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anyanavicka [17]

3 years ago

15

An electron and a proton are each accelerated from rest through a potential difference of 100 V. Afterward, which particle has t

he larger de Broglie wavelength?

1 answer:

Lena [83]3 years ago

3 0

Explanation:

The De-Broglie wavelength in terms of potential difference is given by:

$\lambda=\dfrac{h}{\sqrt{2meV} }$

Where,

h is Planck's constant

m is mass of charged particle

V is potential difference

e is the amount of charge

It means that the De-Broglie wavelength is inversely proportional to the mass.

Since, the mass of the proton is more than the mass of the electron. So, the De- Broglie wavelength of the electron is larger than proton.

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Help please physics !!

malfutka [58]

Answer:

Option A. 1 bar = 1 atm

Explanation:

Pressure has various units of measurement. Each unit of measurement can be converted to other units of measurement. For example:

1 atm = 1 bar

1 atm = 760 mmHg

1 atm = 760 torr

1 atm = 1×10⁵ N/m²

1 atm = 1×10⁵ Pa

With the above conversion scale we can convert from one unit to the other.

Considering the question given above, it is evident from the coversion scale illustrated above that only option A is correct.

Thus,

1 bar = 1 atm

5 0

3 years ago

A camcorder has a power rating of 10 watts. If the output voltage from its battery is 3 volts, what current does it use?

Gemiola [76]

The power is calculated using the following rule:
Power = voltage * current
We are given that:
power = 10 watts
voltage = 3 volts

Substitute with the givens in the above equation to get the value of the current as follows:
P = V*I
10 = 3*I
current (I) = 10/3 amperes

7 0

3 years ago

a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t

Zolol [24]

Answer: $0.146 m/s^{2}$

Explanation:

The <u>centripetal acceleration</u> $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$V$ is the tangential velocity

$r=1.05 m$ is the radius of the circle

On the other hand, the tangential velocity is expressed as:

$V=\omega r$ (2)

Where $\omega$ is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

$\omega=\frac{5 rev}{13.4 s}=0.37 rev/s$ (3)

Substituting (3) in (2):

$V=(0.37 rev/s)(1.05 m)$ (4)

$V=0.39 m/s$ (5)

Substituting (5) in (1):

$a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}$ (6)

Finally:

$a_{c}=0.146 m/s^{2}$

6 0

3 years ago

Read 2 more answers

A 16 kg block is dragged over a rough, hor-

Arlecino [84]

Explanation:

So what's the question here?

3 0

3 years ago

A small segment of wire contains 10 nC of charge. The segment is shrunk to one-third of its original length. A proton is very fa

Alik [6]

To solve this problem we will apply the concepts related to the electric field, linear charge density and electrostatic force.

The electric field is

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Here,

$\lambda$ = Linear charge density

$\epsilon_0$ = Permittivity of free space

r = Distance

The linear charge density can be written as,

Linear charge density is given as

$\lambda = \frac{q}{L}$

Replacing,

$E = \frac{\frac{q}{L}}{2\pi \epsilon_0 r}$

$E = \frac{q}{2\pi \epsilon_0 rL}$

The initial and final electric Force can be written as function of the charge and the electric field as

$F_i = E_i q$

$F_f = E_f q$

If we replace the value for the electric field we have,

$F_i = (\frac{q}{2\pi \epsilon_0 rL})q = (\frac{q^2}{2\pi \epsilon_0 rL})$

Length is one third at the end, then

$F_f = (\frac{q}{2\pi \epsilon_0 r(L/3)})q = (\frac{3q^2}{2\pi \epsilon_0 rL})$

The ratio of the force is

$\frac{F_f}{F_i} = \frac{(\frac{3q^2}{2\pi \epsilon_0 rL})}{(\frac{q^2}{2\pi \epsilon_0 rL})}$

$\frac{F_f}{F_i} = 3$

Therefore the required ratio is 3

7 0

3 years ago