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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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A nautical mile is 6076 feet, and 1 knot is a unit of speed equal to 1 nautical mile/hour. How fast is a boat going 8 knots goin
mezya [45]

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1  nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

 8 knots = 8 nautical mile /hour

               =8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}

               = 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

5 0
3 years ago
The normal force equals the magnitude of the gravitational force as a roller coaster car crosses the top of a 58-m-diameter loop
dedylja [7]

Answer:

16.87 m/s

Explanation:

To find the speed of the car at the top, when the normal force is equal the gravitational force, we just need to equate both forces:

N = P

m*a_c = mg

a_c is the centripetal acceleration in the loop:

a_c = v^2/r

So we have that:

mv^2/r = mg

v^2/r = g

v^2 = gr

v = \sqrt{gr}

So, using the gravity = 9.81 m/s^2 and the radius = 29 meters, we have:

v = \sqrt{9.81 * 29}

v = \sqrt{284.49} = 16.87\ m/s

The speed of the car is 16.87 m/s at the top.

5 0
3 years ago
Problem: The circular blad on a radial arm saw is turning at262
kobusy [5.1K]

Answer:

Net torque, \tau=-0.033\ N-m

Explanation:

It is given that,

Initial angular speed of the blade, \omega_i=262\ rad/s

Final angular speed of the blade, \omega_f=85\ rad/s

Time, t = 18 s

Radius of the disk, r = 0.13 m

Mass of the disk, m = 0.4 kg

We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,

\tau=I\times \alpha

The moment of inertia of the disk, I=\dfrac{mr^2}{2}

\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f-\omega_i}{t}

\tau=\dfrac{0.4\times (0.13)^2}{2}\times \dfrac{85-262}{18}

\tau=-0.033\ N-m

Negative sign shows that the net torque is acting in the opposite direction of its motion. Hence, this is the required solution.

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3 years ago
Difference between elastic and inelastic collisions
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3 years ago
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vova2212 [387]

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