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a_sh-v [17]
3 years ago
5

. A non-uniform positive line charge of length 2 m is put along the x-axis as shown in the figure, where x​0​=1.5 m. The linear

charge density is given by λ(x)=4x​2 C/m​3​. Find the magnitude of the total electric field, E, created by the line charge at the origin using integration. (Take k=9x10​9 ​N m​2​ /C​2​)

Physics
1 answer:
PilotLPTM [1.2K]3 years ago
5 0

Answer:

E = 7.2*10^{10}N/C

Explanation:

The differential electric field dE due to differential charge dQ at distance x from the origin is

dE = k\dfrac{dQ}{x^2}

but since dQ = \lambda dx = 4x^2dx we have

dE = k\dfrac{4x^2dx}{x^2}

dE = 4k\: dx

integrating this from x_0 to x_0+L we get

$E = \int^{x_0+L}_{x_0} {4k} \, dx $

E = 4k[(x_0+L)-x_0]

E =4kL

putting in k = 9*10^9Nm^2/C^2 and L =2m we get

\boxed{E = 7.2*10^{10}N/C.}

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You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then lo
IgorC [24]

ANSWER:

F(h)= 230 N is the horizontal force you will need to move the pickup along the same road at the same speed.

STEP-BY-STEP EXPLANATION:

F(h) is Horizontal Force = 200 N

V is Speed = 2.4 m/s

The total weight increase by 42%

coefficient of rolling friction decrease by 19%

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

F(h) = F(f)

F(h) = mg* u

m is mass

g is gravitational acceleration = 9.8 m/s^2

200 = mg*u

Since weight increases by 42% and friction coefficient decreases by 19%

New weight = 1+0.42 = 1.42 = (1.42*m*g)

New friction coefficient = μ = 1 - 0.19 = 0.81 = 0.81 u

F(h) = (0.81μ) (1.42 m g)

       = (0.81) (1.42) (μ m g)

       = (0.81) (1.42) (200)

       = 230 N

4 0
3 years ago
If you double the pressure on the surface of a can of water, the buoyant force on a stone placed in that water will
Bingel [31]
The buoyant force won't change.
4 0
3 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of n
Oduvanchick [21]

Answer:

15.0 m/s^2

Explanation:

3 0
3 years ago
Read 2 more answers
A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s
lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0
3 years ago
What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?
eimsori [14]

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

4 0
3 years ago
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