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IgorC [24]
3 years ago
13

The closest stars are 4 light years away from us. How far away must you be from a 854 kHz radio station with power 50.0 kW for t

here to be only one photon per second per square meter? Assume that the photons spread out spherically. The area of a sphere is 4????????2.
Physics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0

Answer:

The distance from the radio station is 0.28 light years away.

Solution:

As per the question:

Distance, d = 4 ly

Frequency of the radio station, f = 854 kHz = 854\times 10^{3}\ Hz

Power, P = 50 kW = 50\times 10^{3}\ W

I_{p} = 1\ photon/s/m^{2}

Now,

From the relation:

P = nhf

where

n = no. of photons/second

h = Planck's constant

f = frequency

Now,

n = \frac{P}{hf} = \frac{50\times 10^{3}}{6.626\times 10^{- 34}\times 854\times 10^{3}} = 8.836\times 10^{31}\ photons/s

Area of the sphere, A = 4\pi r^{2}

Now,

Suppose the distance from the radio station be 'r' from where the intensity of the photon is 1\ photon/s/m^{2}

I_{p} = \frac{n}{A} = \frac{n}{4\pi r^{2}}

1 = \frac{8.836\times 10^{31}}{4\pi r^{2}}

r = \sqrt{\frac{8.836\times 10^{31}}{4\pi}} = 2.65\times 10^{15}\ m

Now,

We know that:

1 ly = 9.4607\times 10^{15}\ m

Thus

r = \frac{2.65\times 10^{15}}{9.4607\times 10^{15}} = 0.28\ ly

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A 1.0-kg block of aluminum is at a temperature of 50 Celsius. How much thermal energy will it lose when it’s temperature is redu
Ipatiy [6.2K]

Answer:

The lose of thermal energy is, Q = 22500 J

Explanation:

Given data,

The mass of aluminium block, m = 1.0 kg

The initial temperature of block, T = 50° C

The final temperature of the block, T' = 25° C

The change in temperature, ΔT = 50° C - 25° C

                                                     = 25° C

The specific heat capacity of aluminium, c = 900  J/kg°C

The formula for thermal energy,

                             <em>Q = mcΔT</em>

                                 = 1.0 x 900 x 25

                                 = 22500 J

Hence, the lose of thermal energy is, Q = 22500 J

7 0
3 years ago
A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m
Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

a=2.92164g

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s

Angular velocity is 1.3823 rad/s

Linear velocity is given by

v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g

a=2.92164g

Centripetal force is given by

F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0
4 years ago
A car initially traveling 7 m/s speeds up uniformly at a rate of 3 m/s2 until it reaches a velocity of 22 m/s. How much time did
dezoksy [38]

Answer:

t = 5 s

Explanation:

Data:

  • Initial Velocity (Vo) = 7 m/s
  • Acceleration (a) = 3 m/s²
  • Final Velocity (Vf) = 22 m/s
  • Time (t) = ?

Use formula:

  • \boxed{t=\frac{Vf - Vo}{a}}

Replace:

  • \boxed{t=\frac{22\frac{m}{s} -7\frac{m}{s}}{3\frac{m}{s^{2}}}}

Solve the subtraction of the numerator:

  • \boxed{t=\frac{15\frac{m}{s}}{3\frac{m}{s^{2}}}}

It divides:

  • \boxed{t=5\ s}

How much time did it take the car to reach this final velocity?

It took a time of <u>5 seconds.</u>

8 0
3 years ago
4. O/N 15/P11/Q11
anygoal [31]

Answer:

B) 4500 Pa

Explanation:

As pressure is force per unit area,

P = F/A

It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.

The possible areas are

0.30(0.40) = 0.12 m²

0.30(0.50) = 0.15 m²

0.40(0.50) = 0.20 m²

The pressure when the face with the largest area (0.20 m²) is down is

P = 900 / 0.20 = 4500 N/m² or 4500 Pa

the other possible pressures would be

900/0.15 = 6000 Pa

900/0.12 = 7500 Pa

which are both larger than our solution.

5 0
3 years ago
MATHPHYS HELP
rusak2 [61]

Answer:

16.4287

Explanation:

The force and displacement are related by Hooke's law:

F = kΔx

The period of oscillation of a spring/mass system is:

T = 2π√(m/k)

First, find the value of k:

F = kΔx

78 N = k (98 m)

k = 0.796 N/m

Next, find the mass of the unknown weight.

F = kΔx

m (9.8 m/s²) = (0.796 N/m) (67 m)

m = 5.44 kg

Finally, find the period.

T = 2π√(m/k)

T = 2π√(5.44 kg / 0.796 N/m)

T = 16.4287 s

3 0
3 years ago
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