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blsea [12.9K]
4 years ago
8

A bicycle traveled 150 meters west from point A to point B. Then it took the same route and came back to point A. It took a tota

l of 2 mins for the bicycle to return to point A. What is the average speed and average velocity of the bicycle?
A.
The average speed is 2.5 meters/second, and the average velocity is
2.5 meters/second east.

B.
The average speed is 0 meters/second, and the average velocity is
2.5 meters/second east.
C.
The average speed and average velocity are both 0 meters/second,
D.
The average speed is 2.5 meters/second, and the average velocity is
0 meters/second​
Physics
1 answer:
Katen [24]4 years ago
6 0
B i really
hope i helped
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An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and
Svetradugi [14.3K]

Answer:

Total contraction on the Bar  = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm  

Diameter for aluminum bar  = 40 mm

Hole diameter  = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180  × 10³ N

Calculate the total contraction on the bar = ???

The relation used in  calculating the contraction on the bar is:

\delta L = \dfrac{P *L }{A*E}

The relation used in  calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = \dfrac{P *L_1 }{A_1*E} +  \dfrac{P *L_2 }{A_2 *E}

Total contraction on the Bar  = \dfrac{180 *10^3 \N  }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]

Total contraction on the Bar  = 2.117( 0.398 + 0.182)

Total contraction on the Bar  = 2.117*(0.58)

Total contraction on the Bar  = 1.22786 mm

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<h3><em>Al(OH)3 = 78.01 g/mol</em></h3>
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</span>
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