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andreyandreev [35.5K]
2 years ago
10

A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above

Physics
1 answer:
zlopas [31]2 years ago
8 0

Answer:

330.5  m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

h=\frac{v^2_isin^2\alpha }{2g}

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5  m

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Answer:

Autobus porusza się ze stałą prędkością 54 km/h Oblicz w jakim czasie przebędzie drogę 450 m

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2 years ago
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

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3 years ago
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