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Ostrovityanka [42]
3 years ago
6

Hola puntos gratis !!!!!!!!!!!!!!!!!

Physics
2 answers:
mina [271]3 years ago
7 0

Answer:

gracias

Explanation:

Sloan [31]3 years ago
4 0

Hello.

Brainly points can be used for better situations, brainly points are experience that you receive by answering well.

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( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration
Talja [164]

Answer:

4.725 m/s^{2}

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then a=\frac {F}{m}  where a is acceleration, F is force and m is mass

Also making mass the subject of the formula m=\frac {F}{a}

For m1= \frac {F}{13.5} and m2=\frac {F}{3.5} hence F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}

4 0
3 years ago
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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Answer:

It depends on the size and density but No.

Explanation:

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