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3 years ago

What is the difference between a rock that shows porphyry and a rock called tuff?

Physics

2 answers:

Olin [163]3 years ago

8 0

the rock is caled tuff bc tuff = big tuff

Anna [14]3 years ago

6 0

Who knows. Noone does. That question is way to hard I'm only in 6th grade

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Suppose you need your silicon circuit element to run continuously for 3 minutes before it shuts off long enough to cool back dow

Zigmanuir [339]

The maximum rate at which energy can be added to the circuit element mathematically given as

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$

<h3>What is the maximum rate at which energy can be added to the circuit element?</h3>

Generally, the equation for P is mathematically given as

$P=\ln s \frac{\Delta T}{\Delta t}$

Therefore

$Rate\ of\ Change\ of\ Temp =\frac{p}{lnS}$

$\frac{p}{lnS}=\frac{7.4 \times 10^{-3}}{23 \times 10^{-6} \times 705}$

$\frac{p}{lnS}=0.456^{\circ \mathrm{c}} / \mathrm{sec}$

Max temp Change

$MaxT=5.6^{\circ} \mathrm{C}$

$\text { time }=3 \times 60$

t=180s

In conclusion, Max Energy Rate

$MER =23 \times 10^{-6} \times \frac{301 \times 5.6}{180}$

$MER=5.044 \times 10^{-4} \mathrm{~J} / \mathrm{sec}$

Read more about Energy

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3 0

1 year ago

How much work is required to accelerate a proton from rest up to a speed of 0.993 c? Express your answer with the appropriate un

Keith_Richards [23]

Answer:

(a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

Explanation:

Given that,

Speed = 0.993 c

We need to calculate the work done

Using work energy theorem

The work done is equal to the kinetic energy relative to the proton

$W=K.E$

$W=\dfrac{m_{p}c^2}{\sqrt{1-\dfrac{v^2}{c^2}}}-m_{p}c^2$

Put the value into the formula

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(\dfrac{0.993c}{c})^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=\dfrac{1.67\times10^{-27}\times(3\times10^{8})^2}{\sqrt{1-(0.993)^2}}-1.67\times10^{-27}\times(3\times10^{8})^2$

$W=1.122\times10^{-9}\ J$

$W=7001\ MeV$

(b). We need to calculate the momentum of this proton

Using formula of momentum

$p=\dfrac{m_{0}v}{\sqrt{1-\dfrac{v^2}{c^2}}}$

Put the value into the formula

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(\dfrac{0.993c}{c})^2}}$

$p=\dfrac{1.67\times10^{-27}\times0.993c}{\sqrt{1-(0.993)^2}}$

$p=1.404\times10^{-26}c$

$p=4.20\times10^{8}\ kg-m/s$

Hence, (a). The work done is 7001 MeV.

(b). The momentum of this proton is $4.20\times10^{8}\ kg-m/s$ .

4 0

3 years ago

A car with mass 450 kg has a kinetic energy of 16,256 j. What is the speed of the car?

Vikentia [17]

Do you still need the answer ?

7 0

2 years ago

A clothes dryer uses about 7 amps of current from a 240 volt line. How much power does it use?

Effectus [21]

Electrical power = (voltage) x (current)

Electrical power = (240 volts) x (about 7 Amperes)

<em>Electrical power = about 1,680 watts</em>

4 0

2 years ago

A monoenergetic beam of electrons is incident on a single slit of width 0.500 nm. A diffraction pattern is formed on a screen 25

Dmitry_Shevchenko [17]

Answer:

KE=1.366×10⁻¹⁶J or

KE=854ev

Explanation:

single slit diffraction pattern occur where sinФ=mλ/a for m=±1,±2,±3,.......

Here λ is wavelength

The spacing between successive minima is then

Δy= $y_{m+1}-y_{m}$

Δy=λ(L/a)

as given Δy=2.10cm, L=25cm and a=0.500nm

so for λ

λ=Δy(a/L)

$=(2.10*10^{-2}m )(\frac{(0.500*10^{-9}m)}{25*10^{-2}m} )$

λ=4.2×10⁻¹¹m

The momentum of one of these electrons is then

p=h/λ

$p=\frac{6.63*10^{-34} }{4.2*10^{-11}}\\ p=1.578*10^{-23}kgm/s$

Assuming the electron is non-relativistic, its kinetic energy is

$KE=p^{2}/2m_{e}\\ KE=\frac{(1.578*10^{-23} )^{2} }{2(9.11*10^{-31})}\\ KE=1.366*10^{-16}J$

Convert joules into electronvolt we get

KE=854ev

6 0

3 years ago