I need help finding moment

In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 27

Vera_Pavlovna [14]

Answer:

9.74 x 10^7 m/s

Explanation:

V = 27000 V

energy of electrons = e x V

K = 1.6 x 10^-19 x 27000 = 43200 x 10^-19 J

Energy = 1/2 m v^2

43200 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v^2 = 9.495 x 10^15

v = 9.74 x 10^7 m/s

4 0

3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

nlexa [21]

Answer:

The maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

Explanation:

To solve this problem it is necessary to apply the concepts on maximum electromotive force.

For definition we know that

$\epsilon_{max} = NBA\omega$

Where,

N= Number of turns of the coil

B = Magnetic field

$\omega =$ Angular velocity

A = Cross-sectional Area

Angular velocity according kinematics equations is:

$\omega = 2\pi f$

$\omega = 2\pi*61.5$

$\omega =123\pi rad/s$

Replacing at the equation our values given we have that

$\epsilon_{max} = NBA\omega$

$\epsilon_{max} = NB(\pi (\frac{d}{2})^2)\omega$

$\epsilon_{max} = (1)(1*10^{-3})(\pi (\frac{7.2*10^{-6}}{2})^2)(123\pi)$

$\epsilon_{max} = 1.5732*10^{-11}V$

Therefore the maximum emf that can be generated around the perimeter of a cell in this field is $1.5732*10^{-11}V$

6 0

3 years ago

Where is the deepest cave in the world? How far down is it located?

Sergeeva-Olga [200]

Answer:

Geology Notes

1 Veryovkina Cave 2212[1] 13.5 km (8.4 mi)[1] Abkhazia / Georgia 43°23′52″N 40°21′37″E.

6 0

3 years ago

The density of an object is 1.23 g/cm3. In which of the following materials will it float

Leviafan [203]

Citric Acid is the correct answer because it contains a density of 1.66 g/cm3, whereas water= 1.00 g/cm3, Olive oil= 0.93 g/cm3, Ethyl alcohol= 0.79 g/cm3

5 0

3 years ago

Read 2 more answers

One ring of radius a is uniformly charged with charge +Q and is placed so its axis is the x-axis. A second ring with charge –Q i

kati45 [8]

Answer:

The force exerted on an electron is $7.2\times10^{-18}\ N$

Explanation:

Given that,

Charge = 3 μC

Radius a=1 m

Distance = 5 m

We need to calculate the electric field at any point on the axis of a charged ring

Using formula of electric field

$E=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

$E_{1}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{1}=\dfrac{9\times10^{9}\times3\times10^{-6}\times5}{(1^2+5^2)^{\frac{3}{2}}}$

$E_{1}=1.0183\times10^{3}\ N/C$

Using formula of electric field again

$E_{2}=\dfrac{kqx}{(a^2+x^2)^{\frac{3}{2}}}\hat{x}$

Put the value into the formula

$E_{2}=\dfrac{9\times10^{9}\times(-3\times10^{-6})\times5}{((0.5)^2+5^2)^{\frac{3}{2}}}$

$E_{2}=-1.064\times10^{3}\ N/C$

We need to calculate the resultant electric field

Using formula of electric field

$E=E_{1}+E_{2}$

Put the value into the formula

$E=1.0183\times10^{3}-1.064\times10^{3}$

$E=-0.045\times10^{3}\ N/C$

We need to calculate the force exerted on an electron

Using formula of electric field

$E = \dfrac{F}{q}$

$F=E\times q$

Put the value into the formula

$F=-0.045\times10^{3}\times(-1.6\times10^{-19})$

$F=7.2\times10^{-18}\ N$

Hence, The force exerted on an electron is $7.2\times10^{-18}\ N$

8 0

3 years ago