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1 year ago

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what are the speeds of (a) a proton that is accelerated from rest through a potential difference of −1000 v−1000 v and (b) an el

ectron that is accelerated from rest through a potential difference of 1000 v?

1 answer:

Evgen [1.6K]1 year ago

5 0

Answer:

This is the answer: The speed of a proton is about 5.0 × 10⁵ m/s

Explanation:

Because of the speeds of protons! :D

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A steel bar that is at 10 ° c is 5 meters long, a bar for heated to 120 ° c, how long is that bar? Α = 1.2.10- ° c

svet-max [94.6K]

Answer:

1) 5.0066 m

2A) β = 3×10⁻⁷ / °C

2B) 2500.045 cm²

3A) γ = 8.1×10⁻⁵ / °C

3B) 1618.144 cm³

Explanation:

1) Linear thermal expansion is:

ΔL = α L₀ ΔT

where ΔL is the change in length,

α is the linear thermal expansion coefficient,

L₀ is the original length,

and ΔT is the change in temperature.

Given L₀ = 5 m, ΔT = 110°C, and α = 1.2×10⁻⁵ / °C:

ΔL = (1.2×10⁻⁵ / °C) (5 m) (110°C)

ΔL = 0.0066 m

The length increases by , so the new length is:

L = L₀ + ΔL

L = 5 m + 0.0066 m

L = 5.0066 m

2A) The surface expansion coefficient is:

β = 2α

β = 2 (1.5×10⁻⁷ / °C)

β = 3×10⁻⁷ / °C

2B) The change in area is:

ΔA = β A₀ ΔT

ΔA = (3×10⁻⁷ / °C) (50 cm × 50 cm) (60°C)

ΔA = 0.045 cm²

So the new area is:

A = A + ΔA

A = 2500 cm² + 0.045 cm²

A = 2500.045 cm²

3A) The volumetric expansion coefficient is:

γ = 3α

γ = 3 (2.7×10⁻⁵ / °C)

γ = 8.1×10⁻⁵ / °C

3B) The change in volume is:

ΔV = γ V₀ ΔT

ΔV = (8.1×10⁻⁵ / °C) (1600 cm³) (140°C)

ΔV = 18.144 cm³

So the new area is:

V = V + ΔV

V = 1600 cm³ + 18.144 cm³

V = 1618.144 cm³

6 0

3 years ago

A 15.0 mW laser puts out a narrow beam 2.00 mm indiameter.

Murljashka [212]

Answer:

1341.03 V/m

Explanation:

The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.

$S = \frac{P}{A}$ = cε₀ $E^{2} _{rms}$

⇒ $\frac{P}{A}$ = cε₀ $E^{2}_{rms}$

Where;

P is the power output

A is the area of the beam

c is speed of light

ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m

$E_{rms}$ is the average (rms) value of electric field

Making electricfield $E_{rms}$ the subject of the equation

$E^{2}_{rms}$ = P / Acε₀

$E_{rms}$ = √(P / Acε₀)

But area A = πr²

$E_{rms}$ = √(P / πr²cε₀)

Given:

Output power, P = 15 mW = 0. 015 W

Diameter, d = 2 mm = 0.002 m

⇒ Radius, $r = \frac{d}{2} = \frac{0.002}{2} = 0.001 m$

Solving for average (rms) value of electric field;

$E_{rms} = \sqrt{\frac{0.015 W}{\pi * (0.001 m)^2 * (3 * 10^8 m/s) * (8.85 * 10^-12) C^2/Nm^2} }$

$E_{rms}$ = 1341.03 V/m

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3 years ago

In Ørsted’s observation, the current-carrying wire acted like a

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B, C. Also literally a quick search yielded these results, roughly half the time to type this out.

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The fact that Voyager 10 continues to speed out of the solar system, even though its rockets have no fuel, is an example of Grou

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Answer:

The universal law of gravitation.

PE = m * G M / R^2 potential energy of mass m due to attractive forces

If the kinetic energy of mass m is greater than the energy due to the attractive masses then then mass m can continue indefinitely away from the attracting masses.

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