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2 years ago

12

a new planet is discovered that has twice the earth’s mass and twice the earth’s radius. on the surface of this new planet, a pe

rson who weighs 500 n on earth would experience a gravitational force of

1 answer:

Irina18 [472]2 years ago

8 0

Answer: 250n

Explanation:

The formula for gravitational force is: F = (gMm)/r^2

There are two factors at play here:

1) The mass of the planet 'M'

2) The radius 'r'

We can ignore the small M and the g, they are constants that do not alter the outcome of this question.

You can see that both M and r are double that of earth. So lets say earth has M=1 and r=1. Then, new planet would have M=2 and r=2. Let's sub these two sets into the equation:

Earth. F = M/r^2 = 1/1

New planet. F = M/r^2 = 2/4 = 1/2

So you can see that the force on the new planet is half of that felt on Earth.

The question tells us that the force on earth is 500n for this person, so then on the new planet it would be half! So, 250n!

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A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

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3 years ago

Star A and Star B have different apparent brightnesses but identical luminosities. Star A is 10 light years away from earth and

STALIN [3.7K]

intensity of a star is inversely depends on the square of the distance from the star

we can say it is given as

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

here we know that

$\frac{I_1}{I_2} = 36 times$

also we know that

$r_1 = 10 Ly$

now we will have

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$36 = \frac{r_2^2}{10^2}$

$r_2 = 60 Ly$

so other star is at distance 60 Light years

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3 years ago

A 5kg bag falls a verticle height of 10m before hitting the ground.

g100num [7]

Answer:

$u = 7m {s}^{ - 1}$

Explanation:

We know that when we don't have air friction on a free fall the mechanical energy (I will symbololize it with ME) is equal everywhere. So we have:

$me(1) = me(2)$

where me(1) is mechanical energy while on h=10m

and me(2) is mechanical energy while on the ground

Ek(1) + DynamicE(1) = Ek(2) + DynamicE(2)

Ek(1) is equal to zero since an object that has reached its max height has a speed equal to zero.

DynamicE(2) is equal to zero since it's touching the ground

Using that info we have

$m \times g \times h = \frac{1}{2} \times m \times u {}^{2} \\$

we divide both sides of the equation with mass to make the math easier.

$9.8 \times 10 = \frac{1}{2} \times u {}^{2} \\ \frac{98}{2} = u {}^{2} \\ u { }^{2} = 49 \\ u = 7$

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3 years ago

Find the speed of light in each of the following materials. (a) gallium phosphide m/s (b) carbon disulfide m/s (c) benzene

Oksanka [162]

Explanation:

We need to calculate the speed of light in each materials

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Using formula of speed of light

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Where, $\mu$ = index of refraction

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Put the value into the formula

$v=\dfrac{3\times10^{8}}{3.50}$

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(II) Carbon disulfide,

The index of refraction of Gallium phosphide is 1.63

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.63}$

$v=1.8\times10^{8}\ m/s$

(III). Benzene,

The index of refraction of Gallium phosphide is 1.50

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.50}$

$v=2\times10^{8}\ m/s$

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