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3 years ago

During a new moon, the moon is between the

Physics

1 answer:

mylen [45]3 years ago

5 0

The moon is between the sun and earth.

The side where the light from the sun hits the moon is facing away from earth.

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A river has a steady speed of 0.566 m/s. A student swims upstream a distance of 1.57 km and returns (still swimming) to the star

MArishka [77]

Answer:

3503.72 seconds

819.96 seconds

Explanation:

$V_r$ =Velocity of river = 0.566 m/s

$V_s$ =Velocity of student = 1.17 m/s

Distance to travel = 1.57 km = 1570 m

So,

Time = Distance / Speed

$\frac{1570}{V_s-V_r}+\frac{1570}{V_s+V_r}=t\\\Rightarrow t=\frac{1570}{1.17-0.566}+\frac{1570}{1.17+0.566}\\\Rightarrow t=3503.72\ s$

Time taken by the student to complete the trip is 3503.72 seconds

In still water

$\frac{1570}{V_s}+\frac{1570}{V_s}=t\\\Rightarrow t=\frac{1570}{1.17}+\frac{1570}{1.17}\\\Rightarrow t=2683.76\ s$

The difference in time between moving water and still water is 3503.72-2683.76 = 819.96 seconds

5 0

3 years ago

Which of the following is least likely to result from seafloor spreading

Assoli18 [71]

Answer:

An reversal in the magnetic fields of the north and south pole. This would be the most logical option for me...correct me if I'm wrong.

Explanation:

New seafloor is formed when magma is forced upward toward the surface at a mid-ocean.

7 0

3 years ago

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

2 years ago

What difficulty will you encounter if you only have data from two recording station?

denpristay [2]

<span>If you have only two data from two recording stations then you will be having a hard time finding the correct location of the epicenter. This is because triangulation method requires 3 recording station. If you have 2 recording station, the 2 circles will intersect at 2 points giving you 2 locations that could possibly be the epicenter.</span>

6 0

3 years ago

Read 2 more answers

10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.

oksano4ka [1.4K]

The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1

8 0

3 years ago