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3 years ago

Calculate the total resistance in the circuit

Physics

2 answers:

vazorg [7]3 years ago

4 0

Answer:

The total resistance of a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drops. The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor.

SOVA2 [1]3 years ago

4 0

If you know the total current and the voltage across the whole circuit, you can find the total resistance using Ohm's Law: R = V / I. For example, a parallel circuit has a voltage of 9 volts and total current of 3 amps. The total resistance RT = 9 volts / 3 amps = 3 Ω.

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The Supreme Court of the US shall have original jurisdiction in what 3 areas?

svetoff [14.1K]

Clause 2 of Section 2 provides that the Supreme Court has original jurisdiction in cases affecting ambassadors, ministers and consuls, and also in those controversies which are subject to federal judicial power because at least one state is a party

8 0

3 years ago

Read 2 more answers

A cart with mass m vibrating at the end of a spring has an extra block added to it when its displacement is x=+A. What should th

Pavel [41]

Answer:

The block's mass should be $3m$

Explanation:

Given:

Cart with mass $m$

From the conservation of energy before mass is added,

$\frac{1}{2} mv^{2} = \frac{1}{2} kA^{2}$

Where $A =$ amplitude of spring mass system, $k =$ spring constant

$A = v\sqrt{\frac{m}{k} }$

Now new mass $M$ is added to the system,

$\frac{1}{2} (m +M ) v^{2} = \frac{1}{2} k A^{2}$

$A = v \sqrt{\frac{m +M }{k} }$

Here, given in question frequency is reduced to half so we can write,

$f' = \frac{f}{2}$

Where $f =$ frequency of system before mass is added, $f' =$ frequency of system after mass is added.

$\omega ' = \frac{\omega}{2}$

$\sqrt{\frac{k}{m +M} } = \frac{\sqrt{\frac{k}{m} } }{2}$

$\frac{k}{m +M } = \frac{k}{4m}$

$M = 3m$

Therefore, the block's mass should be $3m$

8 0

3 years ago

A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of

Mrac [35]

Answer:

If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.

Explanation:

By momentum conservation, and given the bullit and the recoil are in a straight line, the momentum analysis will be unidimentional. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so

$0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}$

now we clear $v_{r}$ and use the given data to get that

$v_{r}=-8.5\frac{m}{s}$

But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s, then if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity $\bold{v_{r}}$ too.