All categories

3 years ago

HELPPPPPOPP

Physics

2 answers:

tigry1 [53]3 years ago

7 0

Answer:

0.08

Explanation:

10/120 = 0.08

I hope this helps!

victus00 [196]3 years ago

3 0

Explanation:

v(speed)=10/120=0.08m/s

You might be interested in

Does research lessen its value for its being cyclical? Enumerate Two Possible advantages of a cyclical research

Tanya [424]

Answer:

Research usually is a cyclical process because it starts with a problem and ends with a problem.

This is bad or lessens the value of the research?

No, because there are problems that in nature are cyclical, (for example the related ones to action research).

And it also may be a good thing, solving one "problem" leads to another problem, but in the process of solving the first one we may win a lot of knowledge, and the same happens with the next one, and so on.

Two possible advantages of cyclical research are:

Knowing beforehand that the research will be cyclical, will allow us to estimate better the amount of time and money needed because we already know (more or less) where to aim.

It also may lead to a better end product, as we already know that we must focus in solving one thing and then we can focus in the next one.

Another possible advantage may be that we know that after the work, there will be a new thing to research, and it is fun, so if you are curious enough this may be a good thing (especially in scientific areas, physics, chemistry, etc).

7 0

4 years ago

A year 11 pupil with a mass of 55kg swinging back on their chair and falling off it at a speed of 0.6m/s. What is his kinetic en

posledela

Answer:

Uk = 9.9 J

Explanation:

To calculate the kinetic energie (Uk), you can make use of this formula:

Uk = 0.5 * m * v²

given m = 55 kg and v = 0.6 m/s

Substituting in the formula gives:

Uk = 0.5 * 55 * (0.6)²

Uk = 0.5 * 55 * 0.36

Uk = 9.9 J

Extra:

Now let's examine the formula in relation to the SI units. If you understand the following, it will give you great insight in how smart Phisics is inter twained by looking at formulas and their standard units. It will save you time in future to convert formulas, if you use the right standard units.

The formula for kinetic energie is:

Uk = 0.5 * m * v²

Standard SI unit for mass m is kg.

Standard SI unit for speed v is m/s.

So v * v = v² and therefore v² must have the standard SI unit of m²/s².

From the formula, you see that the unit of Uk must be kg*m²/s² and since Uk is normally given in J, these both forms must be the same !

The main unit for Uk is the Joule. Now please see the picture, which shows the relation between the J and other SI units. Please understand that you can construct your 'own' formulas based these units. Now here is the time saver:

Because almost always the right units are given in a question, or because sometimes you can look up a constant in a table with an exotic and seemingly complicated unit, but that says a lot about the formula which must have been some how involved!

By this, I hope you now understand the implication of using the right standard SI units and how that can help you figure out what formula is needed.

3 0

3 years ago

Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

pochemuha

Explanation:

Formula for calculating the area of a rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm² is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

7 0

3 years ago

A 110 V power line is protected by a 15 A fuse. What is the maximum number of 400 W lamps that can be simultaneously operated in

Nitella [24]

Answer:

Total number of lamps will be 4

Explanation:

We have given power of the lamp W = 400 watt

Potential difference across the lamp V=110 volt

We know that power is equal to $P=VI$

So $400=110\times I$

$I=3.636A$

Total current is given 15 A

As it is given that lamps are connected in parallel so total current is the sum of current through each lamp

So number of lamp will be $n=\frac{15}{3.636}=4.125$

As the lamp can not be in negative

So total number of lamps will be 4

5 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

Read 2 more answers