If the force acting on an object is multiplied by ____

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago

Determine weather or not the equation below is balanced.

Liula [17]

Answer:

Yep.. It's balanced and its a combination reaction

Explanation:

Reactants : S₈ + 24F₂

Product 8SF₆

8 0

3 years ago

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of

HACTEHA [7]

Answer:

$7.1$ ms⁻¹

Explanation:

$d$ = diameter of merry-go-round = 4 m

$r$ = radius of merry-go-round = $\frac{d}{2}$ = $\frac{4}{2}$ = 2 m

$I$ = moment of inertia = 500 kgm²

$w_{i}$ = angular velocity of merry-go-round before ryan jumps = 2.0 rad/s

$w_{f}$ = angular velocity of merry-go-round after ryan jumps = 0 rad/s

$v$ = velocity of ryan before jumping onto the merry-go-round

$m$ = mass of ryan = 70 kg

Using conservation of angular momentum

$Iw_{i} - m v r = (I + mr^{2})w_{f}$

$(500)(2.0) - (70) v (2) = (I + mr^{2})(0)$

$1000 = 140 v$

$v = 7.1$ ms⁻¹

5 0

3 years ago

How fast did Newton decide a cannon ball shot out in order to escape Earth’s gravity?

arsen [322]

Plugging in for the Earth's mass and for G, we have 11.2 km/s for the escape velocity for an object launched from the Earth's surface. This is about 25,000 miles per hour

7 0

3 years ago

Which statement is true for a car that first goes around a curve of radius r at a constant speed v, and then goes around the sam

sasho [114]

Centripetal force is equal to (mv^2)/r
The way I use to answer these question is to set every variable to 1
m=1
v=1
r=1
so centripetal force =1
then change the variable we're looking at
and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
((1)×(2)^2)/1=4
So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around

3 0

4 years ago

If the force acting on an object is multiplied by _____, the product is called work