What is the name for a wave that requires energy to produce it and a medium to travel through?

2. An auditorium has 58 seats in the first row, 62 seats in the second row, 66 seats in the third row, and so

evablogger [386]

The answer is (a) hope it helped!<3

5 0

2 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:

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

7 0

3 years ago

Calculate the average times it took the car to travel 0. 25 and 0. 50 meters. Record the averages, to two decimal places, in Tab

Illusion [34]

The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

$S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22$

For the 0.50 m, the average speed will:

$S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg} = 3.13\rm \ s$

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

Learn more about Average speed:

brainly.com/question/26386984

6 0

2 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

3 0

2 years ago

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

Leona [35]

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

$wsin\theta-\mu F_n=ma$ That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

6 0

3 years ago