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3 years ago

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Two facing surfaces of two large parallel conducting plates separated by 7.0 cm have uniform surface charge densities such that

are equal in magnitude but opposite in sign. The difference in potential between the plates is 560 V.Is the positive or the negative plate at the higher potential?

1 answer:

Semmy [17]3 years ago

7 0

Explanation:

Given that,

The distance between two conducting plates is 7 cm.

The difference in potential between the plates is 560 V.

To find,

We need to tell is the positive or the negative plate at the higher potential?

Solution,

We know that the electric potential is defined as the work done in moving a test charge from one position to another. So, the positive plate is always at the higher potential.

Hence, this is the required solution.

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The magnetic force of a magnet is stronger at its poles than in the middle.<br> true or false

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The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center.

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Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f

musickatia [10]

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

$E_1_x = 0V/m\\E_1_y=100V/m$

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

$E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m$

∑x: $E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m$

∑y: $E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m$

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

$||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees$

or 27.23° to the left of E1.

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3 years ago

A torsion pendulum consists of an irregularly-shaped object of mass 20.0 kg suspended vertically by a wire of torsion constant 0

netineya [11]

Answer:

Rotational inertia of the object is, $I=0.023\ kg-m^2$

Explanation:

Given that,

Mass of the object, m = 20 kg

Torsion constant of the wire, K = 0.85 N-m

Number of cycles, n = 69

Time, t = 66 s

To find,

The rotational inertia of the object.

Solution,

There exists a relationship between the moment of inertia, time period and the torsion constant of the spring is given by :

$T=2\pi\sqrt{\dfrac{I}{K}}$

Here I is the moment of inertia

T is the time period, and it is equal to the number of cycles per unit time

$I=\dfrac{T^2K}{4\pi ^2}$

$I=\dfrac{(69/66)^2\times 0.85}{4\pi ^2}$

$I=0.023\ kg-m^2$

So, the rotational inertia of the object is $0.023\ kg-m^2$ .

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4 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

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4 years ago

A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s

bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

$u_y = 0$ (initial vertical velocity is zero)

s = 20 m

$a=g=9.8 m/s^2$ (acceleration of gravity)

And solving for t, we find the time of flight:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

$v_x = 7 m/s$

Therefore the distance covered after time t is given by

$d=v_x t$

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

$d=(7)(2.02)=14.1 m$

C) 21.0 m/s at $70.5^{\circ}$ below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

$v_x = 7 m/s$

Instead, the vertical velocity is given by

$v_y = u_y +at$

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

$v_y = 0+(9.8)(2.02)=19.8 m/s$

So, the magnitude of the final velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s$

And the angle is given by

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}$

below the horizontal

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3 years ago