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2 years ago

A car is initially travelling at 12.1m/s It accelerates at a rate of -3.9m/s^2 for 7 seconds. What is it’s final velocity?

Physics

1 answer:

rosijanka [135]2 years ago

8 0

It will be traveling in the reverse direction it was originally going at 15.2 m/s

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Mrs. Miller is acting a little strange and is pushing as hard as she can against a wall. Using Newton's Laws, what could be say

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Newton’s 3rd Law of Motion

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2 years ago

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For questions 1 - 4 complete the representations for the four patterns below. Provide the mathematical

larisa [96]

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2 years ago

A certain resistance thermometer read 14.5 ohms in pure melting ice and 18.5 ohms in steam at standard atmospheric pressure what

Vadim26 [7]

The resistance of the thermometer at room temperature is 15.04 ohms.

<h3 />

<h3>What is a resistance thermometer?</h3>

A resistance thermometer is a type of thermometer that measures temperature through a change in resistance.

To calculate the resistance of the thermometer at room temperature, we use the formula below.

Formula:

100/27 = 2/(x-14.5)..............Eqquation 1

Where:

x = Resistance of the thermometer at room temperature

Make x the subject of the equation

x = [(27×2)/100]+14.5
x = (54/100)+14.5
x = 0.54+14.5
x = 15.04 ohms.

Hence, The resistance of the thermometer at room temperature is 15.04 ohms.

Learn more about thermometers here: brainly.com/question/1531442

3 0

1 year 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:

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

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2 years ago

In which one of the following situations is zero net work done? A) A ball rolls down an inclined plane. B) A physics student str

lidiya [134]

Answer:

Option D

Explanation:

The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.

In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.

In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

6 0

2 years ago

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