Can someone explain what is force constant k in Hooke’s law <br><br> F = kx

Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit such that

$P=VI=V^2/R$

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

$Current2=\frac{V}{2R}$

Power2=Voltage * Current2

$Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2$

$TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}$

6 0

2 years ago

The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

$F_1 = \frac{kq_1q_2}{r_1^2}$ --- Coulomb's law

Substitute 2 for r1

$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

$F_2 = \frac{F_1}{4}$

The new force is 1/4 of the previous force.

3 0

2 years ago

The ratio of output power to input power, in percent, is called?

Harrizon [31]

That's efficiency. There's no law that it must be stated in percent.

4 0

2 years ago

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From the 1780s to the late 1800s, people thought the amount of land and resources in the West was limited

Sonbull [250]

Answer:true

Explanation:

7 0

2 years ago

Aluminum has a density of

agasfer [191]

<h3><u>Volume is 0.1848 m³</u></h3><h3 />

Explanation:

<h2>Given:</h2>

m = 49.9 kg

ρ = 270 kg/m³

<h2>Required:</h2>

volume

<h2>Equation:</h2>

$ρ \:= \:\frac{m}{v}$

where: ρ - density

m - mass

v - volume

<h2>Solution:</h2>

Substitute the value of ρ and m

$ρ \:= \:\frac{m}{v}$

$270\: kg/m³\:= \:\frac{49.9\:kg}{v}$

$(v)\:270\: kg/m³\:= \:49.9\:kg$

$v\:= \:\frac{49.9\:kg}{270\: kg/m³}$

$v\:= \:0.1848\:m³$

<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>

3 0

2 years ago

Read 2 more answers

Can someone explain what is force constant k in Hooke’s law F = kx