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Zinaida [17]
3 years ago
14

18. The displacement of an object moving 330 km North for 2 hours and an additional 220

Physics
1 answer:
motikmotik3 years ago
3 0

Answer:

Explanation:

1) Displacement of the object will be gotten using simply adding the distances since they are goinf in the same direction (positive y direction)

Displacement= 330km + 220km

Displacement = 550km North

2) For a falling body, the body will possess positive acceleration due to gravity because it is falling under the influence of gravitational force. If a body is not under the influence of gravity and is thrown up, such body wont come back down. Hence the value of acceleration due to gravity of a body falling to the ground is +9.8 m/s² (note that the unit of acceleration is m/s²)

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What mass will accelerate at 3 m/s² when a net force of 150 N acts on it?
vodomira [7]

Answer:

\huge\boxed{\sf m = 50\ kg}

Explanation:

<h3><u>Given data:</u></h3>

Acceleration = a = 3 m/s²

Force = F = 150 N

<h3><u>Required:</u></h3>

Mass = m = ?

<h3><u>Formula:</u></h3>

F = ma

<h3><u>Solution:</u></h3>

Put the givens in the formula

150 = m (3)

Divide 3 to both sides

150/3 = m

50 kg = m

m = 50 kg

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By what potential difference must a proton [m_0 = 1.67E-27 kg) be accelerated to have a wavelength lambda = 4.23E-12 m? By what
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Explanation:

1. Mass of the proton, m_p=1.67\times 10^{-27}\ kg

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\lambda=\dfrac{h}{\sqrt{2m_pq_pV}}

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V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 1.67\times 10^{-27}\times (4.23\times 10^{-12})^2}

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\lambda=\dfrac{h}{\sqrt{2m_eq_eV}}

V=\dfrac{h^2}{2q_em_e\lambda^2}

V=\dfrac{(6.62\times 10^{-34})^2}{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times (4.23\times 10^{-12})^2}

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A thin-walled cylindrical steel water storage tank 30 ft in diameter and 62 ft long is oriented with its longitudinal axis verti
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