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emmainna [20.7K]
3 years ago
14

What is the speed of electron having the same kinetic energy with photon speed 0.006c?

Physics
1 answer:
Blababa [14]3 years ago
5 0

Answer:

if gossipy egg overriding successors spatial

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SI unit of force is <em>Newton</em>

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Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 10 m/s.
abruzzese [7]

Hi there!

Kinetic energy can be calculated using the following:

\large\boxed{KE = \frac{1}{2}mv^2}}

Where:

KE = Kinetic energy (J)

m = mass (kg)

v = velocity (m/s)

Plug in the given values:

KE = \frac{1}{2}(625)(10^2) = \boxed{31250J}

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What is earth revolution estimated time and direction
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How long it takes to earth to revolve completely on its Axis
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4 years ago
An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0
3 years ago
A charge of 8.0 pc is distributed uniformly on a spherical surface (radius = 2.0 cm), and a second charge of â3.0 pc is distribu
loris [4]
According to Gauss' law, the electric field outside a spherical surface uniformly charged is equal to the electric field if the whole charge were concentrated at the center of the sphere.

Therefore, when you are outside two spheres, the electric field will be the overlapping of the two electric fields:
E(r > r₂ > r₁) = k · q₁/r² + k  · q₂/r² = k · (q₁ + q₂) / r²
where:
k = 9×10⁹ N·m²/C²

We have to transform our data into the correct units of measurement:
q₁ = 8.0 pC = 8.0×10⁻¹² C
q₂ = 3.0 pC = 3.0×10<span>⁻¹² C
</span><span>r = 5.0 cm = 0.05 m

Now, we can apply the formula:
</span><span>E(r) = k · (q₁ + q₂) / r²
      = </span>9×10⁹ · (8.0×10⁻¹² + 3.0×10⁻¹²) / (0.05)²
      = 39.6 N/C

Hence, <span>the magnitude of the electric field 5.0 cm from the center of the two surfaces is E = 39.6 N/C</span>
4 0
4 years ago
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