Ask question

Ask question

All categories

3 years ago

6

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

are 1.00 cm and 1.90 cm. The trajectories are perpendicular to a uniform magnetic field of magnitude 0.0500 T. Determine the energy (in keV) of the incident electron.

1 answer:

Tamiku [17]3 years ago

7 0

Answer:

101.19keV

Explanation:

r1 = 1.0cm = 0.01m

r2 = 1.90cm = 0.019m

B = 0.050T

q = 1.60*10^-19C

m = 9.11 * 10^-31 kg

Mv /r = qB

v = rqB / m

v1 = (0.01 * 1.60*10^-19 * 0.05) / 9.11 * 10^-31

v1 = 8.78 * 10⁷ m/s

V2 = (0.019 * 1.60*10^-19 * 0.05) / 9.11 * 10^-31

V2 = 1.67 * 10⁸ m/s

Kinetic energy of the system K.E = K.E₁ + K.E₂

K.E₁ = ½mv₁² = ½ * 9.11 *10⁻³¹ * (8.78*10⁷)²

k.e₁ = 3.511 * 10⁻¹⁵ J

k.e₂ = 1/2 mv₂² = 1/2 * 9.11*10⁻³¹ * (1.67*10⁸)²

k.e₂ = 1.27*10⁻¹⁴J

K.E = k.e₁ + k.e₂

K.E = 3.511*10⁻¹⁵ + 1.27*10⁻¹⁴

K.E = 1.6211*10⁻¹⁴J

1eV = 1.602 * 10⁻¹⁹J

xeV = 1.6211*10⁻¹⁴J

x = 101,192.23eV = 101.192keV

You might be interested in

A pool ball of 0.17 kg moves at 10 m/s and strikes a stationary ball (at rest) also at 0.17 kg.

Elenna [48]

The final velocity of the second ball is +10 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system must be conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 0.17 kg$ is the mass of the first ball

$u_1 = 10 m/s$ is the initial velocity of the first ball (we take its direction as positive direction)

$v_1 = 0$ is the final velocity of the first ball

$m_2 = 0.17 kg$ is the mass of the second ball

$u_2 = 0$ is the initial velocity of the second ball

$v_2$ is the final velocity of the second ball

Re-arranging the equation and substituting the values, we find:

$v_2 = \frac{m_1 u_1}{m_2}=\frac{(0.17)(10)}{0.17}=10 m/s$

And since the sign is positive, the direction is the same as the initial direction of the first ball.

Learn more about momentum here:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

5 0

4 years ago

Your grandfather clock's pendulum has a length of 0.9930 m. if the clock gains 80 seconds per day, should you increase or decrea

salantis [7]

By Gucci gang gang gang

6 0

3 years ago

How is energy conserved in a transformation?

Aleksandr [31]

This depends entirely on the type of transformation of energy that is taking place, mechanical to thermal, thermal to mechanical, etc. But energy is always conserved, although it may take a different form.

4 0

3 years ago

Read 2 more answers

) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th

Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

Now, we will substitute 0 for both $z_{1}$ and $z_{2}$ , 0 for w, 334.9 kJ/kg for $h_{1}$ , 2726.5 kJ/kg for $h_{2}$ , 5 m/s for $V_{1}$ and 220 m/s for $V_{2}$ .

Putting the given values into the above formula as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

$1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0$

q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

6 0

4 years ago

A horizontal vinyl record of mass 0.10 kg and radius 0.10 m rotates freely about a vertical axis through its center with an angu

Tresset [83]

Answer:

The angular speed of the record is 3.36 rad/s.

Explanation:

Given that,

Mass of record= 0.10 kg

Radius = 0.10 m

Angular speed = 4.7 rad/s

Moment of inertia $I=5.0\times10^{-4}\ kgm^2$

Mass of putty = 0.020 kg

We need to calculate the angular speed

Using law of conservation of momentum

$L_{i}=L_{f}$

$I\omega_{i}=(I+mr^2)\omega_{f}$

$\omega_{f}=\dfrac{I\omega_{i}}{(I+mr^2)}$

Put the value into the formula

$\omega_{f}=\dfrac{5.0\times10^{-4}\times4.7}{5.0\times10^{-4}+0.020\times(0.10)^2}$

$\omega_{f}=3.36\ rad/s$

Hence, The angular speed of the record is 3.36 rad/s.

3 0

4 years ago