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3 years ago

In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find

Physics

1 answer:

katrin2010 [14]3 years ago

5 0

Answer:

a = 1.055 x 10¹⁷ m/s²

Explanation:

First, we will find the force on electron:

$E = \frac{F}{q}\\\\F = Eq\\$

where,

F = Force = ?

E = Electric Field = 6 x 10⁵ N/C

q = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

$F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\$

F = 9.6 x 10⁻¹⁴ N

Now, we will calculate the acceleration using Newton's Second Law:

$F = ma\\a = \frac{F}{m}\\$

where,

a = acceleration = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

therefore,

$a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\$

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Let A^=6i^+4j^_2k^ and B= 2i^_2j^+3k^. find the sum and difference of A and B

andreyandreev [35.5K]

Explanation:

Let $\textbf{A} = 6\hat{\textbf{i}} + 4\hat{\textbf{j}} - 2\hat{\textbf{k}}$ and $\textbf{B} = 2\hat{\textbf{i}} - 2\hat{\textbf{j}} + 3\hat{\textbf{k}}$

The sum of the two vectors is

$\textbf{A + B} = (6 + 2)\hat{\textbf{i}} + (4 - 2)\hat{\textbf{j}} + (-2 + 3)\hat{\textbf{k}}$

$= 8\hat{\textbf{i}} + 2\hat{\textbf{j}} + \hat{\textbf{k}}$

The difference between the two vectors can be written as

$\textbf{A - B} = (6 - 2)\hat{\textbf{i}} + (4 - (-2))\hat{\textbf{j}} + (-2 - 3)\hat{\textbf{k}}$

$= 4\hat{\textbf{i}} + 6\hat{\textbf{j}} - 5\hat{\textbf{k}}$

8 0

3 years ago

Can someone please helpppp meeeeeee

kaheart [24]

1.) Pitch

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3.)Density/Elastic Properties-b. Two of the above

4.)Liquids

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6.) Frequency Increases

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3 years ago

Read 2 more answers

Homework help plz it would help a lot

blagie [28]

#82

here we know that

acceleration = 2 m/s/s

time = 5 s

initial speed = 4 m/s

now we can use kinematics to find the final speed

$v_f = v_i + at$

$v_f = 4 + 2(5)$

$v_f = 14 m/s$

So correct answer will be option D)

#83

here we know that

acceleration = 3 m/s/s

time = 4 s

initial speed = 5 m/s

now we can use kinematics to find the final speed

$v_f = v_i + at$

$v_f = 5 + 3(4)$

$v_f = 17 m/s$

So correct answer will be option C)

#84

here we know that

acceleration = 7 m/s/s

time = 3 s

initial speed = 8 m/s

now we can use kinematics to find the final speed

$v_f = v_i + at$

$v_f = 8 + 7(3)$

$v_f = 29 m/s$

So correct answer will be option C)

6 0

3 years ago

If an object falling freely were somehow equipped with an odometer to measure the distance it travels, then the amount of distan

ioda

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

$h=ut+\frac{1}{2}gt^2$

here u=0 and t=time taken

$h=\frac{1}{2}gt^2$

for $t=1 s$

$h_1=\frac{1}{2}g$

for $t=2 s$

$h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g$

distance traveled in 2 nd sec $=2g-\frac{1}{2}g=\frac{3}{2}g$

for $t=3 s$

$h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g$

distance traveled in 3 rd sec $=\frac{9}{2}g-2g=\frac{5}{2}g$

so we can see that distance traveled in each successive second is increasing

5 0

3 years ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

3 years ago