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Darya [45]
3 years ago
15

The initial speed of a body is 4.24 m/s. What is its speed after 2.91 s if it accelerates uniformly at -1.76 m/s2

Physics
1 answer:
shutvik [7]3 years ago
6 0
  • Initial velocity=u=4.24m/s
  • Time=t=2.91s
  • Acceleration=a=-1.76m/s^2
  • Final velocity=v

Using 1st equation of kinematics

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=4.24+2.91(-1.76)

\\ \sf\longmapsto v=|4.24-5.12|

\\ \sf\longmapsto v=0.8m/s

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\large \boxed{42\, \mu \text{C}}$

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The charges are identical, so we can write the formula as

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\begin{array}{rcl}\text{4.0 N}& = & 8.988 \times 10^{9}\text{ N$\cdot$m$^{2}$C$^{-2}$} \times \dfrac{q^{2}}{\text{(2.0 m)}^{2}}\\\\4.0 & = & 2.25 \times 10^{9}\text{ C$^{-2}$} \times q^{2}\\\\q^{2} & = & \dfrac{4.0}{2.25 \times 10^{9}\text{ C$^{-2}$}}\\\\& = & 1.78 \times 10^{-9} \text{ C}^{2}\\q & = & 4.2 \times 10^{-5} \text{ C}\\& = & 42\, \mu \text{C}\\\end{array}\\\text{Each charge has a value of $\large \boxed{\mathbf{42\, \mu }\textbf{C}}$}

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3 years ago
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we know that

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distance = 0.624c × 1.59 × 10⁻⁷ s

distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s

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3 years ago
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Neporo4naja [7]

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