If the normal force exerted on an object increases, the coefficient of sliding friction will also increase. True False

Physics

2 answers:

lbvjy [14]3 years ago

4 0

Answer:

FALSE

Explanation:

As we know that the friction force between two sliding surfaces is given by the formula

$F_f = \mu F_n$

here

$\mu$ = coefficient of friction

it depends on the the inter molecular force between the molecules of the two surfaces which slide over each other

now if we will increase the normal force then it will increase the force of friction

but if will not change the value of coefficient of friction between two surface as it is property of the surface only

professor190 [17]3 years ago

3 0

The answer is false

I hope I helped

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At resonance, a driven rlcrlc circuit has vcvc = 5.0 vv , vrvr = 8.0 vv , and vlvl = 5.0 vv . part a what is the peak voltage ac

Simora [160]

The real peak voltage is 120/0.707 = 170 V.

The peak voltage is the highest point or voltage value in any voltage waveform. A power quality issue arises when Pulse Width Modulation (PWM) devices, such as variable frequency drives, are added to a power system with a peak voltage equal to the square root of two times the RMS voltage. The peak voltage, for instance, if the RMS voltage is 85 V. The average voltage and maximum voltage of AC power coming from the wall are both about 110 V. Therefore, the real peak voltage is 120/0.707 = 170 V. The sinusoid's amplitude is divided in half by this. Peak-to-peak voltage (Vp-p), also known as the total amplitude, is 340 V, or twice the peak voltage.

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A 3.47-m rope is pulled tight with a tension of 106 N. A wave crest generated at one end of the rope takes 0.472 s to propagate

konstantin123 [22]

Answer:

The mass of the rope is 1.7 kg.

Explanation:

Given;

length of the rope, L = 3.47 m

tension on the rope, T = 106 N

period of the wave, t = 0.472 s

frequency of the is calculated as;

$f = \frac{1}{t} \\\\f = \frac{1}{0.472} \\\\f = 2.1186 \ Hz$

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where;

v is speed of the wave = fλ

λ is the wavelength

μ is mass per unit length

$f\lambda = \sqrt{\frac{T}{\mu} } \\\\f(2l) = \sqrt{\frac{T}{\mu} } \\\\2fl = \sqrt{\frac{T}{\mu} } \\\\(2fl)^2 = \frac{T}{\mu}\\\\4f^2l^2 =\frac{T}{\mu}\\\\\mu = \frac{T}{4f^2l^2}\\\\\frac{m}{l} = \frac{T}{4f^2l^2}\\\\m = \frac{T}{4f^2l}\\\\m = \frac{106}{4(2.1186)^2(3.47)}\\\\m = 1.7 \ kg$