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slava [35]
3 years ago
11

If the temperature of a liquid increases, the density of the liquid decreases because the particles move farther apart. What are

the variables in her hypothesis?
Physics
1 answer:
Lorico [155]3 years ago
8 0

This question is incomplete because part of the content is missing; here is the complete question:

Ingrid wrote the hypothesis below.  If the temperature of a liquid increases, the density of the liquid decreases because the particles move farther apart.

What are the variables in her hypothesis?

A. The independent variable is the temperature, and the dependent variable is the density.

B. The independent variable is the density, and the dependent variable is the temperature.

C. The independent variable is the temperature, and the dependent variable is the distance between particles.

D. The independent variable is the distance between particles, and the dependent variable is the temperature.

The answer to this question is A. The independent variable is the temperature, and the dependent variable is the density.

Explanation:

To begin, the variables in an experiment are the factors being studied or analyzed, which are expressed in the hypothesis. According to this, the two variables in the experiment are the temperature and the density.

Additionally, in experiments, it is common one of the variables is independent, which is the factor manipulated by the researcher, and the other is dependent as this is expected to be changed by the first variable. If this is applied to the experiment described, the temperature is independent because this will be manipulated by Ingrid to prove density changes. Also, the density is expected to be affected by the first variable, and therefore density is the dependent variable.

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Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68
creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is   B= 4.2 *10^ {-6}T , the direction is into the page

Explanation:

From the question we are told that

        The current  is i = 12.0 A

        The radius of arc  bc is r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m

        The radius  of arc da is r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m

        The length of segment cd and ab is = l = 10cm = \frac{10}{100} = 0.10 m

The objective of the solution is to obtain the magnetic field

    Generally magnetic due to the current flowing in the arc is mathematically represented as

             B = \frac{\mu_o I}{4 \pi r}

 Here I is the current

         \mu_o is the permeability of free space with a value of 4\pi *10^{-7}T \cdot m/A

            r is the distance

Considering Arc da

         B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta

Where \theta is the angle the arc da makes with the center  from the diagram its value is  \theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad

     Now substituting values into formula for magnetic field for da

                    B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]

                           = \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]

                   B_{da}= 12.56*10^{-6} T

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

             B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta

Substituting value

          B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]

                B_{bc}= 8.37*10^{-6}T

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of  the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

     Then the net magnetic field would be

                 B = B_{da} - B{bc}

                     = 12.56*10^{-6} - 8.37*10^{-6}

                     = 4.2 *10^ {-6}T

       Since B_{da} > B_{bc} then the direction of the net charge would be into the page

 

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3 years ago
During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.
Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

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Where:

\eta_i = are the viscosities of the concrete before and after the increase

l = Length of the vessel

r_1, R_2 = Radio of the vessel before and after the increase

\Delta P= Change in the pressure

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Our values are given as:

Q_2 = 10Q_1 \rightarrow 10 times her resting rate

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\Delta P_2 = 1.5\Delta P_1 Increase of 50%

Plugging known information to get

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r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}

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r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}

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Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

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Answer:

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Explanation:

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2 years ago
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