All categories

3 years ago

A charge Q is uniformly distributed along the x axis from x = a to x = b. If Q = 45

Physics

1 answer:

luda_lava [24]3 years ago

3 0

Answer:

The potential at the point 8 meters is approximately 48.98 Joules/C

Explanation:

Notice that we need to find the potential at a point aligned to the rod's axis (see attached figure), that is located 6 meters from one end of the rod. Notice as well, that the length (L) of the rod is 5 meters.

Since we have a uniformly distributed charge, the charge density per unit of length is defined as:

$\lambda=\frac{Q}{L} \\\lambda=\frac{45}{5}\,\frac{10^{-9}\,C}{m} \\\lambda=9\,\frac{10^{-9}\,C}{m}$

In order to find the contribution of each little segment dx of charge

$dq=\lambda\,dx$

to the potential at the requested point, we need to perform an integral:

$V=\int\limits^{11}_{6} {} \, dV \\V=\int\limits^{11}_{6} {} \, k\,\frac{dq}{x} \\V=\lambda \,*k\,\int\limits^{11}_{6} {} \, \frac{dx}{x} \\V=8.98*9 \, \,ln(\frac{11}{6}) \,\frac{J}{C} \\V=80.82\,* \,0.606 \,\frac{J}{C}\\V=48.98 \frac{J}{C}$

You might be interested in

A solar panel is used to collect energy from the sun and change it into other forms of energy. The picture below shows some sola

irakobra [83]

C I’m pretty sure!!!!!

4 0

3 years ago

A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s.

Katarina [22]

Option B would be right one
according to momentum conservation
6600*2 = 13200kgm/s
5400*3 = 16200kgm/s
16200-13200 = 3000
now 6600-5400 = 1200 kg
thus 3000/1200 = 2.5 v

5 0

3 years ago

These are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the oz

yarga [219]

Hello! The nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the ozone layer are the Chlorofluorocarbons (CFCs)

These are compounds developed and improved by Thomas Midgley in the late 1920s. They were used as refrigerants and aerosol propellants.

These compounds created a hole in the ozone layer by the following reactions:

CCl₃F → CCl₂F· + Cl· (In the presence of light. Radical Reaction)
Cl· + O₃ → ClO + O₂
ClO + O₃ → Cl· + O₂

The last 2 reactions can repeat in a radical mechanism and explain why these compounds are so harmful to the ozone layer.

4 0

3 years ago

The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee

aniked [119]

Answer: 15m/s

Explanation: <u>Average</u> <u>Velocity</u> is vector describing the total displacement of an object and the time taken to change its position. It is represented as:

$v=\frac{\Delta x}{\Delta t}$

At t₁ = 1.0s, displacement x₁ is:

$x(1)=25+3(1)^{2}$

x(1) = 28

At t₂ = 4.0s:

$x(4)=25+3(4)^{2}$

x(4) = 73

Then, average speed is

$v=\frac{73-28}{4-1}$

v = 15

The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s

5 0

2 years ago

A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu

lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

$F_{frk} = -\mu_{k} * F_{n} (1)$

where μk is the kinetic friction coefficient, and Fn is the normal force.

Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and the weight Fg, downward), we conclude that both must be equal and opposite each other:

$F_{n} = F_{g} = m*g (2)$

We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

$F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)$

According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
In this case, this net force is the friction force which we have already found in a).
Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
We can write the expression for a as follows:

$a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2 (4)$

Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

$a = \frac{-v_{o} }{t} (5)$

Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

$t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)$

From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
So, we can use the following kinematic equation in order to find the displacement before coming to rest:

$v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x (7)$

Since the puck comes to a stop, vf =0.
Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

$\Delta x = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m (8)$

The total work done by the friction force on the object , can be obtained in several ways.
One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

$W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2} = -12.4 J (9)$

By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
$P_{Avg} = \frac{\Delta E}{\Delta t} (10)$
If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

$P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)$

The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

$P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)$

7 0

3 years ago