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Stels [109]
3 years ago
5

Si una bala de 122 g que viaja a 350 m/s golpea a un policía con chaleco antibalas el cual soporta 120 J de energía, ¿la bala at

raviesa o no el chaleco? Explicar brevemente.
Physics
1 answer:
Zanzabum3 years ago
4 0

Answer:

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

Explanation:

Un chaleco antibalas soporta el disparo de una bala disipando la energía de esta última a través de su propio material. Si sabemos que el chaleco antibalas soporta 120 joules de energía, cabe saber si la energía cinética traslacional es igual o inferior a ese límite, significando que la bala no atravesaría el chaleco.

La energía cinética traslacional de la bala (K), in joules, queda expresada con la siguiente fórmula:

K = \frac{1}{2}\cdot m\cdot v^{2} (1)

Donde:

m - Masa de la bala, en kilogramos.

v - Rapidez de la bala, en metros por segundo.

Si sabemos que m = 0.122\,kg y v = 350\,\frac{m}{s}, entonces la energía cinética traslacional de la bala es:

K = 7472.5\,J

Puesto que la energia cinética traslacional es mucho mayor que la capacidad del chaleco antibalas, la bala atravesaría el chaleco antibalas.

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The answer is 8 hours.

Explanation:

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8 0
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Natasha2012 [34]

Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

Wavelength,\lambda=571 nm=571\times 10^{-9} m

1 nm=10^{-9} m

R=75 cm=\frac{75}{100}=0.75 m

1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

6 0
3 years ago
The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it
horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

Where P = Power, I = rms current, V = rms voltage.

make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

V = 1500(4.525)

V = 6787.5 V

Hence the rms voltage = 6787.5 V

7 0
3 years ago
A student connects four AA batteries (1.5 V each) in series to light up a light bulb. The circuit has a resistance of 35 2. How
allsm [11]

Answer:

D

Explanation:

6/35=0.17

8 0
3 years ago
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