Okay! Last question was a warm-up question. And now to get your brain thinking some more, how about another one?:

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work

nika2105 [10]

Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

Q:Heat (J)

Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

Delta U = 658J - 178J = 480J

6 0

3 years ago

Things you can do with an environmental engineering major and a broadcasting minor?

Vika [28.1K]

Answer:

With an Environmental Engineering and a broadcasting minor

You can work as an On Air personality that host programs that provide your audience with documentaries about the environments and project carried out by Environmental Engineer

and also you can work as a journalist that explore the world making research that will preserve the environment and leveraging the media as a broadcaster to provide this research findings as a video for you audience

Explanation:

In order to get a better understanding let define some terms

Environmental Engineer :

Environmental engineers resolve and help prevent environmental problems. They work in many areas, including air pollution control, industrial hygiene, toxic materials control, and land management. The duties of an environmental engineer range from planning and designing an effective waste treatment plant to studying the effects of acid rain on a particular area. An environmental engineer is sometimes required to work outdoors, though most of her work is done in a laboratory or office setting. Career opportunities for environmental engineers exist in consulting, research, corporate, and government positions.

Broadcasting:

Broadcasting is the distribution of audio or video content to a dispersed audience via any electronic mass communications medium, but typically one using the electromagnetic spectrum (radio waves), in a one-to-many model.

7 0

3 years ago

How do you convert Kg's to Newtons?

MA_775_DIABLO [31]

You don't convert kilograms to newtons. By the time you've heard of these units, you know that 'kilogram' is a unit of mass, 'newton' is a unit of force or weight, and that mass and weight are different things.

Mass and force are <u>related</u> by Newton's second law:

Force = Mass x acceleration .

From this simple formula, you can see that in order to relate a mass to a force, you need to know an acceleration. And if the acceleration changes, then the relationship between the force and the mass also changes. So there's no direct conversion.

ON EARTH ONLY, one kilogram of mass <em>weighs</em> 9.8 newtons. The acceleration that connects them is the acceleration of gravity on Earth. In other places, with different gravitational accelerations, 1 kilogram weighs more or less newtons.

But they don't convert directly. That would be like asking "How do you convert miles to miles-per-hour ?"

5 0

3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

5 0

3 years ago

If the car has a mass of 1200 kg, how much force must the engine provide?

vazorg [7]

Force in Newtons =

(1200)•(the rate of acceleration you want)

It depends on how fast you want to accelerate.

6 0

3 years ago