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3 years ago

Compared to gamma rays, X–rays have relatively

2 answers:

8 0

Gamma rays are known to have the greatest frequency and the shortest wavelength. What we also know is that frequency and energy are directly proportional (they are the same), and that the wavelength is the opposite of them - if frequency/energy are high, the wavelength will be short, and vice versa.
Since X-rays are the opposite of gamma rays, then the correct answer is the third option - less energy and long wavelengths.

Eddi Din [679]3 years ago

5 0

C) less energy and long wavelenghts

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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so it's not doing anything but changing the units.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (we can elevate any conversion factor to any number and they still will be 1) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

2 years ago

3. What is the potential energy of a 8 Newton book sitting on a shelf that is 12 meters high?

Svetradugi [14.3K]

Answer:

P = 96 J

Explanation:

Given that,

Weight of the book, W = mg = 8 N

It is placed at a height of 12 m

We need to find the potential energy of the book. The potential energy of an object is given by the formula as follows :

E = mgh

mg = Weight

$E=8\ N\times 12\ m\\E=96\ J$

So, the potential energy of the book is 96 J.

8 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .