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4 years ago

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An object dropped on Planet P falls 144 m in 6 seconds. What is the gravitational acceleration of Planet P ? Gravitational accel

eration of Planet P is m/s2 .

1 answer:

Tju [1.3M]4 years ago

7 0

Answer:

The gravitational acceleration of the planet is, g = 8 m/s²

Explanation:

Given data,

The distance the object falls, s = 144 m

The time taken by the object is, t = 6 s

Using the III equations of motion

S = ut + ½ gt²

∴ g = 2S/t²

Substituting the given values,

g = 2 x 144 /6²

= 8 m/s²

Hence, the gravitational acceleration of the planet is, g = 8 m/s²

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

elena-s [515]

Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

And now that for any shape the gravitational acceleration is given by:

$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

4 0

3 years ago

what is the mechanical advantage of a crowbar when a worker uses 10N of force to pry open a window that has a resistance of 500N

Oksana_A [137]

Answer:

50

Explanation:

The mechanical advantage of a machine is given by

$MA=\frac{F_{out}}{F_{in}}$

where

$F_{out}$ is the output force

$F_{in}$ is the input force

For the crowbar in this problem,

$F_{in}=10 N$ is the force in input applied by the worker

$F_{out}=500 N$ is the force that the machine must apply in output to overcome the resistance of the window and to open it

Substituting into the equation, we find

$MA=\frac{500}{10}=50$

3 0

3 years ago

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Vivian went on a bicycle trip through Germany with her family. One afternoon, she rode her bicycle along a long flat road at a c

natima [27]

d = distance traveled by her on her bicycle on a long flat road = 24 kilometer

t = time taken by her to travel distance "d" on her bicycle on a long flat road = 1.2 hours

v = average speed of vivian = ?

we know that average speed is given as

v = d/t

inserting the values in the above formula

v = 24 kilometer / 1.2 hour

v = 20 kilometer/hour

hence the correct choice is

C) 20 km/h

8 0

3 years ago

Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a

Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

Force applied on camel feet = 4000 N
Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

${\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}$

Substituting all the given values in the formula to find the pressure exerted by camel feet.

$\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\ \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}} \\ \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\ \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\ \\\star \: \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}$

Hence, the pressure exerted by camel feet is 2000 N/m².

$\rule{300}{2.5}$

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2 years ago

When measuring current, always place any type of meter in

Mars2501 [29]

In series with the circuit, so for it pass the current to be mensured.

Letter A

If you notice any mistake in my english, please let me know, because i am not native.

5 0

3 years ago

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