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4 years ago

Using our understanding of the law of gravity. What happens to the gravity as we triple the distance between two objects?

Physics

1 answer:

MA_775_DIABLO [31]4 years ago

3 0

Answer:

1/9

Explanation:

Newton’s Law of Universal Gravitation

Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.

F = GMm/r²

where

F - the gravitatioal force in Newtons,

M and m -two masses in kilograms

r - the separation in meters.

G - the gravitational constant (6.674*10 ⁻¹¹ N (m/kg) ² )

Because of the magnitude of G , gravitational force is very small unless large masses are involved.

So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance

F1 ∝ 1/r² ---------------(1)

F2 ∝ 1(3r)²

F2 ∝ 1/9r²--------------(2)

(2)/(1)

$\frac{F_2}{F_1} =\frac{1}{9}\\ F_2 =\frac{F_1}{9}$

From their you get as the distance tripled, Force reduce by a factor of 9(3³)

for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)

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(a) Suppose that your measured weight at the equator is one-half your measured weight at the pole on a planet whose mass and dia

stealth61 [152]

Answer:

7160.2812 s or 1.988 hours

Explanation:

m = Mass of person

R = Radius of Earth = $6.37\times 10^{6}\ m$

g = Acceleration due to gravity = 9.81 m/s²

$\omega$ = Angular speed

Force at equator would be

$F_e=m(g-\omega^2R)$

Force at pole

$F_p=mg$

From the question

$F_e=\dfrac{1}{2}F_p\\\Rightarrow m(g-\omega^2R)=\dfrac{1}{2}F_p\\\Rightarrow \omega=\sqrt{\dfrac{g}{2R}}$

Time period is given by

$T=\dfrac{2\pi}{\omega}\\\Rightarrow T=2\pi\sqrt{\dfrac{2R}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{2\times 6.37\times 10^6}{9.81}}\\\Rightarrow T=7160.2812\ s=1.988\ hours$

The rotational period of the planet is 7160.2812 s or 1.988 hours

5 0

3 years ago

Assuming that 70 percent of the Earth’s surface is covered with water at average depth of 0.95 mi, estimate the mass of the wate

Alla [95]

Answer:

The mass of the water on earth is $5.4537\times10^{20}\ kg$

Explanation:

Given that,

Average depth h= 0.95 mi

$h=0.95\times1.609$

$h =1.528\ km$

$h=1.528\times10^{3}\ m$

Radius of earth $r= 6.37\times10^{6}\ m$

Density = 1000 kg/m³

We need to calculate the area of surface

Using formula of area

$A =4\pi r^2$

Put the value into the formula

$A=4\pi\times(6.37\times10^{6})^2$

$A=5.099\times10^{14}\ m^2$

We need to calculate the volume of earth

$V = Area\times height$

Put the value into the formula

$V=5.099\times10^{14}\times1.528\times10^{3}$

$V=7.791\times10^{17}\ m^3$

Now, 70 % volume of the total volume

$V= 7.791\times10^{17}\times\dfrac{70}{100}$

$V=5.4537\times10^{17}\ m^3$

We need to calculate the mass of the water on earth

Using formula of density

$\rho = \dfrac{m}{V}$

$m = \rho\times V$

Put the value into the formula

$m=1000\times5.4537\times10^{17}$

$m =5.4537\times10^{20}\ kg$

Hence, The mass of the water on earth is $5.4537\times10^{20}\ kg$

8 0

4 years ago

An arrow is shot at 28.0° above the horizontal. Its initial speed is 50 m/s and it hits the target.

fomenos

We will need to work with the components of the velocity, in the x and the y direction. We will say up is positive so g is -9.81 m/s^2.

Given that the angle was 32 degrees:

Velocity up (in the y direction) is 55 m/s * sin 32 = 29.15 m/s
And

Velocity forward (in the x direction) is 55 m/s * cos 32 = 46.64 m/s

The acceleration of gravity, -9.81 m/s2 continuously decreases the velocity in the y direction. At the maximum height, the velocity will be zero. This should make sense, for as soon as the decreasing velocity becomes negative, the arrow will start to fall.

We have v = v(0) + at

And we set this to zero and solve for t:

0 = 29.15 + -9.81t

9.81t = 29.15

t = 2.97 seconds

To calculate height at this point, we use the equation that calculates position based on time, acceleration, and initial velocity (we could use an alternate too, an equation derived from the one we are now using and v = v(0) + at.

x = x(0) + v(0)t + (1/2)at^2

x = 0 + 29.15 * 2.97 + 0.5 9.81 (2.97)^2

x = 43.30 m

For a projectile, the plot of distance traveled in the upward direction is a parabola, and it takes the same amount of time to come down as it did to go up.

We can double 2.97 to get the time of impact on the target at 2(2.97) = 5.94 seconds

(Alternately, if you like, you can solve

0 = 0 + 29.15t + 0.5 9.81 t^2

And find that the two roots are 0 and 5.94).

http://www.math.com/students/calculators... will do the quadratic for you.

Given a horizontal velocity of 46.64 m/s, we can calculate

46.64 m/s (5.94 s) = 277 m for the distance of the target.

6 0

3 years ago

You are an astronaut in space far away from any gravitational field, and you throw a rock as hard as you can. The rock will:

Nesterboy [21]

Answer:

the rock will continue at the same speed unless it is affected by another force such as gravity and so if you threw it it will continue to move unless affected by a force

Explanation:

this is because Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.

7 0

3 years ago

What factors determine if an earthquake can be classified as an induced earthquake?

Luba_88 [7]

Answer:1 }The state of stress of the earth's crust

2}Existing faults

Explanation:

6 0

1 year ago