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3 years ago

5

Physics

1 answer:

Ronch [10]3 years ago

4 0

Answer:

Current in a parallel circuit = 0.61 amps (Approx)

Explanation:

Given:

Voltage V = 6 volt

Two resistors = 17.2 , 22.4 in parallel circuit

Find:

Current in a parallel circuit

Computation:

1/R = 1/r1 + 1 / r2

1/R = 1/17.2 + 1 / 22.4

R = 9.73 ohms (Approx)

Current in a parallel circuit = V / R

Current in a parallel circuit = 6 / 9.73

Current in a parallel circuit = 0.61 amps (Approx)

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In each cycle of its operation or refrigerator removes 17 J of heat from the inside of the refrigerator and releases 40 J of hea

Lubov Fominskaja [6]

Answer:

23 J

Explanation:

According to the law of conservation of energy, the total heat given off to the surroundings by a refrigerator is equal to the work done by the refrigerator and the heat removed from the inside of the refrigerator.

So we can write:

$Q_{h} = W+Q_{c}$

where

$Q_{c}$ is the heat removed from the cold reservoir (the inside of the refrigerator)

W is the work done

$Q_{h}$ is the work released into the hot reservoir (the surroundings)

Here we have:

$Q_c=+17 J$ (heat removed from the inside of the refrigerator)

$Q_h=+40 J$ (heat released into the surroundings)

Solving the equation for W, we find the work done:

$W=Q_h-Q_c=40-17=23 J$

7 0

3 years ago

a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially

CaHeK987 [17]

Answer:

Velocity of the helium nuleus = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

where $P_{1i}$ : is the proton linear momentum initial, $P_{2i}$ : is the helium nucleus linear momentum initial, $P_{1f}$ : is the proton linear momentum final, $P_{2f}$ : is the helium nucleus linear momentum final

From (1):

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$ : are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$ : are the proton and helium nucleus velocities, respectively, after the collision

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

where $K_{1i}$ and $K_{2i}$ : are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and $K_{2f}$ : are the kinetic energy for the proton and helium, respectively, after the colission