Orange juice with pulp floating around is an example of what type of mixture?

Physics

2 answers:

kolbaska11 [484]4 years ago

8 0

Answer:

I hope this helps

Explanation:

homogeneous mixture

Sonbull [250]4 years ago

4 0

The answer would be heterogeneous mixture

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The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de

spin [16.1K]

Answer:

the spring constant k = $5.409*10^4 \ N/m$

the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

Explanation:

From Hooke's Law

$F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m$

Thus; the spring constant k = $5.409*10^4 \ N/m$

The amplitude is decreasing 37% during one period of the motion

$e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}$

$b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) } }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s$

Therefore; the value for the damping constant $\\ \\b = 1.518 *10^3 \ kg/s$

5 0

4 years ago

Holly puts a box into the trunk of her car. Later, she drives around an unbanked curve that has a radius of 48 m. The speed of t

Troyanec [42]

Answer:

The minimum coefficient of static friction for the box on the floor of the trunk is 0.54.

Explanation:

Given that,

Radius of the unbanked curve, r = 48 m

The speed of the car on the curve, v = 16 m/s

We need to find the minimum coefficient of static friction for the box on the floor of the trunk. It can be calculated by balancing the centripetal force and the force of gravity as :

$\mu mg=\dfrac{mv^2}{r}$