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dem82 [27]
3 years ago
11

The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s2. Neptune has mass 1.02×1026 kg and radiu

s 2.46×104 km and rotates once around its axis in a time of about 16 h.
a)What is the gravitational force on a 4.60-kg object at the north pole of Neptune? Express your answer with the appropriate units.
b)What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.) Express your answer with the appropriate units.

Physics
1 answer:
larisa [96]3 years ago
5 0

Answer: (a) The gravitational force on the object at the North Pole of Neptune is 51.7N

(b) The apparent weight of the object at Neptune's equator is 50.4N

Explanation: Please see the attachments below

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Light of wavelength 550 nm comes into a thin slit and produces a diffraction pattern on a board 8.0 m away. The first minimum da
FrozenT [24]

Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light \lambda =550nm=550\times 10^{-9}m

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So x=3\times 10^{-3}m

We have to find the width of the slit

For the first order wavelength is equal to \lambda =\frac{x}{D}\times a, here a width of slit

So a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm

So width of the slit will be equal to 1.47 mm

5 0
3 years ago
Two light waves, initially emitted in phase, will interfere constructively with maximum amplitude if the path-length difference
777dan777 [17]

Two light waves will interfere constructively if the path-length difference between them is a whole number.

<h3>SUPERPOSITION</h3>

The principle of superposition state that, when two or more waves meet at a point, the resultant displacement at that point is equal to the sum of the displacements of the individual waves at that point.

Interference of waves can either be constructive, or destructive.

The two light waves, initially emitted in phase, will interfere constructively with maximum amplitude if the path-length difference between them is a whole number of wavelenght 1λ, 2λ, 3λ, 4λ etc

The equivalent phase differences between the waves will be 2\pi or 360 degrees, 4\pi or 720 degrees, 6\pi 1080 degrees etc

Therefore, the two light waves, initially emitted in phase, will interfere constructively with maximum amplitude if the path-length difference between them is a whole number.

Learn more about Interference here: brainly.com/question/25310724

8 0
2 years ago
A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

3 0
3 years ago
In normal light conditions, how well do plants grow when there are 10 plants in the
Sati [7]

Answer

when there are ten they don't grow so well but when there is less than 10 they tend to grow

7 0
4 years ago
A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c
GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>
  • Let's draw the free body diagram of the system using the given data.
  • From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
  • For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

                           N_x=86.62N

  • We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

                           N_y=F_V=mg-Tsin59\\

  • To find Ny, we need to find the tension T.
  • For this, we can equate the net horizontal force.

                           F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N

  • Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,

                    N_y= (40*9.8)-(169.8*sin59)=246.4N

  • Thus, the magnitude of the force that the beam exerts on the hi.nge will be,

                 N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N

Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.

Learn more about the tension here:

brainly.com/question/28106871

#SPJ1

4 0
2 years ago
Read 2 more answers
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