Ask question

Ask question

All categories

3 years ago

9

An accelerating voltage of 2.25 103 V is applied to an electron gun, producing a beam of electrons originally traveling horizont

ally north in vacuum toward the center of a viewing screen 36.4 cm away. (a) What is the magnitude of the deflection on the screen caused by the Earth's gravitational field

1 answer:

Lana71 [14]3 years ago

4 0

Answer:

s= 8.28×10⁻¹⁶m

Explanation:

given

V= 2.25×10³V

from conservation of energy

mv²/2=qΔV

v=√(2qΔV/m)

v= √(2×1.6×10⁻¹⁹×2.25×10³/9.1×10⁻³¹)

=√7.9×10¹⁴m/s

=2.8×10⁷m/s

the deflection of electron beam is

S= gt²/2

recall t= d/v

s=g( $\frac{d}{v}$ )²/2

s= $\frac{1}{2}$ ×9.8×(0.364/2.8×10⁷)²

s= 8.28×10⁻¹⁶m

You might be interested in

Transverse waves are generally stronger than longitudinal waves. True or False ?

kondaur [170]

False
If all other factors, such as medium, are kept the same, longitudinal waves tend to be stronger.

8 0

3 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

You are bungee jumping from a bridge. Initially, while you are falling the slack bungee cord isn’t exerting any forces or torque

harina [27]

Answer:

he fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

Explanation:

Let's pose the solution of this problem, to be able to analyze the firm affirmations.

When the person is falling, the weight acts on them all the time, initially the rope has no force, but at the moment it begins to lash it exerts a force towards the top that is proportional to the lengthening of the rope.

The equation for this part is

Fe - W = m a

k x - mg = m a

As the axis of rotation is located at the top where they jump, there is a torque.

What is it

Fe y - W y = I α

angular and linear acceleration are related

a = α r

Fe y - W y = I a / r

In the fall movement we see that both the force is different from zero, and the torque is different from zero.

When analyzing the statements the d is true

4 0

3 years ago

The 1st Law of Thermodynamics is a statement about (A) Temperatures scales (B) Whether a process can proceed in a specific manne

weqwewe [10]

Answer:

Energy conservation.

Explanation:

The 1st Law of Thermodynamics is a statement about energy conservation. It states that $\Delta U=Q-W$ , which means that if we <u>substract the work W done</u> by the system to the <u>heat Q given</u> to the system we get the <u>change in the internal energy</u> $\Delta U$ , so any excess in energy given to the system appears as internal energy, stating that energy is conserved.

3 0

3 years ago

Two separate but nearby coils are mounted along the same axis. A power supply controls the flow of current in the first coil, an

Anit [1.1K]

Answer:

b.only when the current in the first coil changes.

Explanation:

An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.

8 0

3 years ago