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2 years ago

10

when drawing a scatter diagram, along which axis is the explanatory variable placed? along which axis is the response variable p

laced?

1 answer:

eimsori [14]2 years ago

6 0

The explanatory variable is placed on the x-axis and the response variable is placed on the y-axis.

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Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran

alina1380 [7]

Answer:

Datos:

q1 = -50 μC = $50*10^{-6}$

q2 = +30 μC = $30*10^{-6}$

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = $\sqrt\frac{k0 * q1 *q2}{F}$ = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10}$ = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em> $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}$ <em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = $\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50}$ = 0,520m

3 0

2 years ago

Which of the following is true about Viscosity of liquids:

White raven [17]

Viscosity of liquids is essentially the 'thickness' of the liquid. For instance, honey and water have different viscosities. Honey has a higher one and therefore, liquids with high viscosity do not flow as well as liquids with low viscosity (water).

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2 years ago

Read 2 more answers

What is the name of the process in which oceanic crust moves apart at mid-ocean ridges due to convention currents in the mantle?

Katena32 [7]

The answer is D seafloor spreading

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2 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

2 years ago

A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn

jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = $\sqrt{Cx^{2}+Cy^{2} }$

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

brainly.com/question/3184914

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6 0

1 year ago