Will give you brainliest pls help

Physics

1 answer:

ExtremeBDS [4]3 years ago

4 0

1. All the relevant resistors are in series, so the total (or equivalent) resistance is the sum of the resistances of the resistors: 20 Ω + 80 Ω + 50 Ω = 150 Ω [choice A].

2. The ammeter will read the current flowing through this circuit. We can find the ammeter reading using Ohm's law in terms of the electromotive force provided by the battery: I = ℰ/R = (30 V)(150 Ω) = 0.20 A [choice C].

3. The voltmeter will measure the potential drop across the 50 Ω resistor, i.e., the voltage at that resistor. We know from question 2 that the current flowing through the resistor is 0.20 A. So, from Ohm's law, V = IR = (0.20 A)(50 Ω) = 10. V, which will be the voltmeter reading [choice F].

4. Trick question? If the circuit becomes open, then no current will flow. Moreover, even if the voltmeter were kept as element of the circuit, voltmeters generally have a very high resistance (an ideal voltmeter has infinite resistance), so the current moving through the circuit will be negligible if not nil. In any case, the ammeter reading would be 0 A [choice B].

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12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor = how

UkoKoshka [18]

Answer:

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

Explanation:

In the United States, a standard "drink" of beer has 12 ounces, a standard "drink" of wine has 5 ounces and standard drink of liquor has 1.5 ounces. Then, we obtain the quantity of drinks by dividing the total volume of each drink by its respective unit volume and summing each term. That is:

$N = \frac{12\,oz}{12\,\frac{oz}{dr} }+\frac{12\,oz}{5\,\frac{oz}{dr} }+\frac{3\,oz}{1.5\,\frac{oz}{dr} }$

$N = 1\,dr+2.4\,dr+2\,dr$

$N = 5.4\,dr$

$N = 6\,dr$

12 ounces of beer plus 12 ounces of wine plus 3 ounces of liquor are equivalent to 6 drinks.

8 0

3 years ago

S To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volum

erastovalidia [21]

To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be $\frac{4.83598}{c^{\frac{1}{3} } }$ .