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Ulleksa [173]
3 years ago
7

To measure the current through and the voltage across a resistor in a circuit, you should place the ammeter in _____ with the re

sistor and the voltmeter in _____ with the resistor.
Physics
2 answers:
zysi [14]3 years ago
8 0

Answer:

  • ammeter: series
  • voltmeter: parallel

Explanation:

The current is the same through components connected in series. If you want to measure the current in a component, the ammeter must be placed in series.

__

The voltage is the same at the terminals of components connected in parallel. If you want to measure the voltage across a component, the voltmeter must be connected across the component, that is, in parallel with it.

egoroff_w [7]3 years ago
5 0

\huge \mathbb \pink{ANSWER:}

To measure the current through and the voltage across a resistor in a circuit, you should place the ammeter in <u>series</u> with the resistor and the voltmeter in <u>parallel</u> with the resistor.

#CarryOnLearning

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a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long
Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

a=\frac{Vf-Vi}{t}

-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

-9.8  =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2

To solve for "t" in this quadratic equation, we can factor it out as shown:

0= t(20-\frac{9.8}{2} t)

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s

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