All categories

3 years ago

Lab number 14: sudden stops hurt newtons first law

Physics

1 answer:

kompoz [17]3 years ago

7 0

Answer:

you want to know newtons first law here.

Explanation:

An object that is at rest stays at rest or stays in motion.

You might be interested in

A. assuming a perpetual inventory system and using the weighted average method, determine the weighted average unit cost after t

Monica [59]

Assuming a perpetual inventory system and using the weighted average method, the weighted average unit is determined as $11.44 after the October 22 purchase.

<h3>What is Weighted Average Cost (WAC)?</h3>

The Weighted Average Cost (WAC) method of inventory valuation in accounting uses a weighted average to establish the COGS and inventory levels.

The price of the products up for grabs is divided by the quantity of them in the weighted average cost technique.

The WAC technique is appropriate under both GAAP and IFRS accounting. Weighted Average Cost (WAC) Method Formula

<h3>Weighted Average Cost</h3>

Weighted Average Unit Costs = [360 units×$12 + (320-180) ×$10] / [360+(320-180)]units}

Weighted Average Unit Costs = $5720 / 500 units

Weighted Average Unit Costs = $11.44

Costs of goods that are offered for sale are calculated using beginning inventory value plus acquisitions.

Units available for sale are the number of units that can be sold by a company or the total number of units that are in its inventory.

To know more about weighted average cost, Visit

brainly.com/question/13543092

#SPJ4

8 0

2 years ago

What profession is most likely to make use of amphoras and candelas

Monica [59]

Biologists probably

5 0

3 years ago

Night and day have approximately equal length at what time or times of the year?

Thepotemich [5.8K]

There are two every year – in September and March – when the Sun shines directly on the Equator and the length of day and night is nearly equal.

4 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

3 years ago

The rock cycle is driven by internally by heat from

rusak2 [61]

Hi, earth and the sun would be your answer.

4 0

3 years ago

Read 2 more answers