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3 years ago

Dependent variable is the _______ that happens because of the UI independent variable. (No answer choices provided)

1 answer:

4 0

Answer: One is called the dependent variable and the other the independent variable. The independent variable is the variable the experimenter changes or controls and is assumed to have a direct effect on the dependent variable.

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Two stars have the same radius but have very different temperatures. the red star has a surface temperature of 3,000 k and the b

dangina [55]

I would say that insofar as the two stars temperatures are presumably closely related to their luminosity, that the blue star at 156,100 k compared to 3000k for the red star then the blue star has a luminosity of 52 times that of the red star.

4 0

3 years ago

A student makes a model of the sun-Earth system by swinging a ball around her head. Using this model, the student is trying to e

DedPeter [7]

Answer:

c. gravitational attraction between the sun and earth

8 0

2 years ago

Read 2 more answers

A gas is compressed at constant temperature from a volume of 5.68 L to a volume of 2.35 L by an external pressure of 732 torr. C

Vlada [557]

Answer: The work done in J is 324

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

$W=-P\Delta V=-P(V_2-V_1)$

W = amount of work done = ?

P = pressure = 732 torr = 0.96 atm (760torr =1atm)

$V_1$ = initial volume = 5.68 L

$V_2$ = final volume = 2.35 L

Putting values in above equation, we get:

$W=-0.96atm\times (2.35-5.68)L=3.20L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

So, $3.20L.atm=3.20\times 101.3=324J$

The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

5 0

3 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

Explain why it feels more forceful when you catch a baseball with your bare hand, as opposed to catching it with a glove.

Leona [35]

Answer:

your ball exherts more pressure on your hand because your hand is smaller then the glove. The pressure gets distributed a lot more when you catch it with your glove.

Explanation:

PLEASE DO NOT ASK SUCH QUESTIONS AGAIN.

Just kidding :)

5 0

2 years ago