wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength

Answers

Answer 1

Answer:

Complete question:

Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:

a) the midpoint between the two rings?

b) the center of the left ring?

Answer:

a) the electric field strength at the midpoint between the two rings is 0

b) the electric field strength at the center of the left ring is 2712.44 N/C

Explanation:

Given;

distance between the two rings, d = 25 cm = 0.25 m

diameter of each ring, d = 10 cm = 0.1 m

radius of each ring, r = $(0.1)/(2) = 0.05 \ m$

the charge on each ring, q = 20 nC

Electric field strength for a ring with radius r and distance x from the center of the ring is given as;

$E = (kxQ)/((x^2 +r^2)^(3/2))$

The electric field strength at the midpoint;

the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m

$E = (kxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9)*0.125*20*10^(-9))/((0.125^2 + 0.05^2)^(3/2)) \n\nE = 9210.5 \ N/C$

$E_(left) = 9210.5 \ N/C$

The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;

$E_(right) = -9210.5 \ N/C$

The electric field strength at the midpoint;

$E_(mid) = E_(left) + E_(right)\n\nE_(mid) = 9210.5 \ N/C - 9210.5 \ N/C\n\nE_(mid) = 0$

(b)

The distance from the right ring to center of the left ring, x = 0.25 m.

$E = (KxQ)/((x^2 +r^2)^(3/2)) \n\nE = (8.99*10^(9) *0.25*20*10^(-9))/((0.25^2 + 0.05^2)^(3/2)) \n\nE = 2712.44 \ N/C$

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A CO2 gun shoots a 0.2 gram round pellet (bb) at 2800 ft/sec, and as the bb leaves the gun it gets charged by friction . If Earths magnetic field points South to North at an intensity of 20 uT, and the bb is shot W->E. Find the charge the bb would need to stay level by balancing out the force of gravity.

Answers

Answer:

So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Explanation:

Fb = Fg

qvb= mg ⇒ q = mg/vB = 0.2 *10∧-3 * 9.8/853.44 * 20 * 10∧-6

= 0.115C

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So it would need a charge of 0.115C for a upward force to act and cancel the force of gravity.

Two parallel wires carry currents in the same direction. If the currents in the wires are 1A and 4A and the wires are 5 m apart. Calculate the force between the 2 wires. Is it attractive or repulsive? Calculate the magnetic field midway between the wires.

Answers

Answer:

$1.6* 10^(-7)$ N

$2.4* 10^(-7)$ N

Explanation:

$i_(1)$ = 1 A

$i_(2)$ = 4 A

$r$ = distance between the two wire = 5 m

$F$ = Force per unit length acting between the two wires

Force per unit length acting between the two wires is given as

$F = (\mu _(o))/(4\pi )(2i_(1)i_(2))/(r)$

$F = (10^(-7))(2(1)(4))/(5)$

$F = 1.6* 10^(-7)$ N

$r'}$ = distance of each wire from the midpoint = 2.5 m

Magnetic field midway between the two wires is given as

$B = (\mu _(o))/(4\pi ) \left \left ( (2i_(2))/(r') \right - (2i_(1))/(r') \right \right ))$

$B = (10^(-7)) \left \left ( (2(4))/(2.5) \right - (2(1))/(2.5) \right \right ))$

$B = 2.4* 10^(-7)$

Suppose that the speedometer of a truck is set to read the linear speed of the truck, but uses a device that actually measures the angular speed of the tires. If larger-diameter tires are mounted on the truck, will the reading on the speedometer be correct? If not, will the reading be greater than or less than the true linear speed of the truck? Why?

Answers

The measurement will be significantly affected.

Recall that the relationship between linear velocity and angular velocity is subject to the formula

$v = \omega r$ ,

Where r indicates the radius and $\omega$ the angular velocity.

As the radius increases, it is possible that the calibration is delayed and a higher linear velocity is indicated, that to the extent that the velocity is directly proportional to the radius of the tires.

A neutron star has a mass of 2.0 × 1030 kg (about the mass of our sun) and a radius of 5.0 × 103 m (about the height of a good-sized mountain). Suppose an object falls from rest near the surface of such a star. How fast would this object be moving after it had fallen a distance of 0.025 m? (Assume that the gravitational force is constant over the distance of the fall and that the star is not rotating.)

Answers

Answer:

$v=516526.9m/s$

Explanation:

The force to which the object of mass m is attracted to a star of mass M while being at a distance r is:

$F=(GMm)/(r^2)$

Where $G=6.67*10^(-11)Nm^2/Kg^2$ is the gravitational constant.

Also, Newton's 2nd Law tells us that this object subject by that force will experiment an acceleration given by F=ma.

We have then:

$ma=(GMm)/(r^2)$

Which means:

$a=(GM)/(r^2)$

The object departs from rest ( $v_0=0m/s$ ) and travels a distance d, under an acceleration a, we can calculate its final velocity with the formula $v^2=v_0^2+2ad$ , which for our case will be:

$v^2=2ad=(2GMd)/(r^2)$

$v=\sqrt{(2GMd)/(r^2)}$

We assume a constant on the vecinity of the surface because d=0.025m is nothing compared with $r=5*10^3m$ . With our values then we have:

$v=\sqrt{(2GMd)/(r^2)}=\sqrt{(2(6.67*10^(-11)Nm^2/Kg^2)(2*10^(30)Kg)(0.025m))/((5*10^3m)^2)}=516526.9m/s$

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Learn more about the Coefficient of kinetic friction here:

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If you want to play a tune on wine glasses, you’ll need to adjust the oscillation frequencies by adding water to the glasses. This changes the mass that oscillates (more water means more mass) but not the restoring force, which is determined by the stiffness of the glass itself. If you need to raise the frequency of a par- ticular glass, should you add water or remove water?

Answers

Answer:Reducing mass i.e. water

Explanation:

Frequency For given mass in glass is given by

$f=(1)/(2\pi )\sqrt{(k)/(m)}$

where k =stiffness of the glass

m=mass of water in glass

from the above expression we can see that if mass is inversely Proportional to frequency

thus reducing mass we can increase frequency

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