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bija089 [108]
3 years ago
15

The current in the wires of a circuit is 180.0 milliamps. If the resistance of the circuit were doubled ( with no change in volt

age), then its new current would be
Physics
1 answer:
Troyanec [42]3 years ago
7 0

Answer:

I = 0.09[amp] or 90 [milliamps]

Explanation:

To solve this problem we must use ohm's law, which tells us that the voltage is equal to the product of the voltage by the current.

V = I*R

where:

V = voltage [V]

I = current [amp]

R = resistance [ohm]

Now, we replace the values of the first current into the equation

V = 180*10^-3 * R

V = 0.18*R (1)

Then we have that the resistance is doubled so we have this new equation:

V = I*(2R) (2)

The voltage remains constant therefore 1 and 2 are equals and we can obtain the current value.

V = V

0.18*R = I*2*R

I = 0.09[amp] or 90 [milliamps]

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un cilindro uniforme empujado por una tabla desde una posicion inicial mostrada en la figura, rueda hacia delante por el suelo,
Andreyy89

Answer:

L/2.

Explanation:

Hola.

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3 years ago
Read 2 more answers
Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

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