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3 years ago

13

The potential difference between A and B is 5.0 V. A proton starts from rest at A. When it reaches B what is its kinetic energy?

(e = 1.60 x 10^-19 C)

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

total kinetic energy is 8 × $10^{-19}$ J

Explanation:

given data

potential difference = 5 V

e = 1.60 × $10^{-19}$ C

to find out

what is kinetic energy

solution

we will apply here conservation of energy that is

change in potential energy is equal to change in kinetic energy

so

change potential energy is e × potential difference

change potential energy = 1.60 × $10^{-19}$ × 5

change potential energy = 8 × $10^{-19}$ J

so change in kinetic energy = 8 × $10^{-19}$ J

and we know proton start from rest that mean ( kinetic energy is 0 ) so

change in KE is total KE

total kinetic energy is 8 × $10^{-19}$ J

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3 years ago

If a coin has a mass of 29.34g and drops from a height of 14.7 meters, what is its loss in gravitational potential energy? Show

Serga [27]

$U= mgh$
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do not forget to change grams to kilograms and if it were in centimeters convert them to meters.

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3 years ago

Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al

Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire $(m) = 1.10g = 1.10*10^{-3} kg$

Density $(\rho)= 8.92*10^3kg/m^3$

Resistively of copper $(\gamma) = 1.7*10^{-8}\Omega \cdot m$

Resistance (R) = 0.390\Omega

Volume is defined as,

$V= lA \text{ and }\frac{m}{\rho}$

$lA= \frac{1.10*10^{-3}}{8.92*10^3}$

$lA = 1.233*10^{-7} m^3$ (1)

We know that,

$\frac{l}{A} = \frac{R}{\gamma}$

$\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}$

$\frac{l}{A} = 2.2941*10^7 m^{-1}$ (2)

Multiplying equation we have

$l^2 = (1.233*10^{-7})( 2.2941*10^7)$

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3 years ago

Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

a)

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$

e) Recall that the average of a function between two values is the integral (area under the curve) divided by the length of the interval:

Average velocity = $\frac{22}{2} = 11\, \,\frac{m}{s}$

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3 years ago

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nikklg [1K]

Answer:

The wavelength in miles is <u>0.1165 miles</u>.

Explanation:

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We need to convert the wavelength from meters to miles.

In order to convert meters to miles, we have to use their conversion factor.

We know that,

1 meter = $\frac{1}{1609}\ miles$

Therefore, the conversion factor is given as:

$CF=\frac{1}{1609}\ miles\ per\ meter$

So, the wavelength in miles is given as:

$Wavelength=\textrm{Wavelength in meters}\times CF\\\\Wavelength=187.37\ m\times \frac{\frac{1}{1609}\ miles}{1\ m}\\\\Wavelength=\frac{187.37}{1609}\ miles\\\\Wavelength=0.1165\ miles$

Hence, the wavelength in miles is 0.1165 miles.

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3 years ago