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Ulleksa [173]
3 years ago
14

a student drew the following model: volcano cooling crust motion plates tension which landform should the student put next in th

e sequence? a. plain b. cinder cone c. rift valley d. plateau
Physics
2 answers:
Sonbull [250]3 years ago
4 0

Answer:C.rift valley (apex)

Explanation:

Andrei [34K]3 years ago
3 0

Answer:

C) rift valley

Explanation:

A rift valley is a lowland region formed by the interaction of Earth's tectonic plates. This small rift valley has a typical formation—long, narrow, and deep. It was formed by the Thingvellir rift, where the North American and Eurasian tectonic plates are tearing, or rifting, apart over a hotspot on the island of Iceland.

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A positive point charge (q = +5.45 x 10-8 C) is surrounded by an equipotential surface A, which has a radius of rA = 1.49 m. A p
ruslelena [56]

Answer:

r_{B} = 3.34\ m

Solution:

As per the question:

Point charge, q = 5.45\times 10^{- 8}\ C

Test charge, q_{o} = 4.96\times 10^{- 11}\ C

Work done by the electric force, W_{AB} = - 9.05\times 10^{- 9}\ J

Now,

We know that the electric potential at a point is given by:

V = \frac{kqq'}{r}

where

r = separation distance between the charges.

Also,

The work done by the electric force i moving a test charge from point A to B in an electric field:

W_{AB} = kqq_{o}(\frac{1}{r_{B} - \frac{1}{r_{A}}}

- 9.05\times 10^{- 9} = 9\times 10^{9}\times 5.45\times 10^{- 8}\times 4.96\times 10^{- 11}(\frac{1}{r_{B}} - \frac{1}{1.49}}

- 0.3719 = (\frac{1}{r_{B}} - 0.67}

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5 0
3 years ago
A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference
Andre45 [30]

a) 400 \Omega

b) 0.43 V

c) 0.44 %

Explanation:

a)

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

V=E-Ir (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when R_1=550 \Omega, V_1=0.25 V

Using Ohm's Law, the current is:

I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A

2) when R_2=1000 \Omega, V_2=0.31 V

Using Ohm's Law, the current is:

I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A

Now we can rewrite eq.(1) in two forms:

V_1 = E-I_1 r

V_2=E-I_2 r

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega

b)

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

V=E-Ir

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

V=0.25 V is the voltage

I=4.5\cdot 10^{-4}A is the current

r=400\Omega is the internal resistance

Solving for E,

E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V

c)

In this part, we are told that the area of the cell is

A=4.0 cm^2

While the intensity of incoming radiation (the energy received per unit area) is

Int.=5.5 mW/cm^2

This means that the power of the incoming radiation is:

P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W

This is the power in input to the resistor.

The power in output to the resistor can be found by using

P'=I^2R

where:

R=1000 \Omega is the resistance of the resistor

I=3.1\cdot 10^{-4} A is the current on the resistor (found in part A)

Susbtituting,

P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%

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