A walkman uses four standard 1.5 V batteries. How much resistance is in the circuit if it uses a current of 0.02A? *
1 answer:
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Answer:
The circuit is missing attached below is the required circuit
answer :
a) Ic = 1.944 mA
Rp = 288.66 kΩ
b) <em>The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased</em>
Explanation:
Rc = 3.6 kΩ
VBE = 0.8 v
<u>1) predict Ic and specify Rp to establish Vce at 5 V </u>
we will apply Kirchhoff's voltage law to resolve this
solution attached below
b ) The BJT is said to be in Forward reverse bias because <em>The Emitter-base Junction of the BJT is forward biased while its collector-base junction is reverse biased</em>
Answer:
is his final displacement from the point A after 60 seconds.
Explanation:
Given:
Cyclist is moving away from A.
- velocity of cyclist,
- displacement of the cyclist from point A at the time of observation,
- time after which the next observation is to be recorded,
Now as the cyclist is moving away from point A his change in displacement after the mentioned time:
<u>Now the the final displacement from point A after the mentioned time:</u>
Answer:
A) 49,050 N
B) 16 m
Explanation:
Question:
El dibujo de la pregunta se obtiene de un documento titulado "TRABAJO DIVERSO Y ENERGÍA" que se encuentra en línea y se presenta aquí.
La masa dada del vagón, m = 1,000 kg
La altura del punto en el que descansa el vagón, punto 1, h₁ = 12 m
A) El radio en el punto 2, el punto más bajo, R = 6 m
La fuerza, 'N', que la vía ejerce sobre el vagón en el punto 1 viene dada por la siguiente relación;
N = El peso del vagón + La fuerza de movimiento del vagón
∴ N = m × g + m × a
Dónde;
g = La aceleración debida a la gravedad ≈ 9,81 m / s²
a = La aceleración del vagón
Observamos que para el movimiento circular, la fuerza de movimiento del vagón, m × a = La fuerza centrípeta que actúa sobre el vagón = m × v² / R
∴ m × a = m × v² / R
Dónde;
v² = La velocidad del vagón en el punto 2 = 2 · g · h₁
Por lo tanto;
N = m × g + m × a = m × g + m × v² / R = m × g + m × 2 · g · h₁ / R
∴ N = 1000 × 9,81 + 1000 × 2 × 9,81 × 12/6 = 49,050
La fuerza que ejerce el vagón en el punto 2, N = 49,050 N
B) En el punto 3, tenemos;
N = m · g - m · a₃
La fuerza centrípeta en el punto 3, m · a₃ = m · v₃² / R₃
∴ La altura en el punto 3, h₃ = 4 m
El cuadrado de la velocidad en el punto 3, v₃² = 2 · g · (h₁ - h₃)
Para que el vagón esté seguro en el punto 3, la fuerza de la vía sobre el vagón, N = 0 para que el vagón permanezca en la vía actuando
Por lo tanto;
N = m · g - m · a₃ = 0
m · g = m · a₃ = m · v₃² / R₃ = m · (2 · g · (h₁ - h₃)) / R₃
∴ R₃ = (2 · g · (h₁ - h₃)) / g = (2 · (h₁ - h₃)) = 2 × (12 - 4) = 16
El radio de curvatura en el punto 3 para que el punto sea seguro es R₃ = 16 m.
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