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3 years ago

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(physical science) could someone please help me out with this lab? if i’m being honest i did the lab but i lost all of my work :

( if anyone could help at all that would be great . i am in a bind right now

1 answer:

djverab [1.8K]3 years ago

6 0

Explanation:

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Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1200 km/s , measured relative to the

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Answer:

The maximum electrical force is $2.512\times10^{-2}\ N$ .

Explanation:

Given that,

Speed of cyclotron = 1200 km/s

Initially the two protons are having kinetic energy given by

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2$

When they come to the closest distance the total kinetic energy is converts into potential energy given by

Using conservation of energy

$mv^2=\dfrac{kq^2}{r}$

$r=\dfrac{kq^2}{mv^2}$

Put the value into the formula

$r=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{1.67\times10^{-27}\times(1200\times10^{3})^2}$

$r=9.57\times10^{-14}\ m$

We need to calculate the maximum electrical force

Using formula of force

$F=\dfrac{kq^2}{r^2}$

$F=\dfrac{8.99\times10^{9}\times(1.6\times10^{-19})^2}{(9.57\times10^{-14})^2}$

$F=2.512\times10^{-2}\ N$

Hence, The maximum electrical force is $2.512\times10^{-2}\ N$ .

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When an object experiences uniform circular motion the direction of the net force is?

guajiro [1.7K]

In an uniform circular motion, the direction of the net force on the object is radially inward, passing through the center of the circle.

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Explanation:

Inferiority Complex

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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kolezko [41]

Action-reaction pairs.

This is in reference to Newton’s second law of motion.

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