Axis  

there is an acceleration and the initial speed is zero

         v_{y} = v₀ + a t

         v_{y} = a t

         v_{y} = 1.00 t

the velocity vector is

         v = vₓ i ^ + v_{y} j ^

         v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

        d = (6.00 4 i ^ + 0.50 4 2 j⁾

        d = (24.00 i ^ + 8.00 j^ ) m

 

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

        v = (6 i ^ + 1 5 j ^)

         v = (6.00 i ^ + 5 j^ ) m/s

7 0
3 years ago
What is the frequency of a radio wave with an energy of 3.686 × 10−24 j/photon?answer in units of hz?
Makovka662 [10]

To solve this problem, we have to use the formula:

E = h f

where E is total energy, h is Plancks constant 6.626x10^-34 J s, f is frequency

 

f = E / h

f = 3.686 × 10−24 J / (6.626x10^-34 J s)

<span>f = 5.56 x 10^9 Hz</span>

4 0
3 years ago
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