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3 years ago

A machine that has no friction to overcome is called

Physics

2 answers:

Fed [463]3 years ago

8 0

Your answer to the question is d smooth machine<span />

dezoksy [38]3 years ago

4 0

The correct anwser is "D. Smooth machine"

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A simple motor converts _________________ energy into ___________________ energy.

liraira [26]

The answer is C. Electrical;mechanical

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3 years ago

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A Porsche 944 Turbo has a rated engine power of 217 hp. 30% of the power is lost in the drive train, and 70% reaches the wheels.

shutvik [7]

Answer:

a = 6.53 m/s^2

v = 11.5689 m/s

Explanation:

Given data:

engine power is 217 hp

70 % power reached to wheel

total mass ( car + driver) is 1530 kg

from the data given

2/3 rd of weight is over the wheel

w = 2/3rd mg

maximum force

$F = \mu W$

we know that F = ma

$ma = \mu (2/3 mg)$

$a_{max} = 2/3(1.00) (9.8) = 6.53 m/s^2$

the new power is $p = 70\% P_[max} = 0.7 P_{max}$

$P =f_{max} v$

$0.7P_{max} = ma_{max} v$

solving for speed v

$v =0.7 \times \frac{P_{max}}{ma_{max}}$

$v = 0.7 \frac{217 [\frac{746 w}{1 hp}]}{1500 \times 6.53}$

v = 11.5689 m/s

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4 years ago

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3 years ago

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Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$