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viktelen [127]
3 years ago
9

Two objects that may be considered point masses are initially separated by a distance d. The separation distance is then decreas

ed to d/4. How does the gravitational force between these two objects change as a result of the decrease?
Physics
1 answer:
ozzi3 years ago
8 0

Answer:

Increased by 16 times

Explanation:

F = Gravitational force between two bodies

G = Gravitational constant = 6.67408 × 10⁻¹¹ m³/kg s²

m₁ = Mass of one body

m₂ = Mass of other body

d = distance between the two bodies

F=\frac{Gm_1m_2}{d^2}\\ F=\frac{1}{d^2}\quad \text {(as G and masses are constant)}

F_{new}=\frac{1}{\left (\frac{d}{4}\right )^2}\\\Rightarrow F_{new}=\frac{1}{\frac{d^2}{16}}\\\Rightarrow F_{new}={16}\times \frac{1}{d^2}\\\Rightarrow F_{new}=16\times F

∴Force will increase 16 times

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Which of the following statements apply to electric charges?
Gre4nikov [31]

Answer:

The statement "If a positively charged rod is brought close to a positively charged object, the two objects will repel " applies to electric charges.

Explanation:

There are only two types of electric charges. Both having own magnitude but different charge.

1. Positive charge

2. Negative charge

Like charges repel each other and opposite charges always attract each other.

When a positively charged rod is brought close to a positively charged object, the rod and the object will repel.

6 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of
Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

3\times \frac{\lambda}{2} = 0.39

\lambda = 0.26 m

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

d = \frac{\lambda}{4}

d = \frac{0.26}{4} = 0.065 m = 6.5 cm

So the distance will be 6.5 cm

3 0
3 years ago
A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre
butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

8 0
3 years ago
What's saturns rotation
xeze [42]
The way it rotates is Counter-clockwise
4 0
3 years ago
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