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lana66690 [7]
4 years ago
8

Two rings of radius 5 cm are 20 cm apart and concentric with a common horizontal x-axis. The ring on the left carries a uniforml

y distributed charge of +30 nC, and the ring on the right carries a uniformly distributed charge of −30 nC. What is the electric field due to the right ring at a location midway between the two rings?
Physics
1 answer:
Yanka [14]4 years ago
3 0

Answer:

The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

Explanation:

Given that,

Radius of first ring = 5 cm

Radius of second ring = 20 cm

Charge on the left of the ring = +30 nC

Charge on the right of the ring = -30 nC

We need to calculate the electric field due to the right ring at a location midway between the two rings

Using formula of  electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{qx}{(x^2+R^2)^{\frac{3}{2}}}

Put the value into the formula

E=\dfrac{9\times10^{9}\times30\times10^{-9}\times0.1}{((0.1)^2+(0.2)^2)^{\frac{3}{2}}}

E=2.41\times10^{3}\ V/m

Hence, The electric field due to the right ring at a location midway between the two rings is 2.41\times10^{3}\ V/m

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Lina20 [59]

Answer:

The ball impact velocity i.e(velocity right before landing) is 6.359 m/s

Explanation:

This problem is related to parabolic motion and can be solved by the following equations:

x=V_{o}cos \theta t----------------------(1)

y=y_{o}+V_{o} sin \theta t - \frac{1}{2}gt^{2}---------(2)

V=V_{o}-gt ----------------------- (3)

Where:

x = m is the horizontal distance travelled by the golf ball

V_{o} is the golf ball's initial velocity

\theta=0\° is the angle (it was  a horizontal shot)

t is the time

y is the final height of the ball

y_{o} is the initial height of the ball

g is the acceleration due gravity

V is the final velocity of the ball

Step 1: finding t

Let use the equation(2)

t=\sqrt{\frac{2 y_{o}}{g}}

t=\sqrt{\frac{2 (12.5 m)}{9.8 m/s^{2}}}

t=1.597s

Substituting (6) in (1):

67.1 =V_{o} cos(0\°) 1.597-------------------(4)

Step 2:  Finding V_{o}:

From equation(4)

67.1 =V_{o}(1) 1.597

V_0 = \frac{6.71}{1.597}

V_{o}=42.01 m/s (8)  

Substituting V_{o} in (3):

V=42.01 -(9.8)(1.597)

v =42 .01 - 15.3566  

V=26.359 m/s

5 0
3 years ago
Future passive of( win)​
Inessa05 [86]

Answer:

three point charge positioned one x-axis if the charge and corresponding positions are +32Mc x=0 +20Mc x=40cm - 60Mc x=60cm find force 32Mc

Explanation:

7 0
3 years ago
what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be
leonid [27]
F = k \cdot \frac{q_1 q_2}{d} = 9 \cdot 10^{9} \cdot \frac{0.25 \cdot 0.11}{0.88} =  144 \cdot 5^{9} \ N.

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.
6 0
3 years ago
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac
matrenka [14]

Answer:

a) 6.1 m

b) 4.6 s

c) 1.326 m/s

d) 0.325 m

Explanation:

a) The wave length is the distance between 2 crests λ = 6.1m

b) The period of the wave is the time it takes from the lowest point to the next lowest point, which is twice the time it takes from the lowest point to the highest point = 2*2.3 = 4.6 s

c) The speed of the wave is the distance per unit of time, or wave length over period = 6.1 / 4.6 = 1.326 m/s

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6 0
3 years ago
1 point
White raven [17]

Answer:

Waves transfer energy, not motion

8 0
3 years ago
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