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А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?

Answers

Answer 1

Answer:

Answer:

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Explanation:

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Answer 2

Answer:

Final answer:

The maximum expected measurement error for a pressure gauge measuring 0-10 bar with an inaccuracy of 1% of full-scale reading is 0.1 bar. When the gauge measures 1 bar, the expected inaccuracy is 10%.

Explanation:

The inaccuracy mentioned here is related to the full-scale reading which means the error is calculated based on the top measurement value. The pressure gauge range is 0-10 bar, so the inaccuracy is one percent of this. (a) Thus, the maximum measurement error expected for this instrument is 1.0% of 10 bar i.e., 0.1 bar. (b) If the gauge is measuring a pressure of 1 bar, then the relative error expressed as a percentage would be the absolute error (0.1 bar) divided by the observed reading (1 bar) i.e., 10%. It means, when measuring 1 bar pressure, the expected measurement error is 10%. This is an example of how instrument inaccuracy is properly interpreted and employed when working with various measurements.

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In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)

Answers

Answer:

v(t) = 21.3t

v(t) = 5.3t

$v(t) = 48 -48 e ^{ (t)/(9)}$

Explanation:

When no sliding friction and no air resistance occurs:

$m(dv)/(dt) = mgsin \theta$

where;

$(dv)/(dt) = gsin \theta , 0 < \theta < ( \pi)/(2)$

Taking m = 3 ; the differential equation is:

$3 (dv)/(dt)= 128*(1)/(2)$

$3 (dv)/(dt)= 64$

$(dv)/(dt)= 21.3$

By Integration;

$v(t) = 21.3 t + C$

since v(0) = 0 ; Then C = 0

v(t) = 21.3t

ii)

When there is sliding friction but no air resistance ;

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta$

Taking m =3 ; the differential equation is;

$3 (dv)/(dt)=128*(1)/(2) -(√(3) )/(4)*128*(√(3) )/(4)$

$(dv)/(dt)= 5.3$

By integration; we have ;

v(t) = 5.3t

iii)

To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :

$m (dv)/(dt)= mg sin \theta - \mu mg cos \theta - kv$

The differential equation is :

= $3 (dv)/(dt)=128*(1)/(2) - ( √( 3))/(4)*128 *( √( 3))/(2)-(1)/(3)v$

= $3 (dv)/(dt)=16 -(1)/(3)v$

By integration

$v(t) = 48 + Ce ^{(t)/(9)$

Since; V(0) = 0 ; Then C = -48

$v(t) = 48 -48 e ^{ (t)/(9)}$

A force of 200N acts on a body that moves along a horizontal plane in the same direction of movement. The body moves 30m. What is the work done by that force?

Answers

Work = Force times Distance

Work = 200 x 30

Work = 6000

The work done by a force of 200N on a body that moved 30m is 6000J or 6000 Joules.

Tripling the displacement from equilibrium of an object in simple harmonic motion will bring about a change in the magnitude of the object's acceleration by what factor?

Answers

Answer:

acceleration will be tripled.

Explanation:

We know, when an object is performing Simple harmonic motion, the force

experience by it is directly proportional to its displacement from its mean position.

Also, F = ma , therefore, acceleration is also proportional to its displacement .

Now, F = kx

Therefore, $a=(k\ x)/(m)$

If we triple the displacement i.e, 3x.

Acceleration $a'=(k(3x))/(m)=3a.$

Therefore, acceleration is also tripled.

Hence, this is the required solution.

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Answers

Answer: wavelength will reduce

Explanation: The region of low pressure is know as the rarefraction region while the region of high pressure is the compression region.

The distance between 2 successive rarefraction or compression region is known as the wavelength.

Now the question is concerned about what an increase in frequency will cause to wavelength.

The speed of sound in air is a constant and it is approximately 343 m/s.

But recall that v = fλ

By assuming a fixed value for speed (v), we have that

k = fλ

Hence, f = k/ λ

This implies that at a fixed wave speed, the wavelength and frequency have an inverse relationship.

An increase in frequency will bring about a reduction in wavelength.

You are exiting a highway and need to slow down on the off-ramp in order to make the curve. It is rainy and the coefficient of static friction between your tires and the road is only 0.4. If the radius of the off-ramp curve is 36 m, then to what speed do you need to slow down the car in order to make the curve without sliding?