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svetoff [14.1K]
3 years ago
9

For which object is the force of gravity greatest: Earth, Moon, or Sun? Why?

Physics
1 answer:
oksian1 [2.3K]3 years ago
5 0
The sun. The more mass an object has, the greater gravitational pull it will have, as mass attracts other mass.
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Elements have the same number of ______ as you move from left to right
Fiesta28 [93]

Answer:

protons

Explanation:

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An object is 30.0 cm to the left of a convex lens with a focal length of +8.0 cm. Draw a ray diagram of the setup showing the lo
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The distance should be 11 cm and the image will be inverted (and smaller)
I used the Lens Equation:
\frac{1}{obj.}+ \frac{1}{im.}  = \frac{1}{f}
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obj. is the distance of the object
im. is the distance of the image
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When we give the speed of a car what frame of reference are we using?
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Frame of reference, in simplest terms, describes the state of motion of the observer. The frame of reference may also be described by using a set of coordinates, time and motion. We formulate all our equations and solve them using the frame of reference.

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Jonas and his family are moving to another part of the city. As Jonas, his brother, and his Dad were driving one of the trucks f
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i think the answer would be:Jonas’ brother gets out of the cab of the truck and sits in the back of the truck with the furniture. With less mass, they should be able to push the truck to the gas station.

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Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out
PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is  

\lambda _2 = (6.48,6.52) \ cm

Explanation:

From the question we are told that

     The wavelength of the note is \lambda  =  6.50 \ cm = 0.065 \ m

     The difference in beat frequency is \Delta  f = 17.0 \ Hz

     

Generally the frequency of the note played by the guitar that is in tune is  

        f_1 = \frac{v_s}{\lambda}

Where v_s is the speed of sound with a constant value v_s  =  343 \ m/s

       f_1 = \frac{343}{0.0065}

      f_1 = 5276.9 \ Hz

The difference in beat is mathematically represented as

       \Delta  f =  |f_1 - f_2|

Where f_2 is the frequency of the sound from the out of tune guitar

     f_2 =f_1  \pm \Delta f

substituting values

      f_2 =f_1 + \Delta f

      f_2 = 5276.9 + 17.0  

     f_2 = 5293.9 \ Hz

The wavelength for this frequency is

      \lambda_2 = \frac{343 }{5293.9}

     \lambda_2 = 0.0648 \ m

    \lambda_2 = 6.48 \ cm

For the second value of the second frequency

     f_2 =  f_1 - \Delta f

     f_2 = 5276.9 -17

      f_2 = 5259.9 Hz

The wavelength for this frequency is

   \lambda _2 = \frac{343}{5259.9}

   \lambda _2 = 0.0652 \ m

   \lambda _2 = 6.52 \ cm

8 0
3 years ago
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