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yarga [219]
2 years ago
14

The gravitational force between two objects is 100 N.

Physics
1 answer:
defon2 years ago
5 0
200N is the answer (at least thats what I think)
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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
Lubov Fominskaja [6]

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

5 0
2 years ago
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
3 years ago
A block with mass ma = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second
Drupady [299]
<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation. F = Ma F is the net force M is the total mass of the system a is the acceleration a = F / M a = (mb)(g) / (ma + mb) a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg) a = 58.8 N / 20 kg a = 2.94 m/s^2 The magnitude of the acceleration of the system is 2.94 m/s^2</span>
7 0
3 years ago
Which of the following are basic solutions? Check all that apply. A. Laundry detergent B. Vinegar C. Drain cleaner D. Antacids
ratelena [41]
A laundry detergent
B vinegar
4 0
3 years ago
A student drops a ball from the top of a 10-meter tall building. The ball leaves the thrower's hand with a zero speed. What is t
Sergio [31]

Answer:

14 m/s

Explanation:

u = 0, h = 10 m, g = 9.8 m/s^2

Use third equation of motion

v^2 = u^2 + 2 g h

Here, v be the velocity of ball as it just strikes with the ground

v^2 = 0 + 2 x 9.8 x 10

v^2 = 196

v = 14 m/s

7 0
2 years ago
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