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allochka39001 [22]
3 years ago
14

Un cable está tendido sobre dos postes colocados con una separación de 10 m. A la mitad del cable se cuelga un letrero que provo

ca un pandeo, por lo cual el cable desciende verticalmente una distancia de 50 cm. Si la tensión en cada segmento del cable es de 2000 N, ¿cuál es el peso del letrero?
Physics
1 answer:
lisabon 2012 [21]3 years ago
8 0

Answer:

El peso del cartel es 397,97 N

Explanation:

La tensión dada en cada segmento del cable = 2000 N

El desplazamiento vertical del cable = 50 cm = 0,5 m

La distancia entre los polos = 10 m

La posición del letrero en el cable = En el medio = 5

El ángulo de inclinación del cable a la vertical = tan⁻¹ (0.5 / 5) = 5.71 °

El peso del letrero = La suma del componente vertical de la tensión en cada lado del letrero

El peso del signo = 2000 × sin (5.71 grados) + 2000 × sin (5.71 grados) = 397.97 N

El peso del signo = 397,97 N.

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3 years ago
A transformer has a primary coil with 375 turns of wire and a secondary coil with 1,875 turns. An AC voltage source connected ac
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Answer:

The rms voltage (in V) measured across the secondary coil is 459.62 V

Explanation:

Given;

number of turns in the primary coil, Np = 375 turns

number of turns in the secondary coil, Ns = 1875 turns

peak voltage across the primary coil, Ep = 130 V

peak voltage across the secondary coil, Es = ?

\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V

The rms voltage (in V) measured across the secondary coil is calculated as;

V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V

Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V

7 0
3 years ago
Which colors of light represents the lowest visible frequency?
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Answer:

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The nuclear fuel distribution in a nuclear reactor is chosen so that when in operation the wall temperature of the reactor is a
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Explanation:

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aliya0001 [1]

Answer:

Explanation:

Intensity of light is inversely proportional to distance from source

I ∝ 1 /r²  where I is intensity and r is distance from source . If I₁ and I₂ be intensity at distance r₁ and r₂ .

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So intensity will become 16 times less bright .

"16 times " is the answer .

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