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3 years ago

By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged

1 answer:

Nutka1998 [239]3 years ago

4 0

Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:

F = G(Mm / (4(pi)*r^2))

The above expression gives the force that you feel on the earth's surface, as it is today!

Let us now double the mass of the earth and decrease its diameter to half its original size.

This is the same as replacing M with 2M and r with r/2.

Now the gravitational force (F' ) on the new earth's surface is given by:

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F

So:

F' = 8F

This implies that the force that you would feel pulling you down (your weight) would increase by 800%!

You would be 8 times heavier on this "new" earth!

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When humans look at the sky, it appears blue, but the Sun appears yellow. What causes this phenomenon?

Lesechka [4]

I believe that the answer is C. The atmosphere scatters blue light more than yellow light.

8 0

3 years ago

Read 2 more answers

A bystander observes the musicians heading toward each other. When musician #1 is 100 m away, the intensity is 1.24 x 10-8 W/m^2

777dan777 [17]

Explanation:

Given that,

Distance 1, r = 100 m

Intensity, $I_1=1.24\times 10^{-8}\ W/m^2$

If distance 2, r' = 25 m

We need to find the intensity and the intensity level at 25 meters. Intensity and a distance r is given by :

$I=\dfrac{P}{4\pi r^2}$ .........(1)

Let I' is the intensity at r'. So,

$I'=\dfrac{P}{4\pi r'^2}$ ............(2)

From equation (1) and (2) :

$I'=\dfrac{Ir}{r'^2}$

$I'=\dfrac{1.24\times 10^{-8}\times 100}{25^2}$

$I'=1.98\times 10^{-9}\ W/m^2$

Intensity level is given by :

$dB=10\ log(\dfrac{I'}{I_o})$ , $I_o=10^{-12}\ W/m^2$

$dB=10\ log(\dfrac{1.98\times 10^{-9}}{10^{-12}})$

dB = 32.96 dB

Hence, this is the required solution.

7 0

3 years ago

A dart is inserted into a spring-loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is c

bezimeni [28]

Answer:

The second dart leaves the gun two times as faster than the first one.

Explanation:

Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle

$U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}.$

Given an arbitrary $x$ and its double, $2x$ , launch velocities are

$v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}.$

7 0

3 years ago

A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the

vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light λ is 580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D = 1.8 m.

Width of one fringe = $\frac{\lambda\times D}{d}$

Putting the values we get fringe width

= $\frac{580\times10^{-9}\times1.8}{.000315}$

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be

= 4.86 mm.

3 0

3 years ago

1.How does gravity change your weight? 2.How do force work in nature?

kodGreya [7K]

In a stronger gravitational field a given mass will have a larger weight.

3 0

3 years ago