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3 years ago

That minimum coefficient of static friction between the tyres and the road that will allow the car that will round safely.

Physics

1 answer:

liraira [26]3 years ago

3 0

Answer:

The answer is " $0.13562748$ "

Explanation:

Please find the complete question in the attached file.

Using formula:

$\bold{U_s=\frac{v^2}{rg}}$

Given:

$r=1.60 \times 10^2 \ m\\\\g=9.8\\\\v= 52.5\ \frac{km}{hr}= 14.583\ \frac{m}{s}$

put the value into the above-given formula:

$U_s= \frac{14.583^2}{160 \times 9.8}$

$= \frac{212.663889}{1568}\\\\=0.13562748$

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How will you join three resistances, each of 2 ohm so that the effective resistance is 3 ohm?

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First I will parallel two of the resistors, creating a net 1 ohm. Then I will series that with the remaining 2-ohm resistor, resulting in 3 ohms.

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3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

$d_2$ = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

$d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m$

Time

$t=\dfrac{Distance}{Speed}$

Time difference would be

$\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s$

The time difference is 0.0000109261200583 s

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s$

The ratio is

$\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583$

The ratio is 0.0109261200583

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3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago

☆Will Give Brainliest☆

Masja [62]

Answer: :)

1.) The Doppler effect is, the change in sound or light that occurs whenever there is motion between the source and its observer. The siren of the fire engine has a lower pitch as it moves away because the waves are now spread out causing a lower frequency and a lower pitch.

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3 years ago

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. I

oksano4ka [1.4K]

Answer:

The deviation in path is $4.39 \times 10^{-3}$