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gtnhenbr [62]
2 years ago
11

A nonconducting wall carries charge with a uniform density of 8.60μC/cm²(a) What is the electric field 7.00 cm in front of the w

all if 7.00 cm is small compared with the dimensions of the wall? Explain.
Physics
1 answer:
Zina [86]2 years ago
4 0

The electric field at a distance of 7.00 cm is 8.60×10^−2C/m2

The charge per unit area of wall is

σ = (8.60×10^−6C/cm2)( 100cm /m)^2 = 8.60×10^−2C/m2

The electric field at a distance of 2.00 cm is then

E =  σ /7∈0  = (8.60×10^−2C/m^2) / 7(8.85×10−12C2/N⋅m2)

= 1.38×10^9N/C away from the wall.

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Learn more about electric field here:

brainly.com/question/15800304

#SPJ4

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A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
Natasha_Volkova [10]

Answer:

F = 1958.4 N

Explanation:

By volume conservation of the fluid on both sides we can say that volume of fluid displaced on the side of the car must be equal to the volume of fluid on the other side

so we have

L_1A_1 = L_2A_2

1.20(\pi 18^2) = L_2(\pi 5^2)

L_2 = 15.55 m

so the car will lift upwards by distance 1.2 m and the other side will go down by distance 15.55 m

So here the net pressure on the smaller area is given as

P = P_{atm} + \frac{12,000}{\pi (0.18)^2} + \rho g (1.2 + 15.55)

excess pressure exerted on the smaller area is given as

P_{ex} = \frac{12000}{\pi (0.18)^2} + 800(9.81)(16.75)

P_{ex} = 2.49\times 10^5 Pascal

now the force required on the other side is given as

F = P_ex (area)

F = (2.49 \times 10^5)(\pi (0.05)^2)

F = 1958.4 N

3 0
4 years ago
A cyclist rides in a circle with speed 5 m/s. What is his centripetal
Reptile [31]

Answer: 1.32 m/s^2

Explanation:

Centripetal acceleration is given by the formula

a = ( v^2 ) / r

where a is centripetal acceleration, v is velocity and r is radius.

We know that,

v = 5 m/s

r = 19m

Now,

a = ( v^2 ) / r

a = ( 5^2 ) / 19

a = 25 / 19

a = 1.32 m/s^2

6 0
4 years ago
Calculate the velocity of the car between 20 and 30 sec if it is 5/30-4/20​
Bezzdna [24]

Answer:

here you go  4 m/s^2......tadaa

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3 years ago
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A swimmer travels from the south end to the north end of a 30.0 m pool in 15.0 s and makes the return trip to the starting posit
JulsSmile [24]

Answer:

Zero

Explanation:

Average velocity is given by:

v=\frac{d}{t}

where

d is the displacement of the trip

t is the time it takes for the trip to complete

In this problem, the net displacement of the swimmer is zero. In fact:

- First, he swims 30.0 m in the north direction

- Then, he travels back (-30.0 m) in the south direction, to the starting position

Since the final position is equal to the starting position, the displacement is zero:

d = 0

And therefore, the average velocity is also zero.

8 0
4 years ago
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