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3 years ago

Carl measures the temperature of the water at different depths of the lake to see if there is a relationship.

1 answer:

4 0

Independent variable - Water depth

The independent variable is something that is purposely changed in order to see the effects of change on the dependent variable. So since Carl wants to measure the temperature of water at different depths we will purposely change the depth of the water in order to observe how this would affect the temperature.

Dependent variable - Temperature of the water

The dependent variable is something that is dependent on the independent variable and is assumed to change as the independent variable changes. So we can determine that the temperature of the water is the independent variable as Carl expects it to change with the depth of the water. Another reason we can determine this is the dependent variable is because Carl can change the depth of the water measured but not the temperature of the water depths.

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jeyben [28]

Answer:

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3 years ago

A paper-filled capacitor is charged to a potential difference of V0=2.5 V. The dielectric constant of paper is k=3.7 . The capac

Pavlova-9 [17]

Answer:

k = 6.72

Explanation:

K of paper = 3.7

k of air = 1

Given that charge Q on the capacitor is constant because cell is disconnected from the circuit. So

V = Q / C = 2.5

Capacity becomes C / 3.7 in air .

capacity becomes C/3.7 when paper is replaced by air .

V₁ = Q / (C/3.7)

= 3.7 Q/C

3.7 x 2.5

= 9.25 V

In the second case ,

capacitance due to new unknown dielectric k

= C/3.7 x k

= kC / 3.7 ( Capacitance in air is C/3.7 )

V ( new ) = Q / ( kC/3.7 )

= 3.7 Q/kC

.55 x 2.5 = 3.7 x( 2.5 / k )

k = 3.7 / .55

= 6.72

7 0

3 years ago

The average energy per unit time per unit area that reaches the upper atmosphere of the Earth from the Sun, called the solar con

Lynna [10]

Answer: 5.34 MJ

Explanation: The window is a rectangle with area:

A = 1.1(1.5)

A = 1.65 m²

We know that only 0.9 kW/m² reaches Earth so:

P = $0.9\frac{kW}{m^{2}}(1.65)m^{2}$

P = 1.485 x 10³ W

Watts is an unit of Power and it can also be written as J/s.

An hour has 3600s or 3.6 x 10³s, so:

E = 1.485 x 10³ $\frac{J}{s}$ (3.6 x 10³s)

E = 5.346 x 10⁶ J

Mega is equal to 10⁶, then:

E = 5.35 MJ

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3 years ago

A small but measurable current of 1.2 10-10A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a

Rom4ik [11]

Q: A small but measurable current of 1.2×10⁻¹⁰A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a) Calculate the current density.

Answer:

3.82×10⁻⁵ A/m

Explanation:

Current density: This can be defined as the amount of charge passing through a conductor, per unit area, per unit time. The S.I unit of current density is A/m²

From the question above, the expression for current density is given as,

τ = I/A............... Equation 1

Where τ = current density of the copper wire, I = current flowing through the copper wire, A = cross sectional area of the copper wire

But,

A = πd²/4................. Equation 2

Substitute equation 2 into equation 1

τ = 4I/(πd²)............ Equation 3

Given: I = 1.2×10⁻¹⁰ A, d = 2 mm = 2×10⁻³ m, π = 3.14

Substitute into equation 3

τ = 4(1.2×10⁻¹⁰)/[3.14×(2×10⁻³)²]

τ = (4.8×10⁻¹⁰)/(1.256×10⁻⁵)

τ = 3.82×10⁻⁵ A/m

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3 years ago

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Three resistors, of 100, 200, and 300 ohms are connected in parallel. What is their equivalent resistance?

garri49 [273]

Answer:

Equivalent resistanfe in parallel (R):

1/R = 1/100 + 1/200 + 1/300

1/R = (6 + 3 + 2)/600

1/R = 11/600

R = 600/11 = 54.55 ohms

3 0

3 years ago

Read 2 more answers