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2 years ago

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On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr

om a storage room to a machinist.
Part A) Calculate the weight of the rod, w. The acceleration due to gravity, g= 9.81 m/s^2.w= 36.9 N

Part B) Now that you know the weight of the rod, do you think that you will be able to carry the rod without a cart? Yes or No?

1 answer:

slavikrds [6]2 years ago

7 0

Answer:

48.4293354946 N

Yes

Explanation:

d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

$\rho$ = Density of rod = 7800 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Volume of rod

$V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}$

Mass is given by

$m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}\\\Rightarrow m=4.93673144695\ kg$

Weight is given by

$W=mg\\\Rightarrow W=4.93673144695\times 9.81\\\Rightarrow W=48.4293354946\ N$

The weight of the rod is 48.4293354946 N

The mass of the rod is 4.93673144695 kg which is light. So, I will be able to carry the rod without a cart.

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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:

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

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2 years ago

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Firdavs [7]

Initial velocity (u) = 2 m/s

Acceleration (a) = 10 m/s^2

Time taken (t) = 4 s

Let the final velocity be v.

By using the equation,

v = u + at, we get

or, v = 2 + 10 × 4

or, v = 2 + 40

or, v = 42

The final velocity is 42 m/s.

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Where coastal rock is softer, waves erode the land faster

prohojiy [21]

Answer:

Headlands and bays are created where there are bands of hard and soft rock which meet the coastline at right angles. Softer rock is eroded more quickly and erodes backwards to form bays (which may have beaches). The harder rocks are more resistant to erosion and jut out into the sea to form exposed headlands

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What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Alexandra [31]

600/3 = 200
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OR

600/ (3/60) =
600 x 60/3 =
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another way, speed is a scalar value, while reloolg isa Which ONE of the following represents a toy car that moves at a higher a

erica [24]

Car A will have highest speed is 83.3m/s .

<h3>What is speed ? </h3>

The rate of change of position of an object in any direction.

The S.I unit is m/s . Speed is a scalar quantity it defines only magnitude not direction

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speed = distance /time

In case of Car A ,

We have given distance 150Km in 3 min ,

First we have convert the distance km to m

150×1000m

then conversion of min to sec

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speed = 15000/180

speed = 83.3m/sec

In case of Car B

we have given 800m in 150 min

lets convert the time into second

150×60

Speed = 800/150×60

speed = 0.88m/ s

In case of Car C

We have given here distance 250 Km and time in 8 hours

convert km to m

25000

and time into sec

88×60×60

speed = 0.86m/ s

Hence ,Car A has highest speed amongst them .

To learn more about speed click here

brainly.com/question/7359669

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11 months ago