4. The result is that a(n)<br> electric current is induced<br> PLA

Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strengt

jeka57 [31]

If the field strength between the plates is 9 volts per cm, then over a path of 1/2cm parallel to the field, the potential changes by (1/2) x (9 volts) = 4.5 volts.

5 0

3 years ago

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An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu

ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, $qE \times S + mg \times S = 0.5 \times mv^{2}$

$1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}$

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight and the above equation will be as follows.

$(1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}$

v = $sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}$

= 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0

3 years ago

A gamma ray photon has an energy of 0.91 GeV. (1 GeV = 109 eV.) What is the wavelength of the gamma ray in fm? (1 fm = 10-15 m)?

anastassius [24]

Answer:

$\lambda=1.37 fm$

Explanation:

The Planck Eistein relation, states that the energy of a photon is proportional to its frequency:

$E=h\nu(1)$

h is the Plank constant.The frequency of a photon is defined as the speed of light over its wavelength:

$\nu=\frac{c}{\lambda}(2)$

Replacing (2) in (1):

$E=\frac{hc}{\lambda}\\\lambda=\frac{hc}{E}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{0.91*10^{9}eV}\\\\\lambda=(1.37*10^{-15}m)*\frac{1fm}{10^{-15}m}\\\\\lambda=1.37 fm$

6 0

3 years ago

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

garri49 [273]

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

6 0

4 years ago

Two plane mirrors stand facing each other, 3.08 m apart, and a woman stands between them. The woman faces one of the mirrors fro

Svetllana [295]

Answer:

A) 1.3m

B) it shows the back of her hand

C) 4.86m

D) it shows the palm of her hand

E) 7.46m

F) it shows the back of her hand.

Explanation:

Detailed explanation is shown in the illustrative diagram below.

6 0

3 years ago

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4. The result is that a(n) electric current is induced PLA