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umka2103 [35]
3 years ago
12

An electron traveling parallel to a uniform electric field increases its speed from 2.0 * 107 m/s to 4.0 * 107 m/s over a distan

ce of 1.2 cm. What is the electric field strength?
Physics
2 answers:
vodka [1.7K]3 years ago
4 0

Answer:

E = 2.84 x 10⁵ N/ C

Explanation:

given,

initial speed of electron,u = 2 x 10⁷ m/s

final speed of electron, v = 4 x 10⁷ m/s

distance = 1.2 cm

electric field strength = ?

using equation of motion for acceleration of electron

v² = u² + 2 a s

(4\times 10^7)^2 = (2\times 10^7)^2 + 2 \times 0.012 \times a

a = \dfrac{12\times 10^{14}}{2\times 0.012}

    a = 5 x 10¹⁶ m/s²

now,

Force of electric field is equal to , F = q E

and we also know that F = m a

now,

E = \dfrac{ma}{q}

where q is the charge of electron = 1.6 x 10⁻¹⁹ C

m is mass of electron, m = 9.1 x 10⁻³¹ Kg

so,

E = \dfrac{9.1\times 10^{-31}\times 5\times 10^{16}}{1.6\times 10^{-19}}

       E = 2.84 x 10⁵ N/ C

the electric field strength is equal to E = 2.84 x 10⁵ N/ C

Aneli [31]3 years ago
4 0

Answer:

The electric field strength is 2.8\times10^{5}\ N/C

Explanation:

Given that,

Initial velocity v_{i}= 2.0\times10^{7}\ m/s

Final velocity v_{f}=4.0\times10^{7}\ m/s

Distance = 1.2 cm

We need to calculate the acceleration

Using equation of motion

v^2=u^2+2as

Put the value into the formula

(4.0\times10^{7})^2=(2.0\times10^{7})^2+2\times a\times1.2\times10^{-2}

a=\dfrac{(4.0\times10^{7})^2-(2.0\times10^{7})^2}{2\times1.2\times10^{-2}}

a=5\times10^{16}\ m/s^2

We need to calculate the electric field strength

Using formula of electric force

F=qE

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{9.1\times10^{-31}\times5\times10^{16}}{1.6\times10^{-19}}

E=2.8\times10^{5}\ N/C

Hence, The electric field strength is 2.8\times10^{5}\ N/C

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A train moves with a constant velocity for 15s for 155m. How fast is the train moving?
bonufazy [111]

Answer: 10.3m/s

Explanation:

In theory and for a constant velocity the physics expression states that:

Eq(1):  distance  = velocity times time <=> d = v*t    for v=constant.

If we solve Eq (1) for the velocity (v) we obtain:

Eq(2):  velocity  = distance divided by time <=> v = d/t

Substituting the known values for t=15s and d=155m we get:

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3 years ago
What are the units, if any, of the particle in a box wavefunction. What does this mean?
SIZIF [17.4K]

Answer:

  • [\psi]= [Length^{-3/2}]
  • This means that the integral of the square modulus over the space is dimensionless.

Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \,  dy \,  dx

must be dimensionless, as represents a probability.

As the differentials has units of length

[dx]=[dy]=[dz]=[Length]

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

[\psi]^2 = [Length^{-3}]

taking the square root this gives us :

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5 0
3 years ago
In a lab, four balls have the same velocities but different masses.
olya-2409 [2.1K]

Answer:

New Momentum of Ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

<u>Explanation:</u>

Given:

Mass of Ball A=1kg

Mass of Ball B= 2kg

Mass of Ball C=5kg

Mass of Ball D=7kg

Velocities of A=B=C=D=2.2\frac{m}{s}

Momentum of Ball A=2.2\frac{k g m}{s}

Momentum of Ball B=4.4 \frac{k g m}{s}

Momentum of Ball C=11\frac{k g m}{s}

Momentum of Ball D=15\frac{k g m}{s}

To Find:

Change in Momentum When of Ball B gets tripled

Solution:

Though all balls have same velocity, thus we get

Velocities of A=B=C=D=2.2\frac{m}{s}

Initial Momentum of Ball B=4.4\frac{k g m}{s}

If the Mass of Ball B gets tripled;

We get New Mass of Ball B=3×Actual Mass of the ball

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Thus we get Mass of Ball B=6kg

According to the formula,  

Change in momentum of Ball B \Delta p=m \times \Delta v

Where \Delta p=change in momentum

          m=mass of the ball B

         \Delta v=change in velocity ball B

And \Delta v=v, since all balls, have same velocity

Thus the above equation, changes to

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Substitute all the values in the above equation we get

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Result:

 Thus the New Momentum of ball B=13.2 \frac{\mathrm{kgm}}{\mathrm{s}}

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