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3 years ago

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You and your 3 friends are driving down the road. All the wheels go flat at the exact same time, that's not good. You have to us

e a 2,100 N force to push the car 4 meters to the side of the road. How much work did you do on the car?

1 answer:

mel-nik [20]3 years ago

7 0

Answer:

Workdone = 8400 Nm

Explanation:

Given the following data;

Force = 2100 N

Distance = 4 meters

To find the work done;

Workdone = force * distance

Workdone = 2100 * 4

Workdone = 8400 Nm

Therefore, the amount of work done on the car is 8400 Newton meters.

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Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu

Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density $D$ appears as a physical characteristic property of matter that establishes a relationship between the mass $m$ of a body or substance and the volume $V$ it occupies:

$D=\frac{m}{V}$ (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is $m=0.155g$ and th volume $V=0.000235L=0.235mL$

Solving (1) with these values:

$D=\frac{0.155g}{0.235mL}$ (2)

$D=0.660g/mL$ (3)

<h2>Answer B:</h2>

In this case the mass of a sample is $m=4.71g$ and its density is $D=3.63g/mL$ .

Isolating $V$ from (1):

$V=\frac{m}{D}$ (4)

$V=\frac{4.71g}{3.63g/mL}$ (5)

$V=1.297mL$ (5)

<h2>Answer C:</h2>

In this case the volume of a sample is $V=0.293mL$ and its density is $D=0.930g/mL$ .

Isolating $m$ from (1):

$m=D.V$ (6)

$m=(0.930g/mL)(0.293mL)$ (7)

$m=0.272g$ (8)

4 0

3 years ago

Read 2 more answers

What kind of friction is occurring between a pencil and desktop when you flick the pencil?

DanielleElmas [232]

The third one sliding friction
Explanation:

8 0

2 years ago

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago

If a piece of iron is placed in pure nitrogen, nothing happens. If the in is exposed to air, it begins to rust. What conclusion

kari74 [83]

Based on this, you can tell that Iron(Fe) does not react with Nitrogen(N), but does with something in normal air. The element the Iron is reacting with Oxygen(O), and the compound that is formed is called iron oxide.

6 0

3 years ago

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave. What is the smallest separation distance b

AveGali [126]

Answer:

The smallest separation distance between the speakers is 0.71 m.

Explanation:

Given that,

Two speakers, one directly behind the other, are each generating a 240-Hz sound wave, f = 240 Hz

Let the speed of sound is 343 m/s in air. The speed of sound is given by the formula as :

$v=f\lambda\\\\\lambda=\dfrac{v}{f}\\\\\lambda=\dfrac{343\ m/s}{240\ Hz}\\\\\lambda=1.42\ m$

To produce destructive interference at a listener standing in front of them,

$d=\dfrac{\lambda}{2}\\\\d=\dfrac{1.42}{2}\\\\d=0.71\ m$

So, the smallest separation distance between the speakers is 0.71 m. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers