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Irina-Kira [14]

1 year ago

10

Type the correct answer in the box.use numerals instead of words. if necessary, use / for the fraction bar. δabc is a right tria

ngle in which ∠b is a right angle, ab = 1, ac = 2, and bc = . cos c × sin a =

1 answer:

NNADVOKAT [17]1 year ago

8 0

From the diagram The value of cos C × sin A = $\frac{3}{4}$

<h3>Determine the numerical value of cos C × sin A</h3>

First step : determine the values of cos C and sin A

cos C = adjacent / hypotenuse

= a / b

= $\sqrt{3} / 2$

= √3/2

sin A = sin 60⁰

= √3/2

Therefore the numerical value of cos C * sin A = $\frac{3}{4}$

In conclusion From the diagram The value of cos C × sin A = $\frac{3}{4}$

Learn more about right angle : brainly.com/question/24323420

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An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelengt

nadya68 [22]

Answer:

I = 4.189 mA V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1] (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1] (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1 = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

6 0

3 years ago

Mercury has one of the lowest specific heats. This fact added to its liquid state at most atmospheric temperatures make it effec

UNO [17]

The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

$\Delta T=\frac{Q}{mC_s}$

where

$\Delta T$ is the increase in temperature

m is the mass of the sample

$C_s$ is its specific heat capacity

For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

$\Delta T = 4.09 K$

Therefore, by re-arranging the equation we find the mercury's specific heat:

$C_s = \frac{Q}{m\Delta T}=\frac{275}{(0.450)(4.09)}=149.4 J/(kgK)$

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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:

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

7 0

3 years ago

A jet on an aircraft carrier can be launched from rest to 40 m/s in 2 seconds. What is the acceleration of the aircraft? Show st

Alex_Xolod [135]

Answer:

$a=20\ m/s^2$

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 40 m/s

Time, t = 2 s

We need to find the acceleration of the aircraft. We know that, acceleration is equal to the rate of change of velocity. So,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{40-0}{2}\\\\=20\ m/s^2$

So, the acceleration of the aircraft is $20\ m/s^2$ .

5 0

2 years ago

A sample of gas has an initial volume of 4.5 L at a pressure of 754 mmHg . Part A If the volume of the gas is increased to 8.5 L

Firdavs [7]

Answer:

The pressure will be of 399.17 mmHg.

Explanation:

p1= 754 mmHg

V1= 4.5 L

p2= ?

V2= 8.5 L

p1*V1 = p2*V2

p2= (p1*V1)/V2

p2= 399.17 mmHg

6 0

3 years ago