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3 years ago

6

An electron moving parallel to a uniform electric field increases its speed from 2.0 × 10^7 m/s to 4.0 × 10^7 m/s over a distanc

e of 1.2 cm. What is the electric field strength?

1 answer:

julia-pushkina [17]3 years ago

8 0

Answer:

$E = 2.84 * 10^5 N/C$

Explanation:

The speed increased from 2.0 * 10^7 m/s to 4.0 * 10^7 m/s over a 1.2 cm distance.

Let us find the acceleration:

$v^2 = u^2 + 2as$

$(4.0 * 10^7)^2 = (2.0 * 10^7)^2 + 2 * a * 0.012\\\\(4.0 * 10^7)^2 - (2.0 * 10^7)^2 = 0.024a\\\\1.2 * 10^{15}= 0.024a\\\\a = 1.2 * 10^{15} / 0.024\\\\a = 5 * 10^{16} m/s^2$

Electric force is given as the product of charge and electric field strength:

F = qE

where q = electric charge

E = Electric field strength

Force is generally given as:

F = ma

where m = mass

a = acceleration

Equating both:

ma = qE

E = ma / q

For an electron:

m = 9.11 × 10^{-31} kg

q = 1.602 × 10^{-19} C

Therefore, the electric field strength of the electron is:

$E = \frac{9.11 * 10^{-31} * 5 * 10^{16}}{1.602 * 10^{-19}} \\\\E = 2.84 * 10^5 N/C$

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Isla made a diagram to compare X-rays and radio waves.

pickupchik [31]

A: Has a higher frequency --> X-rays

B: Travels at the same speed --> Both

C: Has a longer wavelength --> Radio waves

Explanation:

Electromagnetic waves are waves consisting of oscillations of electric and magnetic field in a direction perpendicular to the direction of motion of the wave.

Electromagnetic waves travel at the speed of light ( $c=3.0\cdot 10^8 m/s$ ) in a vacuum, and they are transverse in nature (the direction of vibration is perpendicular to the direction of propagation).

Electromagnetic waves are classified into different types according to their wavelength and frequency. In order from shortest to longest wavelength (and so, from highest to lowest frequency, since frequency is inversely proportional to the wavelength), we have:

Gamma rays

X-rays

Ultraviolet radiation

Visible light

Infrared radiation

Microwaves

Radio waves

Therefore, we can no says that:

X-rays: has a higher frequency than Radio waves

Radio waves: have a longer wavelength than x-rays

Both: they travel at the same speed

Therefore, the correct pairing is:

A: Has a higher frequency --> X-rays

B: Travels at the same speed --> Both

C: Has a longer wavelength --> Radio waves

Learn more about electromagnetic waves:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

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2 years ago

Radio waves travel 300,000,000 m/s. The frequency is 101,700,000. ehats the wavelength

Nesterboy [21]

Answer:

we have formula of frequency :

frequency(f)= speed of sound(c)/wavelength(λ)

for wavelength we swipe it with frequency as follows

λ=c/f

λ=300,000,000/101,700,000

λ=2.949

8 0

3 years ago

A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an

Mademuasel [1]

Answer:

The angular acceleration is $0.209\ rad/s^2$

Explanation:

Given that,

Angular velocity, $\omega_{i} = 30.0\ rpm$

Angular velocity, $\omega_{f} = 50.0\ rpm$

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}$

$\alpha=\dfrac{50.0-30.0}{10.0}$

Now, we change the angular velocity in rad/s.

$\omega=20\times\dfrac{2\pi}{60}$

$\omega=2.09\ rad/s$

$\alpha=\dfrac{2.09}{10.0}$

$\alpha=0.209\ rad/s^2$

Hence, The angular acceleration is $0.209\ rad/s^2$

5 0

3 years ago

Read 2 more answers

A research submarine has a 30-cm-diameter window that is 8.1 cm thick. The manufacturer says the window can withstand forces up

malfutka [58]

The pressure at a certain depth underwater is:

P = ρgh

P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth

The pressure exerted on the submarine window is:

P = F/A

P = pressure, F = force, A = area

The area of the circular submarine window is:

A = π(d/2)²

A = area, d = diameter

Set the expressions for the pressure equal to each other:

F/A = ρgh

Substitute A:

F/(π(d/2)²) = ρgh

Isolate h:

h = F/(ρgπ(d/2)²)

Given values:

F = 1.1×10⁶N

ρ = 1030kg/m³ (pulled from a Google search)

g = 9.81m/s²

d = 30×10⁻²m

Plug in and solve for h:

h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)

h = 1540m

5 0

3 years ago

A hot-air balloon and its basket are accelerating upward at 0.265 m/s2, propelled by a net upward force of 688 N. A rope of negl

Sergeu [11.5K]

Question:<em> </em><em>Find, separately, them mass of the balloon and the basket (incidentally, most of the balloon's mass is air)</em>

Answer:

The mass of the balloon is 2295 kg, and the mass of the basket is 301 kg.

Explanation:

Let us call the mass of the balloon $m_1$ and the mass of the basket $m_2$ , then according to newton's second law:

$(1). \:F = (m_1+m_2)a$ ,

where $a =0.265m/s^2$ is the upward acceleration, and $F = 688N$ is the net propelling force (counts the gravitational force).

Also, the tension $T$ in the rope is 79.8 N more than the basket's weight; therefore,

$(2). \:T = m_2g+79.8$

and this tension must equal

$T -m_2g =m_2a$

$(3). \:T = m_2g +m_2a$

Combining equations (2) and (3) we get:

$m_2a = 79.8$

since $a =0.265m/s^2$ , we have

$\boxed{m_2 = 301.13kg}$

Putting this into equation (1) and substituting the numerical values of $F$ and $a$ , we get:

$688N = (m_1+301.13kg)(0.265m/s^2)$

$\boxed{m_1 = 2295 kg}$

Thus, the mass of the balloon and the basket is 2295 kg and 301 kg respectively.

8 0

3 years ago