What quantity is measured on the x-axis of a position vs. time graph.

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

2 years ago

The SAF operates the M113 Ultra APC.

HACTEHA [7]

The volume of the object must be no larger than $11.15 m^3$ .

Explanation:

In order for an object to be able to float in water, its density must be equal or smaller than the water density.

The density of water is:

$\rho = 1000 kg/m^3$

This means that the density of the object must be no larger than this value.

We also know that the density of an object is given by

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

where

m is the mass of the object

V is its volume

For the object in this problem, the mass is

$m=1.115\cdot 10^4 kg$

Therefore, we can re-arrange the equation to find its volume:

$V=\frac{m}{\rho}=\frac{1.115\cdot 10^4}{1000}=11.15 m^3$

So, the volume of the object must be no larger than $11.15 m^3$ .

Learn more about density:

brainly.com/question/5055270

brainly.com/question/8441651

#LearnwithBrainly

8 0

2 years ago

Complete the passage to describe the relationship between kinetic energy, internal energy, thermal energy, and

butalik [34]

Answer:

Increases

Explanation:

I don't know if you answered your own question but I'll just answer this for others confused ahh

4 0

2 years ago

Read 2 more answers

An astronaut on the moon places a package on a scale and finds its weight to be 10. N. () What would the weight of the package

jeka94

Answer:

(a) 61.25 N

(b) 6.25 kg

(c) 6.25 Kg

Explanation:

Weight on moon = 10 N

Acceleration due to gravity on moon = 1.6 m/s^2

Acceleration due to gravity on earth = 9.8 m/s^2

Let m be the mass of the package.

(a) Weight on earth = mass x acceleration due to gravity on earth

Weight on earth = 6.25 x 9.8 = 61.25 N

(b) Weight on moon = mass x acceleration due to gravity on moon

10 = m x 1.6

m = 6.25 kg

(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

8 0

2 years ago

Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.

valentina_108 [34]

The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics

5 0

2 years ago