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Mandarinka [93]

3 years ago

14

A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin

ation is

1 answer:

sveticcg [70]3 years ago

7 0

Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Parallel Connection of Resistances</u>

If resistances R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

$\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}$

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

It's known the 10 parts or resistance R/10 were connected in parallel, thus the electric resistance is:

$\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}+...+\frac{1}{R/10}$

Note the sum consists of 10 equal terms. Operating on each term:

$\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}$

The sum of 10 identical fractions yields 10 times each fraction:

$\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}$

Solving for Re needs to take the reciprocal of both sides of the equation:

$R_e=R/100$

The equivalent resistance of the combination is R/100

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A(n) ________________________ is a molecule that is soluble in both polar and nonpolar solvents.

Alex

Answer: ethanol which will dissolve in both water (a polar solvent) and gasoline (a non-polar solvent)

Explanation:

3 0

3 years ago

A hockey puck sliding at 60.0m/s slows uniformly to 20.0m/s while travelling 800.0m.

julia-pushkina [17]

A) -3.75 meters/second
A=(20^2-80^2)/(2x800)
=(400-6400)/1600
=-6000/1600
=-3.75
B) 16 seconds
t=(20-80)/-3.75
=-60/-3.75
=16

3 0

3 years ago

Two blocks with rough surfaces are stacked on top of a slippery horizontal surface. The top block has a mass of m, and the botto

lapo4ka [179]

Answer:

Please refer to the attached schematic for free-body diagrams of the blocks.

The contact forces N1 = mg, N2 = 3mg.

The friction force fs = mg/2.

The tension force T = 3mg/2.

The distance y = (gt^2)/4

Explanation:

Starting with the third block which is declining a time t. Newton's Second Law implies that

$F = (3m)a\\3mg - T = 3ma\\T = 3mg - 3ma$

There are two unknowns in one equation: T and a. Therefore, we need to find more equations to solve these unknown variables.

Let's look at free-body diagram 1:

Newton's Second Law:

$F = ma\\f_s = ma$

That is another equation with one more unknown: fs.

Free-body diagram 2:

$F = (2m)a\\T - f_s = 2ma$

That is our third equation. We now have three equations with three unknowns. Combining equation 2 and 3 gives:

$T - ma = 2ma\\T = 3ma$

Plugging this equation into the first equation gives

$T = 3mg - 3ma\\3ma = 3mg - 3ma\\3mg = 6ma\\a = g/2$

Following this, we can find the other unknowns:

$T = 3ma = 3m(g/2) = \frac{3mg}{2}$

$f_s = ma = mg/2$

The contact forces are N1 and N2.

$N_1 = mg\\N_2 = 2mg + N1 = 3mg$

Finally, using the acceleration and equation of kinematics, we can find the distance the hanging weight drops in a time t.

$y - y_0 = v_0t + \frac{1}{2}at^2\\y = \frac{1}{2}(\frac{g}{2})t^2 = \frac{gt^2}{4}$

6 0

4 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

3 years ago

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

bija089 [108]

This question is incomplete, the complete question is;

Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength 1.5 m from the center of the circle is 7 mV/m.

At what rate is the magnetic field changing?

Answer:

the magnetic field changing at the rate of 9.33 m T/s

Explanation:

Given the data in the question;

Electric field E = 7 mV/m

radius r = 1.5 m

Now, from Faraday law of induction;

∫E.dl = d∅/dt

E∫dl = A( dB/dt )

E( 2πr ) = πr² ( dB/dt )

( 0.007 ) = (r/2) ( dB/dt )

( 0.007 ) = 0.75 ( dB/dt )

dB/dt = 0.007 / 0.75

dB/dt = 0.00933 T/s

dB/dt = ( 0.00933 × 1000) m T/s

dB/dt = 9.33 m T/s

Therefore, the magnetic field changing at the rate of 9.33 m T/s

5 0

3 years ago