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Tatiana [17]
3 years ago
11

Eric has a mass of 80 kg. He is standing on a scale in an elevator that is accelerating downward at 1.7 m/s2. What is the approx

imate reading on the scale? Eric has a mass of 80 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate reading on the scale? 0 N 650 N 780 N 920 N
Physics
1 answer:
Yuki888 [10]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to the Force of Gravity and the Normal Force that act on the body.

The Gravity Force on Eric would be given by

F_g = mg

F_g = (80)(9.8)

F_g = 784N

To find the normal Force we must also consider downward acceleration therefore

F_N = M(-g+a)

F_N = (80)(-9.8+1.7)

F_N = -648N

Since the Normal Force is the Reaction Force the approximate reading on the scale is 650N

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76.8 kPa to mm of Hg
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Explanation:

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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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