How to find the magnitude of a centripetal force? (formula)

What is the mathematical relationship between voltage, resistance, and current?

Alex73 [517]

The formula for Voltage is V = IR or in words
voltage equals current times the resistance

5 0

4 years ago

What happens to light when it is shone through concave lenses? (1 point)

Nadusha1986 [10]

It would be a.,.,.,.,.

3 0

3 years ago

Read 2 more answers

Two electrons are at rest and separated by a distance of 4.32 × 10-10 m. When they are released they accelerate away from each o

sasho [114]

Answer:

Speed of electron when their separation increased by a factor of 4.10 is 9.41 x 10⁵ m/s .

Explanation:

The electric potential energy is given by the relation :

$U = \frac{kq_{1}q_{2} }{r}$

Here q₁ and q₂ are the two charge particles and r is the distance between them and k is electric constant.

In this case, there are two electrons which are separated by the distance 4.32 x 10⁻¹⁰ m.

Let e be the electron charge and r₁ be the distance between them. Then, the initial electric potential energy is :

$U_{1} = \frac{ke^{2} }{r_{1} }$

Now, the distance between the electrons increases by the factor of 4.10. Let r₂ be the new distance between them i.e. r₂ = 4.10 r₁.

Thus, the new electric potential energy is :

$U_{2} = \frac{ke^{2} }{r_{2} }=\frac{ke^{2} }{4.10r_{1} }$

Applying law of conservation of energy :

ΔU = ΔK

Here ΔU is change in electric potential energy and ΔK is change in kinetic energy.

( U₁ - U₂ ) = ( K₂ - K₁ )

Here K₂ and K₁ are initial and final kinetic energy of electron.

Since, the electron initially is at rest, so its initial kinetic energy is zero. Thus, the above equation becomes:

K₂ = U₁ - U₂

$\frac{1}{2}mv^{2}=\frac{ke^{2} }{r_{1} }- \frac{ke^{2} }{4.10r_{1} }$

Here m and v are the mass and final speed of electron respectively.

$v^{2}=\frac{2}{m} \frac{ke^{2} }{r_{1} }(1- \frac{1 }{4.10 })$

Substitute 9.1 x 10⁻³¹ kg for m, 9 x 10⁹ N m² C⁻² for k, 1.6 x 10⁻¹⁹ C for e and 4.32 x 10⁻¹⁰ m for r₁ in the above equation.

$v^{2}=\frac{2}{9.1\times10^{-31} } \frac{9\times10^{9}\times(1.6\times10^{-19})^{2} }{4.32\times10^{-10} }(1- \frac{1 }{4.10 })$

$v^{2}=8.86\times10^{11}$

v = 9.41 x 10⁵ m/s

5 0

3 years ago

So we were working on some pulley problems but this one has kinda left me scratching my head, please help! My question is for pa

USPshnik [31]

Explanation:

(c) I assume we're looking for mA.

Sum of forces on B in the -y direction:

∑F = ma

mBg − T = mBa

Sum of forces on A in the +x direction:

∑F = ma

T = mAa

Substitute:

mBg − mAa = mBa

mBg − mBa = mAa

mA = mB (g − a) / a

Plug in values:

mA = (5 kg) (10 m/s² − 0.01 (10 m/s²)) / (0.01 (10 m/s²))

mA = 495 kg

The answer key seems to have a mistake. It's possible they meant mB = 1 kg, or they changed mB to 5 kg but forgot to change the answer.

8 0

4 years ago

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.301 m to the right of Q1. Q3 is located 0.

Alexxx [7]

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

$F=\frac{k.q.Q}{r^{2}}$

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

$F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}$

$F_{12}=536.02.10^{-3}$ N

Force between 2 and 3:

$F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}$

$F_{23}=2711.63.10^{-3}$ N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

$F_{T}=2711.63.10^{-3}-536.02.10^{-3}$

$F_{T}=2175.61.10^{-3}$ N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, $F_{13}=F_{23}$ . Suppose distance from 1 to 3 is x, then from 2 to 3 is $x-0.301$

Calculating:

$\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}$

$\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}$

$\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}$

$x^{2}=0.67x^{2}-0.40x+0.061$

$0.33x^{2}+0.40x-0.061=0$

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

4 0

3 years ago

How to find the magnitude of a centripetal force?(formula)​

How to find the magnitude of a centripetal force?(formula)