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3 years ago

What potential difference is needed to accelerate a he+ ion (charge +e, mass 4u) from rest to a speed of 1.5×106 m/s ? e?

Physics

1 answer:

Nataly_w [17]3 years ago

8 0

We can solve the problem by using conservation of energy.

In fact, the electric potential energy lost by the charge when moving through the potential difference is equal to the kinetic energy acquired:

$\Delta U=\Delta K =K_f -K_i =K_f$

where $K_f$ is the final kinetic energy, and $K_i$ is the initial kinetic energy, which is zero since the particle starts from rest.

We can rewrite the equation above as:

$q \Delta V=\frac{1}{2}mv^2$

where

$q=e=1.6 \cdot 10^{-19}C$ is the charge of the ion

$\Delta V$ is the potential difference

$m=4u=4 (1.67 \cdot 10^{-27} kg)=6.7 \cdot 10^{-27} kg$ is the mass of the ion

$v=1.5 \cdot 10^6 m/s$ is the final speed of the ion

Substituting the numbers and rearranging the equation, we can find the potential difference needed:

$\Delta V=\frac{mv^2}{2q}=\frac{(6.7 \cdot 10^{-27}kg)(1.5 \cdot 10^6 m/s)^2}{2(1.6 \cdot 10^{-19}C)}=4.7 \cdot 10^4 V=47 kV$

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The amount of heat needed to raise the temperature of a 2kg object from 15°C to 25°C is 34000J.

HOW TO CALCULATE SPECIFIC HEAT CAPACITY:

The amount of heat absorbed by an object can be calculated by using the following expression:

Q = m.c.∆T

Where;

Q = amount of heat absorbed or released (J)
m = mass of object
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)

According to this question, 2 kg object has a specific heat capacity of 1,700J/kg°C and was raised from a temperature of 15 Celsius to 25 Celsius. The heat absorbed is calculated as follows:

Q = 2 × 1700 × {25 - 15}

Q = 3400 × 10

Q = 34000J

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Learn more about how to calculate heat absorbed at: brainly.com/question/11194034?referrer=searchResults

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Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.

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Answer:

$g=13.42\frac{m}{s^2}$

Explanation:

1) Notation and info given

$\rho_{center}=13000 \frac{kg}{m^3}$ represent the density at the center of the planet

$\rho_{surface}=2100 \frac{kg}{m^3}$ represent the densisty at the surface of the planet

r represent the radius

$r_{earth}=6.371x10^{6}m$ represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of the radius

$r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}$

$r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}$

So we can create a linear model in the for y=b+mx, where the intercept b= $\rho_{center}=13000 \frac{kg}{m^3}$ and the slope would be given by $m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}$

So then our linear model would be

$\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r$

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be $dV=r^2 sin\theta d\phi d\theta dr$ .

And the total mass would be given by the following integral

$M=\int \rho (r) dV$

Replacing dV we have the following result:

$M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)$

We can solve the integrals one by one and the final result would be the following

$M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})$

Simplyfind this last expression we have:

$M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})$

$M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]$

And replacing the values we got:

$M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg$

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$g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}$

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