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Orlov [11]
3 years ago
13

Pls help i’m so confused

Physics
1 answer:
Shtirlitz [24]3 years ago
3 0
Okay! Let’s list what we know and what we want to know.

Initial velocity: 0 (since the truck is starting from rest)
Final velocity: ? (what we want to know)
Acceleration: 4.07
Distance: 6.31

Since we want to find the final velocity and the information gave us the distance travelled and the acceleration of the vehicle, we’re going to want to use this equation of motion:

Final Velocity^2 = Initial Velocity^2 + 2 * Acceleration * Distance

Let’s substitute what we know and work out the answer! Since the initial velocity of the truck is zero, it cancels out, so I’m going to take it out of the equation within the next couple of steps, leaving only (2 * acceleration * distance)

Final Velocity^2 = 0^2 + 2 * 4.07 * 6.31
Final Velocity^2 = 2 * 4.07 * 6.31
Final Velocity^2 = 51.3634

This gave us final velocity squared, in order to get the final velocity of the truck, we must square root the answer. Therefore:

Final Velocity = 7.167m/s^2

:)
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In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
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Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

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Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

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The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

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Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

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  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

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