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3 years ago

Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?

2 answers:

5 0

<span>suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2</span>

snow_tiger [21]3 years ago

4 0

Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2

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The base of a right prism is a rhombus with diagonals of 6 and 8. If the altitude of the prism is 12, what is the total surface

Agata [3.3K]

Answer:

total surface area is 432

Explanation:

Given data

base = 6

diagonals = 8

altitude = 12

to find out

total surface area

solution

we know total surface area of prism is

total surface area = lateral surface area + 2base area ..............1

first we calculate base perimeter i.e = 2 length + 2 width

so perimeter = 2(8) + 2(6) = 25

and area = length * width = 8*6 = 48

so lateral surface area is perimeter * height i.e

lateral surface area = 28* 12

lateral surface area = 336

put this value in equation 1 we get

total surface area = lateral surface area + 2base area

total surface area = 336 + 2(48)

total surface area is 432

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3 years ago

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erastovalidia [21]

Answer:

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Explanation:

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3 years ago

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A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?

kupik [55]

Answer:

$workdone = force \times distance \\ = 65000 \times 75 \\ = 4,875,000 \: J$

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3 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

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3 years ago

Why do the constellations seem to move around in the sky?.

Gnoma [55]

Answer: As Earth spins on its axis, we, as Earth-bound observers, spin past this background of distant stars. As Earth spins, the stars appear to move across our night sky from east to west, for the same reason that our Sun appears to “rise” in the east and “set” in the west.

Explanation:

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3 years ago