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3 years ago

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1 answer:

3 0

The correct answer is tissue.

Compression wrapping is used to reduce swelling and assists with sprains and wounds. If the wrap is applied wrong then this can cause impairment in circulation in the skin, it can cause an abnormal amount of swelling and irritate the skin. Therefore it seems likely that the answer would be tissue.

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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

3 years ago

2 ways in which the effects of friction can be minimised in the experiment

Vlad [161]

Answer:

Friction can be minimized by using lubricants like oil and grease and by using ball bearing between machine parts. A substance that is introduced between two surfaces in contact, to reduce friction, is called a lubricant. Fluid friction can be minimized by giving suitable shapes to the objects moving in the fluids.

Explanation:

hope it helps

7 0

3 years ago

A uniform magnetic field passes through a horizontal circular wire loop at an angle 15.1° from the normal to the plane of the lo

Zepler [3.9K]

Answer:

0.5849Weber

Explanation:

The formula for calculating the magnetic flus is expressed as:

$\phi = BAcos \theta$

Given

The magnitude of the magnetic field B = 3.35T

Area of the loop = πr² = 3.14(0.24)² = 0.180864m²

angle of the wire loop θ = 15.1°

Substitute the given values into the formula:

$\phi = 3.35(0.180864)cos15.1^0\\\phi =0.6058944cos15.1^0\\\phi =0.6058944(0.9655)\\\phi = 0.5849Wb$

Hence the magnetic flux Φ through the loop is 0.5849Weber

5 0

3 years ago

What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m

KIM [24]

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

$F=m\frac{v^2}{r}$

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

$F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N$

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

4 0

4 years ago

Read 2 more answers

The sun is to solar energy as uranium is to ______ energy

nadya68 [22]

Explanation:

The sun is to solar energy as uranium is to <u>NUCLEAR</u> energy.

4 0

3 years ago

Read 2 more answers