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bulgar [2K]
4 years ago
5

An electron is to be accelerated from a velocity of 1.50×106 m/s to a velocity of 9.00×106 m/s . through what potential differen

ce must the electron pass to accomplish this?
Physics
1 answer:
levacccp [35]4 years ago
8 0
<span>Velocity of Electron V1 = 1.50 Ă—10^6 m/s Velocity of Electron V2 = 9.00 Ă— 10^6 m/s We know the mass of electron m = 9.11 x 10^-31 Charge of the electron q = 1.60 x 10^-19 Calculating the potential energy at the two stages, K1 = 0.5 x m V1^2 = 0.5 x 9.11 x 10^-31 x (1.50 Ă—10^6)^2 = 10.24 x 10^-19 K2 = 0.5 x m V2^2 = 0.5 x 9.11 x 10^-31 x (9.00 Ă—10^6)^2 = 368.95 x 10^-19 Change in Potential energy delta K= K2 - K1 = 358.71 x 10^-19 As we know for total energy conservation total change in energy is 0 so q x delta V + delta K = 0 delta V = - delta K / q = 358.71 x 10^-19 / 1.60 x 10^-19 = 224.19V Potential difference for the electron to pass = 224.19V</span>
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Arturiano [62]

Answer:

5120 m/s

Explanation:

The acceleration due to gravity is:

g = MG / r²

where M is the mass of the earth, G is the universal constant of gravitation, and r is the distance from the earth's center to the object's center.

Here, r = h + R, where h is the height of the chest above the surface and R is the radius of the earth.

g = MG / (h + R)²

Acceleration is the derivative of velocity:

dv/dt = MG / (h + R)²

Using chain rule, we can say:

(dv/dh) (dh/dt) = MG / (h + R)²

(dv/dh) v = MG / (h + R)²

Separate the variables:

v dv = MG / (h + R)² dh

Integrating:

∫₀ᵛ v dv = MG ∫₀ʰ dh / (h + R)²

½ v² |₀ᵛ = -MG / (h + R) |₀ʰ

½ (v² − 0²) = -MG / (h + R) − -MG / (0 + R)

½ v² = -MG / (h + R) + MG / R

½ v² = MGh / (R(h + R))

v² = 2MGh / (R(h + R))

Given:

M = 5.98×10²⁴ kg

R = 6.37×10⁶ m

h = 1.69×10⁶ m

G = 6.67×10⁻¹¹ m³/kg/s²

Plugging in:

v² = 2 (5.98×10²⁴) (6.67×10⁻¹¹) (1.69×10⁶) / ((6.37×10⁶) (1.69×10⁶ + 6.37×10⁶))

v² = 2 (5.98) (6.67) (1.69) / ((6.37) (1.69 + 6.37)) × 10⁷

v ≈ 5120 m/s

Notice that if we had approximated g as a constant 9.8 m/s², we would have gotten an answer of:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2 (9.8 m/s²) (1.69×10⁶ m - 0 m)

v ≈ 5760 m/s

So we know that our calculated velocity of 5120 m/s is a reasonable answer.

3 0
3 years ago
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

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