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FinnZ [79.3K]
2 years ago
5

Calculati rezistenta filamentului unui bec al lanternei de buzunar daca la tensiunea de 3,5 v el este parcurs de un curent cu in

tensitatea de 0,28 A
Physics
1 answer:
Gennadij [26K]2 years ago
3 0

Answer:

Resistencia = 12.5 Ohmios

Explanation:

Dados los siguientes parámetros;

Voltaje = 3.5 Volts

Corriente = 0.28 Amps

Para encontrar el resistencia;

La ley de Ohm establece que a temperatura constante, la corriente que fluye en un circuito eléctrico es directamente proporcional al voltaje aplicado en los dos puntos e inversamente proporcional a la resistencia en el circuito eléctrico.

Matemáticamente, La ley de Ohm viene dada por la fórmula;

Voltaje = corriente * resistencia

Resistencia = voltaje/corriente

Resistencia = 3.5/0.28

Resistencia = 12.5 Ohmios

Por tanto, la resistencia del filamento de una linterna de bolsillo es de 12,5 ohmios.

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What is the resistance (R) when voltage is 179V and current is 5 Amps?
Evgesh-ka [11]

Answer:

R = 35.8 Ω

Explanation:

Recall Ohm's Law:

V = I * R

then R = V / I

in our case:

R = 179 V / 5 A = 35.8 Ω

3 0
2 years ago
A 3.35 kg object initially moving in the positive x direction with a velocity of 4.90 m s collides with and sticks to a 1.88 kg
ahrayia [7]

Answer:

The final components of velocity of the composite object is 3.33 m/s.

Explanation:

Given;

mass of the first object, m₁ = 3.35 kg

initial velocity of the first object, u₁ = 4.90 m/s in positive x-direction

mass of the second object, m₂ = 1.88 kg

initial velocity of the second object, u₂ = 3.12 m/s in negative y-direction

initial momentum of the first object, P₁ = 3.35 x 4.9 = 16.415 kgm/s

initial momentum of the second object, P₂ = 1.88 x 3.12 = 5.8656 kgm/s

The resultant velocity of the two objects is given by;

R² = 16.415² + 5.8656²

R² = 303.858

R = √303.858

R = 17.432 kgm/s

Apply the principle of conservation of linear momentum for inelastic collision;

total initial momentum before = total final momentum after collision

P₁(x) + P₂(y) = Pf

R = Pf

R = v(m₁ + m₂)

17.432 = v(m₁ + m₂)

where;

v is the final components of velocity of the composite object

v = \frac{17.432}{m_1 + m_2} \\\\v = \frac{17.432}{3.35+1.88} \\\\v = 3.33 \ m/s

Therefore, the final components of velocity of the composite object is 3.33 m/s.

8 0
2 years ago
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed po
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Answer:

6.046N

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

F_n_e_t=F_g+F_t

F_t is the tension force.F_g=9.8N/kg is the force of gravity.

F_n_e_t=ma_c=mv^2/r\\

where r is the rope's radius from the fixed point.

From the net force equation above:

F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N

Hence the tension force is 6.046N

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