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julsineya [31]
3 years ago
12

determine the force of gravitational attraction between a 78kg boy sitting 2 meters away from a 65kg girl.

Physics
1 answer:
bixtya [17]3 years ago
8 0

Answer:

F=8.45\times 10^{-8}\ N

Explanation:

Given that,

Mass of a boy is 78 kg

Mass of a girl is 65 kg

We need to find the force of gravitational attraction between them if they are 2 m away.

The formula for the gravitational force is given by :

F=\dfrac{Gm_1m_2}{r^2}\\\\F=\dfrac{6.67\times 10^{-11}\times 78\times 65}{(2)^2}\\\\=8.45\times 10^{-8}\ N

So, the force between them is 8.45\times 10^{-8}\ N.

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Explanation:

well there is nothing there and it could be different by diffrent objects, idk

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a. A lifted parcel of air will be colder (heavier) that the air surrounding it. Because of this fact, the lifted parcel will ten
kvv77 [185]
I think the answer is A because it’s a better explanation
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Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris
telo118 [61]

Answer:

a

   F_A  =425.42 \ N

b

  F_A_H  = 358.58 \ N

Explanation:

From the question we are told that

    The diameter of the Ferris wheel is  d  =  80 \ ft =  \frac{80}{3.281} =  24.383

    The  period of the Ferris wheel is  T  =  24 \ s

     The  mass of the passenger is  m_g  =  40 \ kg

The  apparent weight of the passenger at the lowest point is mathematically represented as

           F_A_L  =  F_c  + W

Where  F_c is the centripetal force on the passenger,  which is mathematically represented as

         F_c  =m *  r *  w^2

Where w is the angular velocity which is mathematically represented as

         w =  \frac{2* \pi   }{T}

substituting values

         w =  \frac{2* 3.142 }{24}

         w =  0.2618 \ rad/s

and  r  is the radius which is evaluated as r =  \frac{d}{2}

   substituting values

         r =  \frac{24.383}{2}

         r = 12.19 \ ft

So

          F_c  = 40 * 12.19* (0.2618)^2

          F_c  =  33.42 \ N

W is the weight which is mathematically represented as

           W =  40 * 9.8

           W =  392 \ N

So

         F_A    =  33.42 + 392

         F_A  =425.42 \ N

The  apparent weight of the passenger at the highest point is mathematically represented as

          F_A_H  =  W- F_c

substituting values

         F_A_H  = 392 -  33.42

         F_A_H  = 358.58 \ N

4 0
3 years ago
A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w
Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

h_1=\frac{L}{2}sin\theta

The center of mass of the ladder ends at a height of:

h_2= \frac{L}{2}sin90 =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = mg(h_2-h_1)

now h_2-h_1= 1-sin\theta

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0
3 years ago
A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
Virty [35]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

               a = 0.003* 6 =0.018\  m/s^2          

     

8 0
3 years ago
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