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3 years ago

What effect would decreasing the distance between objects have on their gravitational attraction to each other?

2 answers:

8 0

Decreasing the distance between objects will result in a greater gravitational attraction between each other. This is because the distance is inversely related to the gravitational attraction. For instance, in the same way, if the distance between two objects was increased, then their gravitational attraction to each other would decrease.

Lady_Fox [76]3 years ago

5 0

Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.

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Can I PLEASE get some help? I REALLY need it!

soldi70 [24.7K]

The answer is C. Hope this helps.

7 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago

How to intercept a slope

Harlamova29_29 [7]

Answer:

The slope intercept form is probably the most frequently used way to express equation of a line. To be able to use slope intercept form, all that you need to be able to do is 1) find the slope of a line and 2) find the y-intercept of a line.

Explanation:

8 0

3 years ago

Read 2 more answers

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$