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1. In a(n) ______________________ circuit, all parts are connected one after another along one path.

poizon [28]

Answer:

1. A series circuit is a closed circuit which has all loads connected in a row and there is only one path for the current to flow.

2. The Ohm's Law state that a current flow through a resistor is directly proportional to the voltage across it $R = \frac{V}{I}$

3. A parallel circuit is a closed circuit divided into branches that it has two o more paths for the current to flow and the loads are parallel to each other which mean the voltage across them is the same for all loads.

3 0

3 years ago

Read 2 more answers

Many web sites describe how to add wires to your clothing to keep you warm while riding your motorcycle. The wires are added to

Tomtit [17]

Answer:

$P=42.075W$

Explanation:

The power provided by a resistor (wire in this case) is given by:

$P=\frac{V^2}{R}$ .

The resistance of a wire is given by:

$R=\frac{\rho L}{A}$

Where for the resistivity the one of the copper should be used: $\rho=1.68\times10^{-8}\Omega m$ .

The area A is that of a circle, which written in terms of its diameter is:

$A=\pi r^2=\pi (d/2)^2=\frac{\pi d^2}{4}$

Putting all together:

$P=\frac{AV^2}{\rho L}=\frac{\pi d^2V^2}{4\rho L}$

Which for our values is:

$P=\frac{\pi (0.00025m)^2(12V)^2}{4(1.68\times10^{-8}\Omega m)(10m)}=42.075W$

7 0

3 years ago

Students run an experiment to determine the rotational inertia of a large spherically shaped object around its center. Through e

Alex73 [517]

Answer: I = 3.6 m3

(C)

Explanation:

moment of inertia for spherically shaped object around it's center is given as

I = (2/5) mr²

substituting the r = 3m²

I = (2/5)*(9) m3

I = 3.6 m3

5 0

4 years ago

A thermodynamic system undergoes a process in which its internal energy decreases by 1,477 J. If at the same time 678 J of therm

Andrews [41]

Answer:2155 J

Explanation:

Given

Change in Internal energy $\Delta U=-1477 J$ i.e. decrease in Internal Energy

Heat added to system $Q=678 J$

According First law for a system

$dQ=dU+dW$

$678=-1477+dW$

$dW=2155 J$

Thus 2155 J of work is done by system

5 0

3 years ago

Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude

DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, $v_p_i$ = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

$W =K.E_f - K.E_i$

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d = (EQ) x d = EQd

$K.E_f =EQd + \frac{1}{2}m_pv_p_i^2$

$m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C$

$K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\$

$K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J$

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0

3 years ago