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Paraphin [41]
2 years ago
6

You and your lab partner are asked to determine the density of an aluminum bar. the mass is known accurately (to four significan

t figures). you use a simple metric ruler to determine its size and calculate the results in
a. your partner uses a precision micrometer and obtains the results in
b. method a (g/cm3 ) method b (g/cm3 ) 2.2 2.703 2.3 2.701 2.7 2.705 2.4 5.811 the accepted density of aluminum is 2.702g/cm3 .
a. calculate the average density for each method. should all the experimental results be included in your calculations? if not, justify and omissions.
b. calculate the percent error for each methodâs average value.
c. which methodâs average value is more precise? which method is more accurate?

Physics
1 answer:
larisa [96]2 years ago
3 0
The given data is tabulated below

Measurement Method A  Method B
--------------------  --------------  --------------
           1                  2.2           2.703
           2                 2.3            2.701
           3                 2.7             2.705
           4                 2.4             5.811

A graph of the data reveals that 5.811 from measurement B is an outlier, and should be rejected.

Part a: Calculate average density.
Method A:
Avg. density = (2.2+2.3+2.7+2.4)/4 = 2.4  g/cm³
Method B:
Avg. density = (2.703+2.701+2.705)/3 = 2.703 g/cm³

Answer:
Average densities are
2.4 g/cm³ by method A
2.703 g/cm³ by method B, with the outlier removed.

Part b: Calculate the percent error.
The true value is 2.702 g/cm³
Method A:
%error = 100*(|2.4 - 2.702|/2.702) = 11.2%
Method B:
%error = 100*(|2.703 - 2.702|/2.702) = 0.04%

Answer:
The percent error is
11.2 by method A,
0.04% by method B.

Part c: 
Method B is more accurate and more precise because its average value is closest to the true value and its %error is extremely small.


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10 cm.

Explanation:

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2 years ago
A charge of 4.9 x 10-11 C is to be stored on each plate of a parallel-plate capacitor having an area of 150 mm2 and a plate sepa
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Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ is the permitivity of a constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

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7 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

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      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

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Explanation:


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Projectile motion is an idealization of the motion under the action of gravity neglecting the influence of the air (no drag force nor friction).


This  kind of motion is the result of two independent motions: vertical motion and horizontal motion.


The observed net velocity is the vectorial sum of the vertical and horizontal velocities.


The horizontal velocity is constant, since there is not any force acting in the horizontal axis. Thi is, the object, following the first Law of Newton (inertia law) tends to continue in uniform rectilinear movement (with zero acceleration).


The vertical velocity, this is the velocity at which the bullet falls toward the Earth, is influenced (accelerated) by the action of the gravity of the Earth. So, the vertical velocity is accelerated by the pull of the Earth.


Vertical and horizontal velocities are independent of each other, which means that the speed or the magnitude of the horizontal velocity does not affect the speed at which an object (the bullet) falls toward the Earth.

6 0
2 years ago
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natta225 [31]
First, we have to calculate the normal forces on different surfaces.The normal force on the 4.00 kg, N1 = (4)(9.8) = 39.2 N. The normal force on the 10.0 kg, N2 = (14)(9.8) = 137.2 N. Looking at the 10.0 kg block, the static forces that counteract the pulling force equals the sum of the friction from the two surfaces. Fc = N1 * 0.80 + N2 * 0.80 = 141.12 N. Since the counter force is less than the pulling force, the blocks start to move and hence, kinetic frictions are considered.


Therefore, f1 = uk * N1 = (0.60)(39.2) = 23.52 N.
4 0
2 years ago
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