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m_a_m_a [10]
3 years ago
7

. Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north.

How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position?
Physics
1 answer:
KiRa [710]3 years ago
7 0

Answer:

H = 30.8 m

\theta =54.25^0

Explanation:

given,

first walks straight west = 18 m

then straight in north = 25 m

the distance from the start point

H² = B² + P²

H² = 18² + 25²

H = \sqrt{18^2+25^2}

H = \sqrt{324+625}

H = 30.8 m

the angle between

tan \theta = \dfrac{P}{B}

tan \theta = \dfrac{25}{18}

\theta = tan^{-1}(\dfrac{25}{18})

\theta =54.25^0

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Determine the energy required to accelerate an electron from (a) 0.500c to 0.900c and (b) 0.900c to 0.990c
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Answer:

0.582 MeV

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Explanation:

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W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.9^2}}-\frac{1}{\sqrt{1-0.5^2}}\right)0.511\\\Rightarrow W=0.582\ MeV

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W=\left(\frac{1}{\sqrt{1-\frac{v_f^2}{c^2}}}-\frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}\right)E_r\\\Rightarrow W=\left(\frac{1}{\sqrt{1-0.99^2}}-\frac{1}{\sqrt{1-0.9^2}}\right)0.511\\\Rightarrow W=2.45\ MeV

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6 0
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Vector A→, having magnitude 2.5m, pointing 37∘ south of east and vector B→ having magnitude 3.5m, pointing 20∘ north of east are
ahrayia [7]

Answer:

Magnitude of the resultant vector is R = 6.81 m

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Vector A having magnitude of 2.5 m

Vector A having direction 37 degree south of east.

Vector B having magnitude of 3.5 m

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Therefore, the angle between the two vectors is, θ = 37+20 = 57 degree

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$ R = \sqrt{A^2+B^2+2AB \cos \theta}$

$ R = \sqrt{2.5^2+3.5^2+2(2.5)(3.5) \cos 57}$

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