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3 years ago

Which of these experiments would make use of qualitative data?

Physics

2 answers:

Umnica [9.8K]3 years ago

6 0

A. A study of different surfaces to compare luster or shine.

xxTIMURxx [149]3 years ago

6 0

Answer: A. a study of different surfaces to compare luster or shine

Explanation:

A qualitative data can be define as the data that can only be observed and recorded but it cannot be counted in numbers. This type of data cannot be measured in units. For example, color.

A is the correct option. This is because of the fact that luster and shine can only be observed it cannot be measured in units and counted in numbers. It just represents the physical observable property of a material.

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Consider a portion of a cell membrane that has a thickness of 7.50 nm and 1.4 µm ✕ 1.4 µm in area. A measurement of the potentia

Alla [95]

Answer:

Part a: The current is 1.49x10⁻¹² A

Part b: The current is decreased by a factor of 4.

Explanation:

Part a

The area is given as

A =(1.4*10⁻⁶)² =1.96*10⁻¹² m²

The resistance is given as where resistivity of the membrane material is 1.30 x 10⁷ ohms*m

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(1.96*10⁻¹²)

R=4.97x10¹⁰ ohms

So the resistance is 4.97x10¹⁰ ohms.

I=V/R =86.0x10⁻³/5.77x10¹⁰

I=1.49x10^⁻¹² A

So the current is 1.49x10⁻¹² A

S=So/2=1.4/2 =0.7μm

A=(0.7*10^-6)^2=4.9*10⁻¹³ m²

R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(4.9*10⁻¹³)

R=1.98x10¹¹ ohms

So the resistance is 1.98x10¹¹ ohms.

I=V/R =86.0x10⁻³/1.98x10¹¹

I2=4.52x10^⁻¹³ A

So the ratio of the new current vs the old current is as

I2/I=4.52x10^⁻¹³/1.49x10⁻¹²=0.25

So the current is decreased by a factor of 4.

7 0

4 years ago

Please I need help its timed-

leva [86]

Answer:

Explanation:

6 0

3 years ago

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?

Xelga [282]

<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

30 cm = 0.3 m (distance)

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

$\small \boxed{work \: done \: by \: Ryan \: = 3 \: Joules}$

4 0

3 years ago

Now open the simulation. Activate ""grid"" and ""show numbers"" to read values. Place a 1 nC positive (red color) charge on the

zmey [24]

Answer:

The divergence on the sensor shows the magnitude of the charges

Explanation:

This will increase as length increases since it is said to be proportional to the length. note that test charge is always positive and charge on the grid is positive as indicated (1 nC)

8 0

3 years ago

Water at 20 C flows through a 5-cm-diameter pipe that has a 180 vertical bend, as in Fig. P3.43. The total length of pipe betwee

Nataliya [291]

Answer:

F = 749 [N]

Explanation:

We must give full information on this problem, as well as the question that needs to be resolved.

Water at 20°C flows through a 5-cm-diameter pipe that has a 180° vertical bend. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s,

P1=165kPa

and

P2=134kPa

Neglecting pipe weight, determine the total force that the flanges must withstand for this flow.

We must make a U-shaped body diagram of the pipe in order to visualize the forces acting according to the pressures and the area of the pipe.

Then by means of the second law of motion of Newton, which says that the sum of the forces must be equal to the product of mass times acceleration, we can find the force Fp

Let us remember that pressure is defined as the divided force over the area, therefore:

F = P *A

where:

P = pressure

A = area

The product of the mass by acceleration, is equal to the product of the speed of the fluid by the mass flow, since we know the weight of the fluid we can find its mass flow.

$W_{flow}=230[N/s]\\W_{flow} =g*m_{flow}\\m_{flow} = W_{flow} / g\\m_{flow} = 230/9.81\\m_{flow}= 23.45[kg/s]$

In the function of the mass flow, we can find the velocity of the fluid, as we also know the diameter of the pipe

$m_{flow} = density*v*A\\where\\density = 1000[kg/m^{3}]\\ v= velocity[m/s]\\A = area [m^{2}]\\v=\frac{m_{flow}}{density*A} \\v=\frac{23.45}{1000*\frac{\pi}{4}*(5*10^{-2})^{2} } \\v= 11.94 [m/s]$

We know that the atmospheric pressure is equal to:

$P_{atm} = 101.325[kpa]$

The value of the pipe area is calculated for a circular section

$A = \frac{\pi}{4} * (0.05)^{2}\\ A = 0.00196[m^{2} ]$

The resultant force is 749 [N]

The solution of the equations and the free body diagram can be seen in the attached picture.

3 0

3 years ago