A circuit with a resistance of 25 ohms is carrying a current of 2

1 answer:

7 0

The heat developed can be calculated using the power rules as follows:

P = I x V = I^2 x R

First solution is to use the given values of I and R as follows:
P = I^2 x R = (2)^2 x 25 = 4 x 25 = 100 watt

Second solution is to use ohm's law to get V, then substitute to get the power as follows:
V = I x R = 2 x 25 = 50 volts
Power = V x I = 50 x 2 = 100 watt

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A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

Ugo [173]

Answer:

$f_{e}$ = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

M = M₀ $m_{e}$

Where M₀ is the magnification of the objective and $m_{e}$ is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

$m_{e}$ = 25 / $f_{e}$

The objective is focused on the distances of the tube (L)

M₀ = -L / f₀

Substituting

M = - L/f₀ 25/ $f_{e}$

1) Let's look for the focal length of the eyepiece (faith)

$f_{e}$ = - L 25 / f₀ M

M = 400X = -400

$f_{e}$ = - 12 25 /0.40 (-400)

$f_{e}$ = 1.875 cm

Let's approximate two significant figures

$f_{e}$ = 1.9 cm

8 0

3 years ago

In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.

alisha [4.7K]

Answer:

$\Delta Q=-230kJ$

Explanation:

Using the first law of thermodynamics:

$\Delta U=\Delta Q-W$

Where $\Delta U$ is the change in the internal energy of the system, in this case $\Delta U=250kJ$ , $\Delta Q$ is the heat tranferred, and $W$ is the work, $W=-480kJ$ with a negative sign since the work is done by the system.