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rodikova [14]
2 years ago
9

I am stumped - please help!

Physics
1 answer:
mestny [16]2 years ago
4 0

Answer:

a It is moving closer to the other sphere.

Explanation:

Hope it helps.

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Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e
Grace [21]

Answer:

The duration  is  T  =72 \  years /tex]Explanation:From the question we are told that     The  distance is  [tex]D  =  35 \ light-years = 35 *  9.46 *10^{15} = 3.311 *10^{17} \  m

  Generally the time it would take for the message to get the the other civilization is mathematically represented as

         t =  \frac{D}{c}

Here c  is the speed of light with the value  c =  3.08 *10^{8} \  m/s

=>     t =  \frac{3.311 *10^{17} }{3.08 *10^{8}}

=>     t =  1.075 *10^9 \ s

converting to years

           t =  1.075 *10^9 *  3.17 *10^{-8}

              t =  1.075 *10^9 *  3.17 *10^{-8}

            t =  34 \ years

Now the total time taken is mathematically represented as

      T  =  2*  t  +  2 + 2

=>   T  =  2* 34  +  2 + 2

=>   [tex]T  =72 \  years /tex]

4 0
3 years ago
HELP!!! HURRY!!!!! 100 POINTS!!!!!!!!!
bagirrra123 [75]

what happens at Point C is sublimation. the increase in temperature affects the Vapour pressure soon as you can see the curve is increasing with increasing pressure there is increase in temperature that is the sublimation Curve

7 0
2 years ago
Read 2 more answers
A 60 kg acrobat is in the middle of a 10 m long tightrope. The center of the rope dropped 30 cm in relation to the ends that are
Zigmanuir [339]

Answer:

The tension in each half of the rope, is approximately 4,908.8 N

Explanation:

The mass of the acrobat, m = 60 kg

The length of the rope, l = 10 m

The extent by which the center dropped = 30 cm = 0.3 m

Let, 'T' represent the tension in each half of the rope

Weight, W = Mass, m × The acceleration due to gravity, g

∴ W = m × g

The acceleration due to gravity, g ≈ 9.8 m/s²

∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N

The angle the dropped rope makes with the horizontal, θ is given as follows;

θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°

At equilibrium, the sum of vertical forces, \Sigma F_y = 0

The vertical component of the tension, T_y, in each half of the rope is given as follows;

T_y = T × sin(θ)

∴ \Sigma F_y = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)

Plugging in the values, with θ = arctan(0.06) for accuracy, we get;

588 N + 2 × T × sin(arctan(0.06) = 0

∴ 2 × -T × sin(arctan(0.06) = 588 N

-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N

The tension in each half of the rope, T ≈ 4,908.8 N.

4 0
2 years ago
Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0
3 years ago
Read 2 more answers
Evaluate tan(249).<br> O A. 1.36<br> B. 0.45<br> C. 0.41<br> D. 0.91
vodomira [7]

Answer:

tan 249 = 2.61

tan 249 = tan (249 - 180) = tan 69 = 2.61

3 0
2 years ago
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