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3 years ago

7

Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go

?

1 answer:

AURORKA [14]3 years ago

8 0

Answer:

123 kilometer

Explanation:

none

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La masa de un camión es de

Nutka1998 [239]

Answer:

45000 kg and 45 tons

Explanation:

The expression in kilograms and tons is shown below;

As we know that

1 gr is 0.001 kg

So, 45000000 = 45,000 kg

And,

1 kg = 0.001 tons

So, 45000 kg = 45 tons

Therefore the same would be considered

7 0

3 years ago

Lol<br>idk what this is

alex41 [277]

Answer:

I know

Explanation:

Physics= hard

3 0

3 years ago

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4. Unlike a gas, plasma can

kap26 [50]

Answer:

Unlike a gas, plasma can conduct electricity and respond to magnetism. That's because plasma contains charged particles called ions.

4 0

3 years ago

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Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo

klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy ( $E_{k}$ ) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

$E_{k}$ = h(f − f₀)

$E_{k}$ = hf - hf₀

E is the energy of the absorbed photons: E = hf

ϕ is the work function of the surface: ϕ = hf₀

$E_{k}$ = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy $E_{k}$ = 4.16×10⁻¹⁷ J

Speed of light c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js

E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

E = 53.8356 x 10⁻¹⁶ J

from $E_{k}$ = E - ϕ ;

ϕ = E - $E_{k}$

ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J

4 0

3 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

3 years ago