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attashe74 [19]
2 years ago
8

You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that

slopes upward at a 12 angle with the horizontal. As you face the window (0.90 m high, 2.0 m wide) in your compartment, the train is moving to the left, as the drawing indicates. The top edge of the wall first appears at window corner A and eventually disappears at window corner B. How much time passes between appearance and disappearance of the upper edge of the wall
Physics
1 answer:
loris [4]2 years ago
5 0

Questions Diagram is attached below

Answer:

T=2.08s

Explanation:

From the question we are told that:

Speed of Train V=3.0m.s

Angle \theta=12\textdegree

Height of window h_w=0.90m

Width of window w_w=2.0m

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

 b=\frac{0.9}{tan12}

 b=4.23

Therefore

Distance from A-A

 d_a=2.0+4.23

 d_a=6.23

Therefore

Time Required to travel trough d is mathematically given as

 T=\frac{d_a}{v}

 T=\frac{6.23}{3}

 T=2.08s

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Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length
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Answer with Explanation:

We are given that

Length of wire 1=L_1

Length of wire 2=L_2

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Resistivity of aluminum wire=\rho_2=2.82\times 10^{-5}\Omega-m

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Wire 2=Aluminum wire

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When diameter of wires are same then their cross section area are also same .

l=\frac{RA}{\rho}

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity  is small then the length of wire will be large.

\rho_1

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Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66

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3 years ago
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Answer:

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