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bulgar [2K]
3 years ago
13

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}

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Why does the total amount of energy before and after<br> any energy transformations remain the same?
snow_tiger [21]

The Law of Conservation of Energy states that, in an isolated system, energy remains constant and can not be created or destroyed, only transferred from one form to another. This law was created by Julius Robert Mayer.

8 0
3 years ago
Read 2 more answers
What object currently has the most gravitational potential energy?<br><br>A, B, C, or D​
eduard

Answer:

A

Explanation:

this because

gravitational potential energy = mass x height x gravitational field strength

so let's assume mass is 2 kg and gravitational field strength is 10 N /kg

so when height is very low, take it as 3 m

gravitational potential energy= 2 x 3 x 10 = 60 j

but when height is 6m

gravitational potential energy = 2 x 6 x 10 = 120 j

so when the height is the greatest, the gravitational potential energy is the highest

so A is the heighest so it has the highest gravitational potential energy.

hope this helps

please mark it brainliest :D

5 0
3 years ago
A ball bearing of radius of 1.5 mm made of iron of density
Serjik [45]

Answer:

\boxed{\sf Viscosity \ of \ glycerine \ (\eta) = 14.382 \ poise}

Given:

Radius of ball bearing (r) = 1.5 mm = 0.15 cm

Density of iron (ρ) = 7.85 g/cm³

Density of glycerine (σ) = 1.25 g/cm³

Terminal velocity (v) = 2.25 cm/s

Acceleration due to gravity (g) = 980.6 cm/s²

To Find:

Viscosity of glycerine (\sf \eta)

Explanation:

\boxed{ \bold{v =  \frac{2}{9}  \frac{( {r}^{2} ( \rho -  \sigma)g)}{ \eta} }}

\sf \implies \eta =  \frac{2}{9}  \frac{( {r}^{2}( \rho -  \sigma)g )}{v}

Substituting values of r, ρ, σ, v & g in the equation:

\sf \implies \eta =  \frac{2}{9}  \frac{( {(0.15)}^{2}  \times  (7.85 - 1.25) \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \frac{(0.0225 \times 6.6 \times 980.6)}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times  \frac{145.6191}{2.25}

\sf \implies \eta =  \frac{2}{9}  \times 64.7196

\sf \implies \eta =  2 \times 7.191

\sf \implies \eta =  14.382 \: poise

6 0
3 years ago
HELP
blagie [28]

Answer:

A

Explanation:

hope its right

3 0
3 years ago
The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
Pie

Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

t=68.94 m

8 0
3 years ago
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